Which is the answer for this question

As a box slides down a ramp, friction does 23.0 joules of work. At the bottom of the ramp, the box has 3.8 joules of kinetic ene

tensa zangetsu [6.8K]

Answer:

The high of the ramp is 2.81[m]

Explanation:

This is a problem where it applies energy conservation, that is part of the potential energy as it descends the block is transformed into kinetic energy.

If the bottom of the ramp is taken as a potential energy reference point, this point will have a potential energy value equal to zero.

We can find the mass of the box using the kinetic energy and the speed of the box at the bottom of the ramp.

$E_{k}=0.5*m*v^{2}\\\\where:\\E_{k}=3.8[J]\\v = 2.8[m/s]\\m=\frac{E_{k}}{0.5*v^{2} } \\m=\frac{3.8}{0.5*2.8^{2} } \\m=0.969[kg]$

Now applying the energy conservation theorem which tells us that the initial kinetic energy plus the work done and the potential energy is equal to the final kinetic energy of the body, we propose the following equation.

$E_{p}+W_{f}=E_{k}\\where:\\E_{p}= potential energy [J]\\W_{f}=23[J]\\E_{k}=3.8[J]\\$

And therefore

$m*g*h + W_{f}=3.8\\ 0.969*9.81*h - 23= 3.8\\h = \frac{23+3.8}{0.969*9.81}\\ h = 2.81[m]$

8 0

3 years ago

What is the last step to take in order to keep food safe?

gtnhenbr [62]

I think it would be clean and store.

6 0

3 years ago

Read 2 more answers

The tip of a tuning fork goes through 440 complete vibrations in a time of 0.510s. Find the angular frequency and the period of

Brilliant_brown [7]

Solve for the time it will take to complete a revolution. That is,
0.510 s / 440 revolutions = 51/44000 s
The frequency in Hertz is the reciprocal of this value thus the frequency is approximately equal to 862.75 Hz.

Angular velocity expressed in radians/second
(440 rev / 0.510 s) x (2π rad / 1 rev) = 5420.787 rad/s

The period is the reciprocal of frequency which is approximately equal to 1.16x10^-3 s.

7 0

3 years ago

Read 2 more answers

Suppose we take a 1 m long uniform bar and support it at the 21 cm mark. Hanging a 0.40 kg mass on the short end of the beam res

Ipatiy [6.2K]

Answer:

The mass of the beam is = 29 kg.

Explanation:

A beam with mass 40 kg is shown in figure. Point S is the support point. Point B is the middle point on the beam where mass of the beam acts.

Taking moment about Point S

40 × 21 = $M_{beam}$ × 29

$M_{beam}$ = 29 kg

Therefore the mass of the beam is = 29 kg.

4 0

3 years ago

A 750 kg car is stalled on an icy road during a snowstorm. A 1000 kg car traveling eastbound at 13 m/s collides with the rear of

scZoUnD [109]

(a) The magnitude of the velocity of the 1000 kg car after the collision is 10.5 m/s.

(b) The direction of the velocity of the 1000 kg car after the collision is 8.2 ⁰ north west.

(c) The ratio of the kinetic energy of the two cars just after the collision to that just before the collision is 0.72.

<h3>Velocity of the 1000 kg after the collision</h3>

Apply the principle of conservation of linear momentum as follows;

<h3>Final velocity in x direction</h3>

m₁u₁ + m₂u₂ = m₁v₁x + m₂v₂x

where;

m₁ is mass of 750 kg car
u₁ is initial velocity of 750 kg mass
m₂ is mass of 1000 kg car
u₂ is initial velocity of 1000 kg mass
v₁ is final velocity of 750 kg mass
v₂ is final velocity of 1000 kg mass

750(0) + 1000(13) = 750(4 cos 30) + 1000v₂x

13000 = 2,598.1 + 1000v₂x

10,401.9 = 1000v₂x

v₂x = 10.4 m/s

<h3>Final velocity in y direction</h3>

m₁u₁ + m₂u₂ = m₁v₁y + m₂v₂y

750(0) + 1000(0) = 750(4 sin 30) + 1000v₂y

0 = 1500 + 1000v₂y

v₂y = -1500/1000

v₂y = -1.5 m/s

<h3>Resultant final velocity</h3>

v = √(v₂ₓ² + v₂y²)

v = √[(10.4)² + (-1.5)²]

v = 10.5 m/s

<h3>Direction of the final velocity of 1000 kg car</h3>

tanθ = v₂y/v₂ₓ

tanθ = -1.5/10.4

tanθ = -0.144

θ = arc tan(-0.144)

θ = 8.2 ⁰ north west

<h3>Kinetic energy of the cars before the collision</h3>

K.Ei = 0.5m₁u₁² + 0.5m₂u₂²

K.Ei = 0.5(750)(0)² + 0.5(1000)(13)²

K.Ei = 84,500 J

<h3>Kinetic energy of the cars after the collision</h3>

K.Ef = 0.5(750)(4)² + 0.5(1000)(10.5)²

K.Ef = 61,125 J

<h3>Ratio of the kinetic energy</h3>

K.Ef/K.Ei = 61,125/84,500

K.Ef/K.Ei = 0.72

Thus, the magnitude of the velocity of the 1000 kg car after the collision is 10.5 m/s.

The direction of the velocity of the 1000 kg car after the collision is 8.2 ⁰ north west.

The ratio of the kinetic energy of the two cars just after the collision to that just before the collision is 0.72.

Learn more about kinetic energy here: brainly.com/question/25959744

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3 0

2 years ago