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Art [367]
3 years ago
10

A box slides down a frictionless ramp.if it starts at rest, what is it’s speed at the bottom?

Physics
1 answer:
zhenek [66]3 years ago
3 0

8.854 m/s is the speed of the box after it reaches bottom of the ramp.

<u>Explanation</u>:

From the figure we came to know that height of the block is 4 m.

We know that,

Total "initial energy of an object" = Total "final energy of an object "

Total "initial energy of an object" is = "sum of potential energy" and "kinetic energy" of an object at its initial position.

\text { "g" acceleration due to gravity is } 9.8 \mathrm{m} / \mathrm{s}^{2}

\text { Total initial energy }=\mathrm{m} \times \mathrm{g} \times \mathrm{h}_{\mathrm{i}}+\frac{1}{2} \mathrm{m} v_{i}^{2}

Initial velocity is “0” as the object does not have starting speed

\text { Height of the block where the object is placed initially }\left(h_{i}\right) \text { is } 4 \mathrm{m} \text { . }

\text { Total initial energy }=\mathrm{m} \times 9.8 \times 4+\frac{1}{2} \mathrm{m} 0^{2}

Total initial energy = 39.2 × m

\text { Total final energy }=\mathrm{m} \times \mathrm{g} \times \mathrm{h}_{\mathrm{f}}+\frac{1}{2} m v_{f}^{2}

\text { We need to find final velocity } v_f

\text { Height of the block where the object is travelled to bottom (h_) is } 0 \mathrm{m} \text { . }

\text { Total final energy }=\mathrm{m} \times 9.8 \times 0+\frac{1}{2} m v_{f}^{2}

Now,  Total initial energy of an object = Total final energy of an object

39.2 \times \mathrm{m}=0.5 \mathrm{m} v_{f}^{2}

\frac{39.2}{0.5}=v_{f}^{2}

v_{f}^{2}=78.4

v_{f}=\sqrt{78.4}

v_{f}=8.854 \mathrm{m} / \mathrm{s}

Final speed is 8.854 m/s.

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Delvig [45]

To solve this problem it is necessary to apply the kinematic equations of motion.

By definition we know that the position of a body is given by

x=x_0+v_0t+at^2

Where

x_0 = Initial position

v_0 = Initial velocity

a = Acceleration

t= time

And the velocity can be expressed as,

v_f = v_0 + at

Where,

v_f = Final velocity

For our case we have that there is neither initial position nor initial velocity, then

x= at^2

With our values we have x = 401.4m, t=4.945s, rearranging to find a,

a=\frac{x}{t^2}

a = \frac{ 401.4}{4.945^2}

a = 16.41m/s^2

Therefore the final velocity would be

v_f = v_0 + at

v_f = 0 + (16.41)(4.945)

v_f = 81.14m/s

Therefore the final velocity is 81.14m/s

8 0
3 years ago
Anyone know the answer ?
dlinn [17]
Momentum = mass x velocity, so 500kg x 2m/s = 1000 kg m/s
5 0
2 years ago
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Light of wavelength 560 nm passes through a slit of width 0. 170 mm. (a) the width of the central maximum on a screen is 8. 00 m
faltersainse [42]

The distance between slit and the screen is 1.214m.

To find the answer, we have to know about the width of the central maximum.

<h3>How to find the distance between slit and the screen?</h3>
  • It is given that, wavelength 560 nm passes through a slit of width 0. 170 mm, and the width of the central maximum on a screen is 8. 00 mm.
  • We have the expression for slit width w as,

                           w=\frac{2*wavelength*d}{a}

where, d is the distance between slit and the screen, and a is the slit width.

  • Thus, distance between slit and the screen is,

                           d=\frac{w*a}{2*wavelength} =\frac{8*10^{-3}*0.17*10^{-3}}{560*10^{-9}*2} \\\\d=1.214m

Thus, we can conclude that, the distance between slit and the screen is 1.214m.

Learn more about the width of the central maximum here:

brainly.com/question/13088191

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3 0
1 year ago
a 1500 kg car traveling at 15 m/s to the south collides with a 4500 kg truck that is intially at rest at a spotlight. The car an
harkovskaia [24]

Answer:

3.75 m/s south

Explanation:

Momentum before collision = momentum after collision

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

Since the car and truck stick together, v₁ = v₂.

m₁ u₁ + m₂ u₂ = (m₁ + m₂) v

Given m₁ = 1500 kg, u₁ = -15 m/s, m₂ = 4500 kg, and u₂ = 0 m/s:

(1500 kg) (-15 m/s) + (4500 kg) (0 m/s) = (1500 kg + 4500 kg) v

-22500 kg m/s = 6000 kg v

v = -3.75 m/s

The final velocity is 3.75 m/s to the south.

4 0
2 years ago
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A rectangular gasoline tank can hold 50.0 kg of gasoline when full. What is the depth of the tank if it is 0.500-m wide by 0.900
sweet-ann [11.9K]

Answer:

0.16 m

Explanation:

A rectangular gasoline tank can hold 50.0 kg of gasoline when full, and the density of gasoline is 6.8 × 10² kg/m³. We can find the volume occupied by the gasoline (volume of the tank).

50.0 kg × (1 m³/6.8 × 10² kg) = 0.074 m³

The volume of the rectangular tank is:

volume = width × length × depth

depth = volume / width × length

depth = 0.074 m³ / 0.500 m × 0.900 m

depth = 0.16 m

3 0
3 years ago
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