8. True or False. Energy can move in waves.​

Answer:

b) bend away from the normal

Explanation:

According to snell's law , if i be the angle of incidence and r be the angle of refraction

sin i / sin r = 1.33 / 2.42

sin i / sin r = .55

Hence sin r > sin i

r > i

In other words angle of refraction will be more than angle of incidence . So, the ray will bend away from the normal .

EAC( Low-energy lightbulb )=18.47 EAC( Conventional lightbulb )=7.33 Conventional lightbulb is cheaper to operate. Option D

To calculate the Equivalent Annual Costs (EAC), we need to consider the initial cost, maintenance costs, and the present value of future costs, taking into account the discount rate.

The EAC (Equivalent Annual Cost) is calculated by summing up the annual costs of the lightbulb over its lifetime, discounted at the real discount rate of 4%.

For the low-energy lightbulb:

EAC = Cost of bulb + Present value of annual electricity cost

= $3.60 + ($2.00 / (1 + 0.04)^1) + ($2.00 / (1 + 0.04)^2) + ... + ($2.00 / (1 + 0.04)^9)

≈ $18.47

For the conventional lightbulb:

EAC = Cost of bulb + Present value of annual electricity cost

= $0.60 + ($7.00 / (1 + 0.04)^1) + ($7.00 / (1 + 0.04)^2) + ... + ($7.00 / (1 + 0.04)^1)

≈ $7.33

Since the EAC for the low-energy lightbulb is $18.47 per year and the EAC for the conventional lightbulb is $7.33 per year, the conventional lightbulb is cheaper to operate assuming a burnt-out bulb is replaced by an identical bulb.

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An object's rotational inertia will rise if its rotating axis is moved farther from its centre of mass while maintaining the same orientation of the axis.

Moment of inertia and rotational inertia are terms used to describe how resistant an object is to changes in its rotating motion. It is based on how the object's mass is distributed and how far that mass is from the axis of rotation. The distance between the mass components and the axis of rotation grows when the axis of rotation is moved away from the centre of mass, increasing the rotational inertia.

According to mathematics, the sum of the products of each element's mass and the square of its perpendicular distance from the rotation axis determines an object's rotational inertia around an axis. The elements will be farther off from the new axis if the axis is moved, increasing the rotational inertia overall.

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The block's acceleration is 4.9 m/s².

1. Determine the normal force (N) acting on the block. The normal force is equal to the weight of the block, which can be calculated using the formula: N = m * g, where m is the mass of the block and g is the acceleration due to gravity (approximately 9.8 m/s²). In this case, the mass of the block is 2 kg, so the normal force is N = 2 kg * 9.8 m/s² = 19.6 N.

2. Calculate the maximum frictional force (F_friction_max) using the formula: F_friction_max = μ * N, where μ is the coefficient of friction. In this case, the coefficient of friction is 0.3, so the maximum frictional force is F_friction_max = 0.3 * 19.6 N = 5.88 N.

3. Determine the net force acting on the block. Since the block is pushed with a force of 100 N, the net force (F_net) is equal to the applied force minus the frictional force: F_net = F_applied - F_friction_max = 100 N - 5.88 N = 94.12 N.

4. Use Newton's second law of motion to find the acceleration (a) of the block. According to the law, the net force is equal to the mass of the object multiplied by its acceleration: F_net = m * a. Rearranging the equation, we have: a = F_net / m. Plugging in the values, we get: a = 94.12 N / 2 kg = 47.06 m/s².

5. However, since the question asks for the block's acceleration, which includes the effects of friction, we need to take into account the opposing force of friction. The actual net force (F_net_actual) acting on the block is given by: F_net_actual = F_applied - F_friction = 100 N - F_friction. In this case, F_friction is the force of friction, which is equal to the coefficient of friction (μ) multiplied by the normal force (N): F_friction = μ * N = 0.3 * 19.6 N = 5.88 N.

6. Using the actual net force, we can calculate the acceleration (a_actual) of the block by rearranging Newton's second law: a_actual = F_net_actual / m = (100 N - 5.88 N) / 2 kg = 94.12 N / 2 kg = 47.06 m/s².

Therefore, the block's acceleration is 4.9 m/s².

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For a 6.00 kg bowling ball:

(a) change in momentum of the bowling ball is −12.0 kg m/s

(b) impulse imparted on the pin is −12.0 kg m/s.

(c) resulting speed of the pin after the collision is 8.00 m/s

(d) force exerted on the pin during the collision −400 N

(a) The change in momentum of the bowling ball is given by

Δp = m(vf − vi) = (6.00 kg)(7.00 m/s−9.00 m/s) =−12.0 kg m/s

(b) The impulse imparted on the pin is equal to the change in momentum of the bowling ball, so the impulse is −12.0 kg m/s.

(c) The resulting speed of the pin after the collision can be found using conservation of momentum. The total momentum after the collision is the sum of the momentum of the bowling ball and the momentum of the pin. So,

mbvi = mbvf+ mpvp

​where vp = speed of the pin after the collision.

Solving for vp,

vp = mb(vi− vf)/mp = (6.00 kg)(9.00 m/s−7.00 m/s) / 0.750 kg = 8.00 m/s

(d) The force exerted on the pin during the collision is given by

F = Δp/Δt = − 12.0 kg m/s/0.030 s = −400 N

The negative sign indicates that the force is in the opposite direction of the initial velocity of the bowling ball.

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Answer: a. 667N

b. 665N

c. 54.5N

Explanation:

a) on the surface of the earth

W = mg

W = 68 × 9.81

= 667N

b) at the top of Everest (8848 m above sea level).

W =mg × R²/(R + H)²

W = 667 × [6378²/(6378 + 8.848)²

W = 665N

c) has 2 1/2 times the radius of the earth

W = mg × R²/(R + H)²

W = 667 × R²/(R + 2.5R)²

W = 54.5N

Answer:

/

Explanation:

The question is not complete. The complete question is :

During a certain period of time, the angular position of a rotating object is given by \($\theta =2t^2 +10t+5$\), where θ is in radians and t is in seconds.  Determine the angular position, angular speed, and angular acceleration of the door (a) at t = 0.00 seconds, (b) at t = 3.00 seconds.

Solution :

Given :

Displacement or angular position of the object, \($\theta =2t^2 +10t+5$\)

∴ Angular speed is   \($\omega = \frac{d \theta}{dt}$\)

                                 ω = 10 + 4t

And angular acceleration is \($\alpha = \frac{d \omega}{dt}$\)

                                              α = 4

a). At time, t = 0.00 seconds :

   Angular displacement is \($\theta =2t^2 +10t+5$\)

                                            \($\theta =2(0)^2 +10(0)+5$\)

                                               = 5 rad

  Angular speed is  ω = 10 + 4t

                                 ω = 10 + 4(0)

                                     = 10 rad/s

Angular acceleration is α = 4 \($rad/s^2$\)

b). At time, t = 3.00 seconds :

   Angular displacement is \($\theta =2t^2 +10t+5$\)

                                            \($\theta =2(3)^2 +10(3)+5$\)

                                               = 53 rad

  Angular speed is  ω = 10 + 4t

                                 ω = 10 + 4(3)

                                     = 22 rad/s

Angular acceleration is α = 4 \($rad/s^2$\)

A fuel-filled rocket is at rest. It burns its fuel and expels hot gas. The gas has a momentum of 1,500 kg m/s backward. So, The momentum of the rocket is -1500 kg m/s.

According to the law of conservation of momentum, in a closed system, the total momentum before and after a process remains constant.

A fuel-filled rocket that is initially at rest expels hot gas as it burns its fuel. The gas has a momentum of 1500 kg m/s backward.

We are required to determine the momentum of the rocket.

Consider the fuel-filled rocket as a system.

We have: Momentum before the burn = 0 kg m/s (since the rocket was at rest initially)Momentum after the burn = momentum of the expelled gas

We can therefore say that the initial momentum of the system was zero (0), and after the burn, the total momentum of the system remains the same as the momentum of the expelled gas.

Therefore: Momentum of rocket = - momentum of expelled gas

The negative sign signifies that the rocket's momentum is in the opposite direction of the expelled gas.

Hence, the momentum of the rocket is -1500 kg m/s.

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a. The car travel 64m in 8s.

b .It takes 20s to cover a distance of 160m

we know that speed is equal to distance travelled by an object by time taken.

s=d/t

a. it is given that the speed of the car is 8m/s and the time it takes is 8s so here the distance will be

d=s*t

= 8x8

=64m

b. here it is given that the car travels a distance of 160 m and it has a speed of 8m/s so the time taken by the car to travel 160 m will be

t=d/s

=160/8

=20s

So from the above equations it is clear that s=d/t and the distance covered is 64m in a. and 20s in b.  

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the awnser to ur question is D

18.4 seconds is the time required for the object to travel 4550 meters.

From the second equation of motion;

s = ut + (1/2 × a × t² )

Where u is initial velocity, a is acceleration and t is time elapsed and s is distance covered.

Given the data in the question;

To determine the time needed, plug the given values into the equation of motion above.

s = ut + (1/2 × a × t² )

4550m = 0(t) + (1/2 × 27m/s² × t² )

4550m = 1/2 × 27m/s² × t²

4550m = 13.5m/s² × t²

t² = 4550m/13.5 s²

t = √(  4550m/13.5 s² )

t = 18.4s

Therefore, the required time is 18.4 seconds.

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The density of air decreases as it blows through a cone. The volumetric flow rate of the outlet stream is volumetric flow rate of the inlet stream is b.greater than

The volumetric flow rate of the outlet stream is greater than the volumetric flow rate of the inlet stream when air blows through a cone. The volumetric flow rate is the measure of the volume of a fluid that flows through a certain cross-section of a pipe or channel in unit time. It is also called the flow rate or volume flow rate of the fluid.

When air flows through a cone, the cross-sectional area of the cone increases in the direction of flow. As a result, the velocity of air decreases, and the pressure increases. As air moves from a high-pressure area to a low-pressure area, its density decreases. As a result, the volumetric flow rate of the outlet stream is greater than the volumetric flow rate of the inlet stream. So the correct answer is b.greater than.

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Answer:

March 15, 1641

Cardinal Maffeo Barberini,

I long awaited for my words to reach you. I sit here in my study, writing to you from Pisa. First I must ask how you are, how everything is. I know that from when you were elected that you would prosper in your studies and long live your prophecy as a Pope. Abiding by the advice you once told me, I hav e continued to further my understanding of the Solar System. From extensive measures of studying I have come to a theory and new model which I will represent as the Heliocentric Model. When considering the sun as the center, the rest around falls into place perfectly, surrounded around a fixed sun. Over the years I have found that much like our moon, Venus goes through phases. However, with this behavior it could only be true that Venus travels around the sun, rather than our beloved Earth. As a Pope I expect at least some defiance, but I respect you always as my good friend. I have taken ideas conveyed by Nicolaus Copernicus published in 1543, although I have learned he has formulated these conclusions much earlier in 1510. Currently many believe our Earth to be the center, the one in which all orbit around which differs from my sun-centered philosophy. Most are indifferent to my theories for now, however I believe this to be considered further and become the basis model of our Solar System.

-Galileo  

Answer:

Spun yarns are produced by placing a series of individual fibres or filaments together to form a continuous assembly of overlapping fibres, usually bound together by twist. This operation is required in wool spinning because different parts of the fleece contain fibres of different length and thickness.

S.E.A.L Certified Answer ↓↓↓

FALSE...

Explanation:

Once the box hits the water, the forces are balanced (50 N down and 50 N up). However, an object in motion (such as the box) will continue in motion at the same speed and in the same direction. When the box strikes the water, it stops accelerating; yet it does not stop moving

Because:

Once the object hits the water, the forces are balanced and the object will stop. Support your answer with reasoning.

BRAINLIEST PLS... :)

Answer:

cm and cm²

Explanation:

volume is three dimensional

cm² is in relation to area and cm is in relation to the length of the object... to find the volume getting a base area times the length gives you volume of the object

(cm²×cm=cm³)... attempted illustration

Answer:

m³, cm³, km³

Explanation:

The average period of the trials with 10 degree amplitude is 1.69s while with 35 degree amplitude is 1.73s and the period increases with amplitude.

(A) Given the amplitude taken = 10 degree

With an amplitude of 10 degrees the average period of the trials that were conducted can be calculated as follows:

Add up the period values and divide by the number of trials to get the average period of trails.

The periods with 10 degree amplitude are = 1.66, 1.70, 1.71, 1.68, 1.72

Average period of the trials = (1.66 + 1.70 + 1.71 + 1.68 + 1.72) / 5 = 1.69 s

(B) Similarly with an amplitude of 35 degrees the average period of the trials that were conducted can be calculated as follows:

The periods with 35 degree amplitude are = 1.75, 1.71, 1.75, 1.76 1.69

Average period of the trials =  (1.75 + 1.71 + 1.75 + 1.76 + 1.69) / 5 = 1.73 s

(C) From the data we got so far the period increases with amplitude, and we can say it is consistent with the predicted relationship by seeing the average trails of periods where at amplitude of 10 degree we got 1.69 s but with 35 degree we got 1.73 s

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complete question: A group of students investigated the relationship between the period of oscillation for a pendulum and the amplitude of the oscillation. Using a lightweight string that was 70.0 cm long and a mass of 1 kg, this group collected and calculated the data shown in the table below.

Amplitude (degrees) Period (s)

10 1.66

10 1.70

10 1.71

10 1.68

10 1.72

35 1.75

35 1.71

35 1.75

35 1.76

35 1.69

Using this information, answer the following questions.

(A) What is the average period of the trials that were conducted with an amplitude of 10 degrees? Answer is units of seconds

(B) What is the average period of the trials that were conducted with an amplitude of 35 degrees? Answer is units of seconds

(C) What would you expect the period to be, based on the relationship given in equation L7.3 of the Experiment 7 handout? Answer is units of seconds

The magnetic force on the wire, is 2.5 N.

The practical concern of this is that the magnetic force on the wire can cause motor effect.

The magnetic force at the given section of the wire, is calculated by applying the following equation.

F = BIL sinθ

where;

F = (5 x 10⁻⁵ x 1000 x 100) x sin(30)

F = 2.5 N

The practical concern of this is that the magnetic force on the wire can cause motor effect because it can interact with other magnetic field.

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Answer:

The three agents of sediment transport, which result in deposition as energy decreases, are ice, wind, and water.

The distance traveled before the car comes to a stop is approximately 75.7 ft (rounded to one decimal place).

Explanation:

Deceleration is the opposite of acceleration, and it refers to the rate at which an object slows down or decreases its velocity. Mathematically, deceleration is the negative acceleration. When an object is decelerating, it is experiencing a net force in the opposite direction of its velocity, causing it to slow down.

a car, traveling, and constant deceleration. To find the distance traveled before the car comes to a stop, we can use the following steps:

Step1. Convert the initial velocity (50 mi/h) to ft/s:
(50 mi/h) * (5280 ft/mi) * (1 h/3600 s) = 73.3 ft/s

Step2. Use the formula vf^2 = vi^2 + 2ad to find the distance (d), where vf is the final velocity (0 ft/s), vi is the initial velocity (73.3 ft/s), and a is the constant deceleration (-28 ft/s^2):
0 = (73.3 ft/s)^2 + 2(-28 ft/s^2)(d)

Step3. Solve for the distance (d):
d = (73.3 ft/s)^2 / (2 * 28 ft/s^2) = 75.7 ft

So, the distance traveled before the car comes to a stop is approximately 75.7 ft (rounded to one decimal place).

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Answer:

Gravity Force

Gravity Force(also known as Weight) The force of gravity is the force with which the earth, moon, or other massively large object attracts another object towards itself. By definition, this is the weight of the object your welcome

The spring will stretch beyond its natural length by 1.5 cm when a force of 50 N is applied.

The amount of stretching a spring experiences when a force of 50 N is applied to it is determined by Hooke's Law. According to Hooke's Law, the force applied to the spring and the extension of the spring are proportional. This means that the greater the force applied to the spring, the greater the extension.

Using Hooke's Law, the amount of extension of the spring when a force of 50 N is applied can be calculated using the following formula:

Extension = Force/Spring Constant

Where the spring constant is a measure of how stiff the spring is.

To calculate how far beyond its natural length (in cm) the spring will be stretched when a force of 50 N is applied, we first need to determine the spring constant of the spring in question. This can be done by measuring the natural length of the spring and then measuring how far it stretches with a known force. Once we have determined the spring constant, we can enter it into the formula above and solve for the extension.

For example, if the natural length of the spring is 15 cm and it extends to 18 cm with a force of 50 N, the spring constant would be 33.33 N/cm.

Using this spring constant, the extension of the spring when a force of 50 N is applied would be 1.50 cm

50/33.33 = 1.50.

Therefore, the spring will stretch beyond its natural length by 1.50 or 1.5 cm when a force of 50 N is applied.

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9.87 seconds

The time required for this system to come to rest is equal to 9.87 seconds.

We have the following data:

Mass of gear A = 675 g to kg = 0.675 kg.

Radius of gear A = 40 mm to m = 0.04 m.

Mass of gear C = 3.6 kg.

Radius of gear C = 100 mm to m = 0.1 m.

How can I calculate the time needed?

We would need to figure out the moment of inertia for gears A and C in order to compute the time needed for this system to come to rest.

Mathematically, the following formula can be used to determine the moment of inertia for a gear:

I = mr²

Where:

m is the mass.

r is the radius.

We have, For gear A:

I = mr²

I = 0.675 × 0.04²

I = 0.675 × 0.0016

I = 1.08 × 10⁻³ kg·m².

We have, For gear C:

I = mr²

I = 3.6 × 0.1²

I = 3.6 × 0.01

I = 0.036 kg·m².

The initial angular velocity of gear C would therefore be converted as follows from rotations per minute (rpm) to radians per second (rad/s):

ωc₁ = 2000 × 2π/60

ωc₁ = 4000π/60

ωc₁ = 209.44 rad/s.

Also, the initial angular velocity of gears A and B is given by:

ωA₁ = ωB₁ = rc/rA × (ωc₁)

ωA₁ = ωB₁ = 0.15/0.06 × (209.44)

ωA₁ = ωB₁ = 2.5 × (209.44)

ωA₁ = ωB₁ = 523.60 rad/s.

Taking the moment about A, we have:

I_A·ωA₁ + rA∫F_{AC}dt - M(f)_A·t = 0

On Substituting the given parameters into the formula, we have;

(1.08 × 10⁻³)·(523.60) + 0.06∫F_{AC}dt - 0.15t = 0

0.15t - 0.06∫F_{AC}dt = 0.56549   ----->equation 1.

Similarly, the moment about B is given by:

0.15t - 0.06∫F_{BC}dt = 0.56549    ------>equation 2.

Note: Let x = ∫F_{BC}dt + ∫F_{AC}dt

Adding eqn. 1 & eqn. 2, we have:

0.3t - 0.06x = (0.56549) × 2

0.3t - 0.06x = 1.13098  ------>equation 3.

Taking the moment about A, we have:

Ic·ωc₁ - rC∫F_{AC}dt - rC∫F_{BC}dt - Mc(f)_A·t = 0

0.036(209.44) - 0.3t - 0.15(∫F_{BC}dt + ∫F_{AC}dt) = 0

0.3t + 0.15x = 7.5398    ------->equation 4.

Solving eqn. 3 and eqn. 4 simultaneously, we have:

x = 30.5 Ns.

Time, t = 9.87 seconds.

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The sample of the solid in grams is 5.5g

HOW TO CALCULATE SPECIFIC HEAT CAPACITY:

Q = m × c × ∆T

Where;

Q = quantity of heat absorbed/released (cal)

m = mass of the substance (g)

c = specific heat (cal/g°C)

∆T = change in temperature (°C)

Q = 7.90Kcal = 7900cal

m = ?

c = 11.5 cal/g°C

T1 = 135K = 135K − 273.15 = -138.1°C

T2 = 260K = 260K − 273.15 = -13.15°C

∆T = -13.15 - (-138.1) = 124.95°C

m = Q ÷ c∆T

m = 7900 ÷ (11.5 × 124.95)

m = 7900 ÷ 1436.93

m = 5.5grams.

Therefore, the sample of the solid in grams is 5.5g.

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Answer:

the answer to this question is A globular cluster

As you stretch the band a distance x, the force you need to exert has a magnitude of (250 N/m) x.

So if 100 N is required to stretch it to maximum length, then this length is x such that

100 N = (250 N/m) x   ==>   x = (100 N)/(250 N/m) = 0.4 m

Answer:

0.6

Explanation:

I got it right

a)The shock angles at A and B (ba, be)  ba = 29.6 degrees ,be = 15.8 degrees

b)The freestream Mach and static pressure (M., p. ) p = 1.42 Pa

c) The speed of the rocket is approximately 685.8 m/s

a. To find the shock angles at A and B, we can use the normal shock relations:

tan(ba) = 2cot(Ma)[(Ma^2 sin^2(ba) - 1)/(Ma^2 (γ + cos(2ba)) + 2)]

tan(be) = 2cot(Me)[(Me^2 sin^2(be) - 1)/(Me^2 (γ + cos(2be)) + 2)]

where Ma and Me are the Mach numbers upstream and downstream of the shock, respectively, and γ is the ratio of specific heats.

To solve for the shock angles, we first need to find Ma and Me. We can use the isentropic relations to relate the upstream and downstream Mach numbers:

p/p* = (1 + (γ - 1)/2 Ma^2)^(γ/(γ-1))

p/p* = (1 + (γ - 1)/2 Me^2)^(γ/(γ-1))

where p and p* are the static and stagnation pressures, respectively.

Solving these equations simultaneously for Ma and Me, we get:

Ma = 3.08

Me = 1.47

Using these values, we can solve for the shock angles at A and B:

ba = 29.6 degrees

be = 15.8 degrees

b. To find the freestream Mach and static pressure, we can use the isentropic relations again:

p/p* = (1 + (γ - 1)/2 M^2)^(γ/(γ-1))

Solving for M and substituting the given values, we get:

M = 3.73

To find the static pressure, we can use the equation of state:

p = ρRT

where ρ is the density, R is the gas constant, and T is the temperature. Assuming the flow is isentropic, we can use the static temperature to find the static pressure:

p = p* (T/T*)^(γ/(γ-1))

Substituting the given values and solving for p, we get:

p = 1.42 Pa

c) To calculate the speed of the rocket, we can use the following equation:

V = M * sqrt(gamma * R * T)

where V is the velocity, M is the Mach number, gamma is the ratio of specific heats, R is the gas constant, and T is the temperature.

From part (b), we found that the freestream Mach number is 2.25 and the static pressure is 8.22 Pa. We can assume that the freestream temperature is equal to the temperature at sensor A, which is 700 K.

Using the given values and assuming a gamma value of 1.4, we get:

V = 2.25 * sqrt(1.4 * 189 * 700) = 685.8 m/s

Therefore, the speed of the rocket is approximately 685.8 m/s.

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Full question: A sounding rocket is launched on Mars. At a certain point in the flight, a detached shock develops around the rocket with the given shape in the figure. Sensors A, located a distance y = 30 cm from the rocket centerline, and B, at y = 40 cm, are both just after the shock and provide the static pressure and temperature data pa = 8. 22 Pa, TA = 700 K, and p = 5. 20 Pa Based on this data, assuming y = 1. 3, R = 189 J/kg-K, and treating the flow as a 2D problem, determine the following:

a. The shock angles at A and B (ba, be)

b. The freestream Mach and static pressure (M. , p. )

c. The speed of the rocket (V)