In order to define speed direction is not important true or false

Answers

Answer 1
That’s true. You don’t need direction if you are only interested in speed
Answer 2
The answer to the question is true

Related Questions

Find the value of n if nc₂ = 15

Answers

nc2 = 15
Solution
Divide each term in nc2= 15 by c and
simplify.
N=15/c2

Please help im down bad

Please help im down bad

Answers

Answer:

Try this on for size.

Explanation:

Please help im down bad

A frictionless plane is 10.0 m long and inclined at 36.0°. A sled starts at the bottom with an initial speed of 6.00 m/s up the incline. When the sled reaches the point at which it momentarily stops, a second sled is released from the top of the incline with an initial speed Vi. Both sleds reach the bottom of the incline at the same moment.
(a) Determine the distance that the first sled traveled up the incline. m
(b) Determine the initial speed of the second sled. m/s

Use the equation for the position of the second sled as a function of time to find the speed that makes it reach the bottom of the slope in the same time that the first sled takes to slide back down.

Answers

If a frictionless plane is 10.0 m long and inclined at 36.0°.

The sled traveled  8.17 m up the incline.The initial speed of the second sled is about  5.68 m/s

How to find the initial speed?

We can use conservation of energy to find the distance that the first sled travels up the incline. The potential energy of the sled at the bottom of the incline is zero, and its kinetic energy is:

KE = (1/2)mv^2

where m is the mass of the sled and v is its speed. At the point where the sled stops, all of its kinetic energy has been converted into potential energy, so we can write:

mgh = (1/2)mv^2

where h is the height that the sled has traveled up the incline. Solving for h, we get:

h = (v^2)/(2g)

where g is the acceleration due to gravity. Using the given values, we have:

h = (6.00 m/s)^2 / (2 * 9.81 m/s^2) = 1.83 m

So the first sled travels a distance of 10.0 m - 1.83 m = 8.17 m up the incline.

b. To find the initial speed of the second sled, we can use conservation of energy again. At the top of the incline, the sled has potential energy:

PE = mgh

where h is the height of the incline. As the sled slides down the incline, its potential energy is converted into kinetic energy:

KE = (1/2)mv^2

where v is the speed of the sled at the bottom of the incline. We can equate these two expressions and solve for v:

mgh = (1/2)mv^2

v = sqrt(2gh)

Using the given values, we have:

v = sqrt(2 * 9.81 m/s^2 * 10.0 m * sin(36.0°)) = 12.2 m/s

So the second sled must be released from the top of the incline with an initial speed of 12.2 m/s.

The position of the sled as a function of time is given by:

y = -0.5gt^2 + Vi*t + h

where y is the vertical position of the sled, t is the time, Vi is the initial speed of the sled, and h is the height of the incline. At the bottom of the incline, y = 0, so we can solve for the time it takes for the second sled to reach the bottom:

0 = -0.5gt^2 + Vi*t + h

t = (Vi ± sqrt(Vi^2 - 2gh)) / g

Since we want both sleds to reach the bottom at the same time, we set the time for the first sled to slide down the incline equal to this expression for t and solve for Vi:

t = sqrt(2h/g) = sqrt(2 * 1.83 m / 9.81 m/s^2) = 0.619 s

0 = -0.5gt^2 + Vi*t + h

Vi = (h - 0.5gt^2) / t

Vi = (1.83 m - 0.5 * 9.81 m/s^2 * (0.619 s)^2) / 0.619 s

Vi = 5.68 m/s

So the initial speed of the second sled is about  5.68 m/s

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People's perceptions of what is important in life are known as:

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Social perception (or person perception) is the study of how people form impressions of and make inferences about other people as sovereign personalities.A real-world example of social perception is understanding that others disagree with what one said when one sees them roll their eyes.

76. Two electric charges -6μC and
+6μC are placed respectively in two
points A and B distant of 1m apart. The
electric field is null at the point C:

A.Located in the middle of the
segment AB
B.Located outside segment AB at
1m from A.
C.Located outside segment AB at
1m from B
D.Outside the line AB
E.No answer is right.

Answers

A. The electric field is null at the point C; located in the middle of the

segment AB.

What is electric field?

Electric field is the region of space where the influence of electric force is felt.

Electric field at the middle of AB

E = kq/r²

where;

r is the middle of AB = 0.5 m

E(+6μC) = (9 x 10⁹ x 6 x 10⁻⁶) / (0.5²)

E(+6μC) = +216,000

E(-6μC) = (9 x 10⁹ x 6 x 10⁻⁶) / (0.5²)

E(-6μC) = -216,000

Sum of the electric field at the middle of AB

E(net) = E(+6μC) + E(-6μC)

E(net) = 216,000 - 216,000 = 0

Thus, the electric field is null at the point C; located in the middle of the

segment AB.

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Two uniform solid spheres, each with mass 0.852 kg

and radius 8.00×10−2 m are connected by a short, light rod that is along a diameter of each sphere and are at rest on a horizontal tabletop. A spring with force constant 153 N/m has one end attached to the wall and the other end attached to a frictionless ring that passes over the rod at the center of mass of the spheres, which is midway between the centers of the two spheres. The spheres are each pulled the same distance from the wall, stretching the spring, and released. There is sufficient friction between the tabletop and the spheres for the spheres to roll without slipping as they move back and forth on the end of the spring.
Assume that the motion of the center of mass of the spheres is simple harmonic. Calculate its period.

Answers

The period of the simple harmonic motion of the center of mass of the two spheres is approximately 0.770 seconds.

To find the period of the simple harmonic motion of the center of mass of the two spheres, we need to use the equation for the period of a mass-spring system:

T = 2π√(m/k)

where T is the period, m is the total mass of the system (two spheres), and k is the spring constant.

First, we need to find the total mass of the system:

m = 2m1 = 2(0.852 kg) = 1.704 kg

where m1 is the mass of one sphere.

Next, we need to find the spring constant:

k = 153 N/m

Now, we can calculate the period:

2π√(1.704 kg/153 N/m) ≈ 0.770 s

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Two separate disks are connected by a belt traveling at 5m/s. Disk 1 has a mass of 10kg and radius of 35cm. Disk 2 has a mass of 3kg and radius of 7cm.
a. What is the angular velocity of disk 1?
b. What is the angular velocity of disk 2?
c. What is the moment of inertia for the two disk system?

Answers

Explanation:

Given that,

Linear speed of both disks is 5 m/s

Mass of disk 1 is 10 kg

Radius of disk 1 is 35 cm or 0.35 m

Mass of disk 2 is 3 kg

Radius of disk 2 is 7 cm or 0.07 m

(a) The angular velocity of disk 1 is :

\(v=r_1\omega_1\\\\\omega_1=\dfrac{v}{r_1}\\\\\omega_1=\dfrac{5}{0.35}\\\\\omega_1=14.28\ rad/s\)

(b) The angular velocity of disk 2 is :

\(v=r_2\omega_2\\\\\omega_2=\dfrac{v}{r_2}\\\\\omega_2=\dfrac{5}{0.07}\\\\\omega_2=71.42\ rad/s\)

(c) The moment of inertia for the two disk system is given by :

\(I=I_1+I_2\\\\I=\dfrac{1}{2}m_1r_1^2+\dfrac{1}{2}m_2r_2^2\\\\I=\dfrac{1}{2}(m_1r_1^2+m_2r_2^2)\\\\I=\dfrac{1}{2}\times (10\times (0.35)^2+3\times (0.07)^2)\\\\I=0.619\ kg-m^2\)

Hence, this is the required solution.

A 25 kg of bananas is being pushed across the floor with a force of 37.5 N. What is the acceleration of the crate?

Answers

the acceleration should be 1.5 because 37.5 divided by 25 is 1.5

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A tiny water droplet of radius . descends through air from a high building.
Calculate its terminal velocity. Given that for air = × − −− and density of the water = -3

Answers

Correct question is;

A tiny water droplet of radius 0.010 cm descends through air from a high building .Calculate its terminal velocity . Given that η of air = 19 × 10^(-6) kg/m.s and density of water ρ = 1000kg/ms

Answer:

1.146 m/s

Explanation:

We are given;

Radius; r = 0.010 cm = 0.01 × 10^(-2) m

η = 19 × 10^(-6) kg/m.s

ρ = 1000 kg/ms

The formula for the terminal velocity is given by;

V_t = 2r²ρg/9η

g is acceleration due to gravity = 9.8 m/s²

Thus;

V_t = (2 × (0.01 × 10^(-2))² × 1000 × 9.8)/(9 × 19 × 10^(-6))

V_t = 1.146 m/s

What is the mass of a toy car if it has 5 J of potential energy and is sitting on top of a track that has a height of 2m?

(PE= m x g x h) (hint g=9.8 m/s2)

Answers

Explanation:

PE=mgh

5=m(9.8)(2)

m=5/19.6

m=0.2251 kg

m=225.1 grams

What types of memories are stored in the collective unconscious?

Answers

The types of memories that are stored in the collective unconscious include instinctive patterns and archetypes from ancestral heritage.

What is the collective unconscious?

The collective unconscious can be defined as  the part of the human mind  that contains distinct types of memories and also instinctive impulses in humans which may shape some of our behaviors regarding specific situations and also environmental contexts (e.g., during danger situations).

In conclusion, the types of memories that are stored in the collective unconscious include instinctive patterns and archetypes from ancestral heritage.

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A car is moving east along a straight, level road at a constant velocity. Which forces form an action/reaction force pair?
A) the force of the tires against the road and the force of the road on the tires
B) the weight of the car and the force of the road pushing up on the car
C) the force of the engine on the car and the force due to friction on the car
D) the normal force on the car and the force of the car pulling on Earth

Answers

Answer:

The force of the tires against the road and the force of the road on the tires

Explanation:

right on edge(2021)

A car is moving east along a straight, level road at a constant velocity. The force of the tires against the road and the force of the road on the tires. Thus option A is correct

What is force?

Force is defined as an outside force that has the power to alter the motion or rest of a body.

It can be defined as a push or pull exerted on an object as a result of the interaction of the object with another object.

It can also be defined as a push or a pull that modifies or tends to modifies an object's uniform motion, state of rest, or modifies the object's direction or shape It either accelerates items or increases their overall pressure.

The joule is the accepted unit of energy in electronics and most other branches of science. The amount of energy expended when a force of one newton is applied across a displacement of one meter is known as a joule.

There are seven types of force.

Applied forceFrictional forceNormal forceGravitational forceAir resistance forceTension forceSpring force

Thus, a car is moving east along a straight, level road at a constant velocity. The force of the tires against the road and the force of the road on the tires. Thus option A is correct

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Select the Moon and use the Info view to determine which of the following statements is correct. The Last Quarter Moon.... rises near noon and sets near midnight. rises at about 6am and sets at about 6pm. rises near midnight and sets near midday. rises and sets at the same time as the Sun. Submit Your Answer

Answers

Answer: The correct statement is that the Last Quarter Moon (rises near midnight and sets near midday).

Explanation:

Phases of the moon also called the LUNAR PHASE can be defined as the different shades of illumination on the moon as seen from the earth. The moon is the natural satellite of the earth that illuminates upon reflection of light from the sun. This means it doesn't have power to shine on its own. When carefully observed, there are times it gets dark and beings to glow brighter over a period of time. This occurs because as the moon completes its four weeks lunar cycle round the earth, how much of its face we see illuminated by sunlight depends on the angle the Sun makes with the Moon.

There are 8 main types of the moon phases these includes:

--> New moon: This is when the moon is not visible to the earth because it's between the earth and the sun. It rises at sunrise and sets at sunset.

--> The waxing crescent: At this phase the moon gets brighter and illuminated from the sun that a crescent shape is seen.

--> First quarter: this occurs one week after the new moon. The moon rises at noon and sets at midnight.

--> The waxing gibbous: This occurs after the first quarter phase where more than half of the lit part of the moon is seen.

--> Full moon: This occurs when the moon and the sun are opposite each other. That is way it is said to rise at sunset and sets at sun rise.

--> The waning gibbous: this occurs when more than half of the lit part of the moon gradually becomes darker

--> Third quarter ( Last Quarter): The moon rises at midnight and sets at noon. This occurs a week after the full moon.

--> The waning crescent: This occurs after the last quarter phase where a very thin fading crescent shaped moon is seen, just before the Moon is invisible again at the start of the cycle, the new moon.

I need help fast please help

I need help fast please help

Answers

A and C Im pretty sure :)

1. Two charges are separated by a distance of 1 cm. One charge has a value of 7 micro Coulombs. The other charge has a value of 10 micro Coulombs. What is the force between them, in pounds. Make sure to include the sign of the force which will be positive if the charges repel each other and negative if they attract each other.

2. 12 gauge copper wire is normally used in house wiring. When aluminum wire is used one needs to use a smaller gauge size to obtain the same resistance, 40 ft of 12 gauge copper wire was calculated. What would the resistance be if 10 gauge aluminum wire were used?

3. A 12 V automobile battery can supply 51 amps for one hour and cost $194. What is the cost of this electricity in cents per kWh?

4. Most of the body's resistance is in its skin. When wet, salts go into ion form, and the resistance is lowered. Thus, the resistance of the skin can go from 100,000 ohms when dry to 300 ohms when wet. What is the current that would be carried through the body, in milliAmperes, if you touched a 240 V power line while dry? Currents over 10 mA are almost always deadly.

Answers

1. The force between the two charges is 1.78 × 10⁻⁵ pounds, with opposite signs indicating attraction between the charges.

2. The resistance of 10 gauge aluminum wire over a 40 ft distance would be 0.506 ohms.

3. The cost of electricity from the automobile battery is 38.6 cents per kWh.

4. The current that would be carried through the body is 0.8 mA if dry.

1. The force between two point charges can be calculated using Coulomb's law, which states that the force is proportional to the product of the charges and inversely proportional to the square of the distance between them.

Using the values given, the force can be calculated as F = (k * q1 * q2) / r², where k is Coulomb's constant, q1 and q2 are the charges, and r is the distance between them. Plugging in the values, the force can be calculated as 1.78 × 10⁻⁵ pounds, with opposite signs indicating attraction between the charges.

2. The resistance of a wire is determined by its length, cross-sectional area, and resistivity. The resistivity of aluminum is higher than that of copper, so a larger cross-sectional area is required to achieve the same resistance. Using the gauge size conversion chart, 10 gauge aluminum wire has a cross-sectional area of 5.26 mm², which is approximately 83% of the cross-sectional area of 12 gauge copper wire.

Thus, the resistance of 10 gauge aluminum wire over a 40 ft distance can be calculated as R = (rho * L) / A, where rho is the resistivity of aluminum, L is the length, and A is the cross-sectional area. Plugging in the values, the resistance can be calculated as 0.506 ohms.

3. To calculate the cost of electricity per kWh, the total cost and the total amount of energy supplied must be known. Since the battery supplies 12 V and 51 A for one hour, the total energy supplied can be calculated as E = V * I * t, where V is the voltage, I is the current, and t is the time.

Plugging in the values, the total energy supplied can be calculated as 612 watt-hours (Wh). Since one kWh is equal to 1000 Wh, the total energy supplied can be converted to 0.612 kWh. Dividing the total cost by the total energy supplied gives the cost per kWh, which is 38.6 cents.

4. The current through the body can be calculated using Ohm's law, which states that current is equal to voltage divided by resistance. Using the values given, the resistance can be either 100,000 ohms or 300 ohms depending on whether the skin is dry or wet.

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The half-life of a radioactive isotope is 210 d. How many days would it take for the decay rate of a sample of this isotope to fall to 0.58 of its initial rate?

Answers

It would take approximately 546 days for the decay rate of the sample of this radioactive isotope to fall to 0.58 of its initial rate.

1. The decay rate of a radioactive isotope is proportional to the number of radioactive atoms present in the sample at any given time.

2. The decay rate can be expressed as a function of time using the formula: R(t) = R₀ * \(e^{(-\lambda t\)), where R(t) is the decay rate at time t, R₀ is the initial decay rate, λ is the decay constant, and e is the base of the natural logarithm.

3. The half-life of a radioactive isotope is the time it takes for half of the radioactive atoms in a sample to decay. In this case, the half-life is given as 210 days.

4. Using the half-life, we can find the decay constant (λ) using the formula: λ = ln(2) / T₁/₂, where ln(2) is the natural logarithm of 2 and T₁/₂ is the half-life.

5. Substituting the given half-life into the formula, we have: λ = ln(2) / 210.

6. Now, we need to find the time it takes for the decay rate to fall to 0.58 of its initial rate. Let's call this time "t".

7. Using the formula for the decay rate, we can write: 0.58 * R₀ = R₀ * e^(-λt).

8. Simplifying the equation, we get: 0.58 = \(e^{(-\lambda t\)).

9. Taking the natural logarithm of both sides, we have: ln(0.58) = -λt.

10. Substituting the value of λ from step 5, we get: ln(0.58) = -(ln(2) / 210) * t.

11. Solving for t, we have: t = (ln(0.58) * 210) / ln(2).

12. Evaluating the expression, we find: t ≈ 546.

13. Therefore, it would take approximately 546 days for the decay rate of the sample of this radioactive isotope to fall to 0.58 of its initial rate.

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h. How is a schematic diagram drawn? What should we be aware of while drawing them?​

Answers

To draw a schematic diagram effectively, there are several key points to consider. First, it is important to understand the system or circuit you are trying to represent. Second, organize the diagram in a logical and readable manner. Third, maintain clarity and simplicity in the diagram. Lastly, follow standard practices and guidelines for drawing schematic diagrams. We should be aware of while drawing them because for understanding the system.

A schematic diagram is a graphical representation that depicts the electrical connections and components of a system. It is commonly used in various fields such as electronics, electrical engineering, and industrial automation. To draw a schematic diagram effectively, there are several key points to consider.

Firstly, it is important to understand the system or circuit you are trying to represent. Analyze the components and their relationships, and determine the appropriate symbols to represent each component accurately. Use standard symbols and conventions to ensure consistency and easy interpretation by others.

Secondly, organize the diagram in a logical and readable manner. Arrange components and connections in a way that reflects their physical arrangement or the flow of signals. Use clear and concise labeling to identify each component and connection point.

Thirdly, maintain clarity and simplicity in the diagram. Avoid cluttering the diagram with unnecessary details. Focus on the key elements and connections that are essential for understanding the system.

Lastly, follow standard practices and guidelines for drawing schematic diagrams. This includes using consistent wire and connection symbols, labeling voltage and current values, and providing necessary annotations or notes for clarification.

Drawing a schematic diagram requires a good understanding of the system, adherence to standard symbols and conventions, logical organization, and clarity. By following these guidelines, you can create schematic diagrams that effectively convey the desired information. Remember, practice and attention to detail are essential for improving your schematic diagram drawing skills.

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Which 5 are examples of locomotor movements? Select all that apply.
Question 1 options:

Hopping

Galloping

Sitting

Sleeping

Walking

Running

Skipping

Eating

Answers

Answer:

Five examples of locomotor movements are:

Hopping

Galloping

Walking

Running

Skipping

Explanation:

I hope it helps ❤❤

Identify the type of chemical reaction:

CaCO3 -->CaO + CO2

Answers

Explanation:

decomposition reaction.....

Answer:

caco₃--cao+co₂

calcium oxide + carbon die oxide gives us calcium carbonate

this is the reaction of Acidic oxide(cao) and Basic oxide (co₂) to form salt.

I’m confused how to start it and I just need help atleast doing one.

Im confused how to start it and I just need help atleast doing one.

Answers

Answer:

CuCl₂   +    H₂S    →    CuS   +    2HCl  

Explanation:

The unbalanced reaction expression is given as;

        CuCl₂   +    H₂S    →    CuS   +    HCl

The problem here involves balancing of chemical equations.

 We use a mathematical approach to solve this problem. Here assign coefficients a, b, c and d as values that will effect the balance;

          aCuCl₂   +    bH₂S    →    cCuS   +    dHCl

Conserving Cu:  a  = c

                     Cl:   2a  = d

                     H: 2b  = d

                     S: b  = c

let a  = 1; c  = 1, b  = 1 and d  = 2

         CuCl₂   +    H₂S    →    CuS   +    2HCl  

what is the pressure of a tank of uniform cross sectional area 4.0m2 when the tank is filled with water a depth of 6m when given that 1 atm=1.013 x 10^5pa density of water=1000kgm-3 g=9.8m/s2​

Answers

The pressure of the tank, when filled with water at a depth of 6 m, is approximately 580.124 atmospheres (atm). To calculate the pressure of the tank, one can use the equation: Pressure (P) = Density (ρ) × g × Depth (h)

Pressure (P) = Density (ρ) × g × Depth (h)

Given: Density of water (ρ) = 1000 kg/m³

Acceleration due to gravity (g) = 9.8 m/s²

Depth (h) = 6 m

Using the given values, one can calculate the pressure:

Pressure = 1000 kg/m³ × 9.8 m/s² × 6 m Pressure

= 58800 kg·m⁻¹·s⁻²

Now, let's convert the units to pascals (Pa) using the conversion 1 atm = 1.013 x \(10^5\) Pa:

Pressure = 58800 kg·m⁻¹·s⁻² × (1 atm / 1.013 x\(10^5\) Pa)

Pressure = 580.124 atm

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7. A particle of mass 3 kg is held in equilibrium by two light unextensible strings. One string is horizontal, as shown in Figure 7.30. The tension in the horizontal string is PN and the tension in the other string is N. Find a) the value of 0 b) the value of P.​

Answers

The tension in the strings are 31.47 and 19.25 N respectively.

Mass of the block, m = 3 kg

From the figure, consider the vertical components,

T₁ sin45° + T₂ sin30° = mg

(T₁/√2) + (T₂/2) = 3 x 9.8 = 29.4

Also, consider the horizontal components,

T₁ cos45° = T₂ cos30°

T₁/√2 = T₂ x√3/2

T₁ = T₂ x √3/2 x √2

So,

T₁ = 0.612T₂

Applying in the first equation,

(T₁/√2) + (T₂/2) = 29.4

(0.612T₂/1.414) + 0.5T₂ = 29.4

0.434 T₂ + 0.5 T₂ = 29.4

0.934 T₂ = 29.4

Therefore, the tension,

T₂ = 29.4/0.934

T₂ = 31.47 N

So, the tension,

T₁ = 0.612 T₂

T₁ = 0.612 x 31.47

T₁ = 19.25 N

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in a young's double-slit experiment the center of a bright fringe occurs wherever waves from the slits differ in the distance they travel by a multiple of

Answers

Answer:

Zero

Explanation:

Because using

Deta X= dsinစ x n(lambda)

But we know that for central maxima

n is zero

So after substituting

Deta x = 0

A 1420-kg car moving east at 17.0 m/s collides with a 1880-kg car moving south at 15.0 m/s, and the two cars connect together.A. What is the magnitude of the velocity of the cars right after the collision? (m/s )B. What is the direction of the cars right after the collision? Enter the angle in degrees where positive indicates north of east and negative indicates south of east. (°)C. How much kinetic energy was converted to another form during the collision? (kJ)

Answers

Given,

The mass of the car moving east, m=1420 kg

The mass of the car moving south, M=1880 kg

The velocity of the car moving east, u₁=17.0 m/s

The velocity of the car moving south, u₂=-15.0 m/s

Here we assume that the eastward direction is the positive x-direction and the southward direction is the negative y-direction.

From the law of conservation of momentum, the momentum is conserved in both directions simultaneously and independently.

Considering the conservation of momentum in the x-direction,

\(mu_1=(m_{}+M)v_x\)

Where v_x is the x-component of the final velocity of the two cars.

On substituting the known values,

\(\begin{gathered} 1420\times17.0=(1420+1880)v_x \\ v_x=\frac{1420\times17.0}{(1420+1880)} \\ =7.32\text{ m/s} \end{gathered}\)

Considering the conservation of momentum in the y-direction,

\(Mu_2=(m+M_{})v_y\)

Where v_y is the y-component of the final velocity of the cars.

On substituting the known values,

\(\begin{gathered} _{}1880\times-15.0=(1420+1880)v_y \\ v_y=\frac{1880\times-15.0}{(1420+1880)} \\ =-8.55\text{ m/s} \end{gathered}\)

A.

The magnitude of the velocity of the cars tight after the collision is given by,

\(v=\sqrt[]{v^2_x+v^2_y}\)

On substituting the known values,

\(\begin{gathered} v=\sqrt[]{7.32^2+(-8.55)^2} \\ =11.26\text{ m/s} \end{gathered}\)

Thus the magnitude of the velocity of the cars right after the collision is 11.25 m/s

B.

The direction of the cars right after the collision is given by,

\(\theta=\tan ^{-1}(\frac{v_y}{v_x})\)

On substituting the known values,

\(\begin{gathered} \theta=\tan ^{-1}(\frac{-8.55}{7.32}) \\ =-49.4^{\circ} \end{gathered}\)

Thus the direction of the cars right after the collision is -49.4°. That is 49.4° south of the east.

C.

The total kinetic energy of the system is before the collision is

\(K_1=\frac{1}{2}mu^2_1+\frac{1}{2}Mu^2_2\)

On substituting the known values,

\(\begin{gathered} K_1=\frac{1}{2}\times1420\times17.0^2+\frac{1}{2}\times1880\times15.0^2 \\ =205.19\times10^3+211.5\times10^3 \\ =416.69\times10^3\text{ J} \end{gathered}\)

The kinetic energy of the system of two cars after the collision is,

\(\begin{gathered} K_2=\frac{1}{2}(m+M)v^2 \\ =\frac{1}{2}\times(1420+1880)11.26^2 \\ =209.2\times10^3\text{ J} \end{gathered}\)

Thus the kinetic energy lost during the collision is,

\(\Delta K_{}=K_1-K_2\)

On substituting the known values,

\(\begin{gathered} \Delta K=416.69\times10^3-209.2\times10^3 \\ =207.49\times10^3\text{ J} \\ \approx207.5\text{ kJ} \end{gathered}\)

Thus the total kinetic energy lost during the collision is 207.5 kJ

A marine weather station reports waves along the shore that are 2 meters high, 8 meters long, and reach the station 8 seconds apart. Determine the speed of these waves.

Answers

One full weather station reports  wave passing a spot every second is equivalent to a frequency of 1 Hz. The frequency is 48=12=0.5 Hz if 4 waves pass a spot in 8 seconds.

What is the frequency of a 30-second ocean wave that hits the shore?

The number of cycles that make up a time unit is the frequency. The frequency of a wave with a 30-second period is therefore 1 30 = 0.033 cycles per second, or 0.033 Hertz (Hz).

Which of the following statements most accurately sums up how a wave crosses a border and enters a new medium?

A wave's speed and wavelength vary as it crosses a border and enters a new medium, but its frequency doesn't.

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Show that the displacement of a body moving with uniform acceleration a is given by
\(s = ut + 1 \div 2at {}^{2} \)
where U is the is the velocity of the body at time (t)=0

Answers

Answer:

I only Know to derive so

Explanation:

We have from first equation,

V = u+at

From second equation

S=u+v/2×t

Now by placing the value of v from equation 1 into 2

or s =u+u+at/2×t

or,s= 2u+at/2×t

or,s=2ut/2+at^2/2

Therefore s=ut+1/2at^2

anaerobic transmission is when you touch a contaminated surface true or false

Answers

true because my mom said to me this morning that i have to take my breakfast

Anaerobic transmission is when you touch a contaminated surface is a false statement.

What is anaerobic respiration?

Anaerobic organisms are the living things that can survive and grow where there is no oxygen in the surrounding environment so we can conclude that Anaerobic transmission is when you touch a contaminated surface is a false statement.

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If the instantaneous voltage at a given moment in the circuit RL is V=VmaxSIN(150), then the instantaneous current at the...... same instant I=Imaxsin​

Answers

The instantaneous current at the same moment in the RL circuit can be expressed as I = Imaxsin(150), where Imax represents the maximum current.

1. Given that the instantaneous voltage at a specific moment in the RL circuit is V = Vmaxsin(150).

2. We can express the current at the same moment using Ohm's Law, which states that V = IR, where V is voltage, I is current, and R is resistance.

3. In an RL circuit, the resistance is represented by the symbol R, and it is typically associated with the resistance of the wire or any resistors in the circuit.

4. However, the given equation does not explicitly mention resistance.

5. Since we are considering an RL circuit, it suggests the presence of inductance (L) along with resistance (R).

6. In an RL circuit, the voltage across the inductor (VL) can be expressed as VL = L(di/dt), where L is the inductance and di/dt represents the rate of change of current.

7. At any given instant, the total voltage across the circuit (V) can be expressed as the sum of the voltage across the resistor (VR) and the voltage across the inductor (VL).

8. Therefore, V = VR + VL.

9. Since the given equation represents the instantaneous voltage (V), we can deduce that V = VR.

10. By comparing V = VR with Ohm's Law (V = IR), we can conclude that I = Imaxsin(150), where Imax represents the maximum current.

The specific values of Vmax, Imax, and the phase angle have not been provided in the question, so we are working with the general expression.

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what determines the magnification of an imagev
measure:what the current values of ​

Answers

Answer:

The magnification of an image is equal to the ratio of the image height to the object height.

5. A Car of mass 1000kg is driven from it's initial
state with uniform acceleration and reaches a
Speed of 30km/h in 10 second find
1 ). Force exerted by the engine.
2). the power in killo watts (kw)

Answers

1). To find the force exerted by the engine, you can use the equation: force = mass x acceleration. If the car has a mass of 1000 kg and accelerates uniformly to a speed of 30 km/h in 10 seconds, you can use the equation: force = 1000 kg x (30 km/h / 10 s) = 3000 N.

2). To find the power in kilowatts, you can use the equation: power = force x velocity. In this case, you can use the force you found above and the final velocity of the car (30 km/h) to calculate the power: power = 3000 N x (30 km/h x 1000 m/km) / (3600 s/h) = 8.33 kW.
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