The __________ field of the earth is produced in part by the flow of hot liquid iron near the earth's core.

Answer:

H = Vy t - 1/2 g t^2       height after time t

H = g t^2 / 2        initial Vy = 0  and take down as positive

t = S / Vx = 280 / 52,7 = 5.31 sec    time to travel horizontally

H = g / 2 * 5.31^2  = 138 m         height of tee above fairway

Total distance travelled is 35m and displacement is 25m.

Distance: Distance is a scalar that expresses the total distance traveled by an object. It's a measure of physical distance that's covered and always good. The distance has nothing to do with the direction, only the magnitude of the change is determined.

Displacement: Displacement is the vector of change in the position of an object. It takes into account the magnitude and direction of change in position from the starting point to the ending point. Perception can be positive, negative or zero depending on the direction of movement.

Total distance = 15m + 20m

Total distance = 35m

Displacement = \(\sqrt{15^{2} +20^{2} \\}\)         (as both are perpendicular to each other)

Displacement = 25m

Therefore, Total distance is 35m and displacement is 25m.

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... then your weight is 25.2 lbf on the moon.

The work done on the box from y=0 to 6.0m is 120 J.

To calculate the work done on the box from y=0 to 6.0m, we need to find the area under the force vs. position graph over that interval.

First, we can find the work done from 0 m to 2 m. Since the force is constant at 40 N over this interval, the work done is simply:

W = F * d = 40 N * 2 m = 80 J

From 2 m to 6 m, the force is constant at -20 N, so the work done is:

W = F * d = (-20) N * 4 m = -80 J

Note that the negative sign indicates that the work is done by the box on the force (since the force is in the opposite direction of the displacement).

Therefore, the total work done on the box from y=0 to 6.0m is:

W_total = 80 J - 80 J = 0 J

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Answer:  

Explanation:

Step one:

given data

period T= 3 milliseconds= 0.003

velocity v= 175m/s

wave lenght λ=?

Step two:

we know that f=1/T

the expression relating period and wave lenght is

v=λ/T

λ=v*T

λ=175*0.002

λ=0.525m

to cm= 0.525*100

=52.5cm

The wavelength of the wave is E) 52.5 cm

Answer:

6 km

Explanation:

Let's first reduce the time travelled from minutes to hours since the speed of the car is given in km/h:

5 minutes = 5 / 60  h = 1/12 h

Now use this value in the formula for distant in uniform motion:

Distance = speed * time

Distance = 72 km/h * 1/12 h = 6 km

Taking into account the Newton's second law, the acceleration of the object is 5 m/s².

Newton's second law states that this force will change the speed of an object because the speed and/or direction will change and these are called acceleration.

Newton's second law states that: "The acceleration of an object is directly proportional to the force acting on it and inversely proportional to the mass."

Mathematically, Newton's second law is expressed as:

F= m×a

where:

In this case, you know:

Replacing in the definition of Newton's second law:

200 N= 40 kg× a

Solving:

a= 200 N÷40 kg

a= 5 m/s²

Finally, the acceleration is 5 m/s².

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Answer:

plis plis pilis plis plis plis plis plis plis plis plks plis plis plis

Answer:

effort arm mean the use of any work by using your hand force motion or by hand power

Answer:

10.4 m/s

Explanation:

I just know it's right!

Answer:

The correct answer is - fight or flight response or reaction.

Explanation:

According to Cannon, the fight or flight reaction is a strong emotion experienced by a person especially people experiences associated with a perceived threat.

This reaction takes place in form of arousal in a perceived threat or stress conditions. It helps in maintaining the body environment in panic or stressed conditions.

Answer:

To solve this problem, you'll need to break the initial velocity of the seed into its horizontal and vertical components, then use the equations of motion to find the time of flight and horizontal distance.

The initial velocity (v) of the seed is 2.65 m/s. The angle it's launched at (θ) is 30.0 degrees below the horizontal. The height (h) it's launched from is 0.460 m.

First, calculate the horizontal (v_x) and vertical (v_y) components of the velocity. Because the seed is launched downward, the vertical component will be negative:

v_x = v * cos(θ) = 2.65 m/s * cos(30.0) = 2.29 m/s

v_y = v * sin(θ) = -2.65 m/s * sin(30.0) = -1.325 m/s

Next, use the equation of motion to find the time it takes for the seed to hit the ground:

h = v_y * t + 0.5 * g * t^2

Where g is the acceleration due to gravity, which is approximately 9.8 m/s². Solving the equation for t gives:

t = (-v_y - sqrt((v_y)^2 - 4 * 0.5 * g * (-h))) / (2 * 0.5 * g)

Plugging in the values:

t = (1.325 + sqrt((-1.325)^2 - 4 * 0.5 * 9.8 * (-0.460))) / (2 * 0.5 * 9.8)

t = 0.182 seconds

Finally, use the horizontal velocity and time of flight to find the horizontal distance the seed covers:

x = v_x * t = 2.29 m/s * 0.182 s = 0.417 m

So, the seed lands after approximately 0.182 seconds and travels approximately 0.417 meters horizontally.

The minimum stopping distance of the car is determined as 8 m.

The minimum stopping distance of the car is calculated as follows;

d = (u²)/(2a)

where;

when the minimum stopping distance = 2 m, initial velocity = 40 km/hr = 11.11 m/s

2 = (11.11²)/(2a)

a = (11.11²)/(2 x 2)

a = 30.86 m/s²

when the speed becomes 80 km/h, the minimum stopping distance is calculated as;

u = 80 km/h = 22.22 m/s

d = (22.22² )/ (2 x 30.86)

d = 8 m

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Answer:

What are your options? also i believe the answer may be ecosystem.

Explanation:

There are both biotic and abiotic factors in an ecosystem. Be cause living organisms Are Biotic. While water and rocks are abiotic which are needed to form an ecosystem.

The minimum entropy change of the heat-supplying source is -0.87 kJ/K.

In this case, given the initial conditions, we first use the 10-% quality to compute the initial entropy.

S₁ = 1.485  +  (0.1)(5.6865) = 2.0537 kJ/kg K

The entropy at the final state given the new 40-% quality:

pressure inside the cooker = 150 kPa

S₂ = 1.4337  +  (0.4)(5.7894) = 3.7495 kJ/kg K

m₁ = 0.026/(0.001057   +  0.1 x 1.002643) = 0.257 kg

m₂ = 0.026/(0.001053   +  0.4 x 1.158347)  = 0.056 kg

ΔS + S₁  - S₂  + S₂m₂  - S₁m₁  -  sfg(m₂  -  m₁)  > 0

ΔS + 2.0537 -  3.7495 + (3.7495 x 0.056)  -   (2.0537 x 0.257)  - 5.6865( 0.056 - 0.257)  >  0

ΔS > -0.87 kJ/K

Thus, the minimum entropy change of the heat-supplying source is -0.87 kJ/K.

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The complete question is below:

A rigid, 26-L steam cooker is arranged with a pressure relief valve set to release vapor and maintain the pressure once the pressure inside the cooker reaches 150 kPa. Initially, this cooker is filled with water at 175 kPa with a quality of 10 percent. determine the exergy.

Heat is now added until the quality inside the cooker is 40 percent. The water is stirred at the same time that it is being heated. Determine the minimum entropy change of the heat-supplying source if 100 kJ of work is done on the water as it is being heated. Use steam tables

Answer:

C. the force applied must cause the movement of the object in the same direction as the force

Explanation:

looked it up

the rope wouldnt move, the force of both dogs pulling us also called tension

It as dangerous to be hit by the gun as by the bullet because of the following;

(A) The butt distributes the recoil force over an area much larger than that of the bullet.

(B) The rifle has a much lower speed than the bullet.

The principle of conservation of linear momentum states that in an isolated system, the total momentum of the system is conserved.

That is the sum of the initial momentum is equal to the sum of the final momentum.

momentum of the gun = momentum of the bullet

Mu = mU

where;

If the forward momentum of a bullet is the same as the backward momentum of the gun, the speed of the gun will be smaller than the speed of the bullet since the mass of the gun is bigger than mass of the bullet.

We cannot conclude on the kinetic energy, since it depends on both mass and velocity.

Finally, the butt distributes the recoil force over an area much larger than that of the bullet, since the butt has a larger surface area and will hit more surface area than the bullet.

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Answer:

Explanation:

b) light

Note that the amplitude of the SHM after the collision is 0.275 m. This  is solved using the conservation of momentum and conservation of energy principles.

To solve this problem, we need to use the conservation of momentum and conservation of energy principles.

Before the collision, the velocity of block 1 is in the direction of the spring's length. Therefore, we can assume that block 1 does not affect the SHM of block 2.

Using conservation of momentum:

m1v1 + m2v2 = (m1 + m2)vf

where

m1 = 5.80 kg (mass of block 1)

v1 = 5.70 m/s (velocity of block 1)

m2 = 2.90 kg (mass of block 2)

v2 = Aω (velocity of block 2 in SHM)

vf = (m1v1 + m2v2)/(m1 + m2)

Substituting the values:

(5.80 kg)(5.70 m/s) + (2.90 kg)(Aω) = (5.80 kg + 2.90 kg)vf

Simplifying:

16.63 + 2.9Aω = 8.7vf

Using conservation of energy:

(1/2)kA^2 = (1/2)m2ω^2A^2

where

k = m2ω^2 (spring constant)

A = amplitude of SHM

Substituting the values:

(1/2)(2.90 kg)(2π/0.018 s)^2 A^2 = (1/2)(2.90 kg)(2π/0.018 s)^2 A^2

Simplifying:

kA^2 = m2ω^2A^2

k = m2ω^2

Substituting the values:

(2.90 kg)(2π/0.018 s)^2 = 1307.6 N/m

Now we can solve for A:

(1307.6 N/m)A^2 = (5.80 kg)(5.70 m/s)2 + (2.90 kg)(Aω)2

Substituting the values:

(1307.6 N/m)A^2 = 99.01 J + (2.90 kg)(0.01 m/s)2A^2

Simplifying:

(1307.6 N/m - 0.01 kg/s^2)A^2 = 99.01 J

A^2 = 0.0757 m^2

A = 0.275 m

Therefore, the amplitude of the SHM after the collision is 0.275 m.

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Answer:

The remains of producers are broken down by decomposers.

The remains of consumers are broken down by soil decomposers.

Animals break down food molecules to obtain energy.

Explanation:

Answer:

It is A. B. and C.

Explanation:

Hope this helps

The force that results in the decrease in speed from the midpoint to the end of the track is friction. The friction force slows down the vehicle because it acts in the opposite direction of the car's motion.

The force that would cause the Hot Wheels car to slow down from the midpoint of the track to the end of the track is friction between the car's wheels and the track.

Friction is a force that opposes motion between two surfaces in contact.

In this case, the wheels of the car and the surface of the track are in contact, and the friction force acts in the opposite direction of the car's motion, which slows it down.

As the Hot Wheels car travels down Track #2 during the Speed Lab activity, its initial velocity decreases due to friction.

Friction is a resistance force that opposes motion.

It is caused by the interaction between the surfaces in contact. In this case, the surface of the track and the wheels of the car are in contact.

When the car is moving, there is friction between the two surfaces.

The direction of the friction force is opposite to the direction of motion of the car.

This means that the friction force slows the car down.

In conclusion, the force that results in the decrease in speed from the midpoint to the end of the track is friction.

The friction force slows down the vehicle because it acts in the opposite direction of the car's motion.

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Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

The bug will cross the sticky region once in each cycle of its motion, where a cycle is defined as one complete round trip from the top of the bowl to the bottom and back to the top.

To find the number of cycles the bug goes through, we can use conservation of mechanical energy. At the top of the bowl, the bug has only potential energy, which is converted to kinetic energy as it slides down the bowl. At the bottom of the bowl, all of the potential energy has been converted to kinetic energy, and as the bug slides up the other side of the bowl, the kinetic energy is converted back into potential energy. At the top of the bowl again, the bug has only potential energy, and the cycle repeats.

Because there is no friction (except for the sticky patch), the total mechanical energy of the system is conserved. Therefore, the potential energy at the top of the bowl is equal to the potential energy at the bottom of the bowl, and the kinetic energy at the bottom of the bowl is equal to the kinetic energy at the top of the bowl.

We can set the potential energy at the top of the bowl to zero, and use the conservation of energy to find the potential energy at the bottom of the bowl:

mgh = (1/2)mv^2

where m is the mass of the bug, g is the acceleration due to gravity, h is the depth of the bowl, and v is the speed of the bug at the bottom of the bowl.

Solving for v, we get:

v = sqrt(2gh)

Plugging in the numbers, we get:

v = sqrt(29.810.12) = 0.775 m/s

The time it takes for the bug to slide from the top of the bowl to the bottom and back up to the top is twice the time it takes to slide from the top to the bottom:

t = 2sqrt(2h/g) = 2sqrt(2*0.12/9.81) = 0.774 s

Therefore, the frequency of the bug's motion is:

f = 1/t = 1/0.774 = 1.29 Hz

Since the bug completes one cycle in each oscillation, the bug will cross the sticky region 1.29 times per second, or approximately once every 0.78 seconds.

The tax rate you will pay is displayed in tax brackets for each category of taxable income.

Thus, For instance, in 2022, the first $10,275 of your taxable income is subject to the lowest tax rate of 10% if you are single.

Up until the maximum amount of your taxable income, the following portion of your income is taxed at a rate of 12%.

As taxable income rises, the tax rate rises under the progressive tax system. Overall, this has the result that taxpayers with higher incomes often pay a greater rate of income tax than taxpayers with lower incomes.

Thus, The tax rate you will pay is displayed in tax brackets for each category of taxable income.

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Answer:10

Explanation: Acceleration of car is given by a= v^2/r, v= 50m/s, r= 250m a= 10 m/s^2 2

ANSWER:

(a) 4.73*10^-30 m

(b) 2.37*10^-19 times smaller

STEP-BY-STEP EXPLANATION:

Given:

Mass of ball = m = 0.10 kg

Speed of ball = v = 1.4x10^-3 m/s

(a)

Since, de Broglie wavelength is given by:

Where, h is the Plank's Constant ( h = 6.626x10^-34 kg m^2/s). Therefore, de Broglie wavelength of the ball will be:

(b)

It means that the wavelength of the ball is 2.37*10^-19 times smaller