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Keith_Richards [23]
3 years ago
13

A 3.00-kg crate slides down a ramp. the ramp is 1.00 m in length and inclined at an angle of 30.08 as shown in the figure. The c

rate starts from rest at the top, experiences a constant friction force of magnitude 5.00 N, and continues to move a short distance on the horizontal floor after it leaves the ramp.
Physics
1 answer:
Dominik [7]3 years ago
4 0

Answer:

2.55 m/s

Explanation:

A 3.00-kg crate slides down a ramp. the ramp is 1.00 m in length and inclined at an angle of 30° as shown in the figure. The crate starts from rest at the top, experiences a constant friction force of magnitude 5.00 N, and continues to move a short distance on the horizontal floor after it leaves the ramp. Use energy methods to determine the speed of the crate at the bottom of the ramp.

Solution:

The work done by friction is given as:

W_f=F_f\Delta S\\\\Where\ F_f\ is\ the \ frictional\ force=-5N(the\ negative \ sign\ because\ it\\acts\ opposite\ to \ direction\ of\ motion),\Delta S=slope\ length=1\ m\\\\W_f=F_f\Delta S=-5\ N*1\ m=-5J

The work done by gravity is:

W_g=F_g*s*cos(\theta)\\\\F_g=force\ due\ to\ gravity=mass*acceleration\ due\ to\ gravity=3\ kg*9.81\\m/s^2, s=1\ m, \theta=angle\ between\ force\ and\ displacement=90-30=60^o\\\\W_g=3\ kg*9.81\ m/s^2*1\ m*cos(60)=14.72\ J\\\\The\ Kinetic\ energy(KE)=W_f+W_g=14.72\ J-5\ J=9.72\ J\\\\Also, KE=\frac{1}{2} mv^2\\\\9.72=\frac{1}{2} (3)v^2\\\\v=\sqrt{\frac{2*9.72}{3} } =2.55\ m/s

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Answer:

1.) h = 164.8 m

2.) U = 49.1 m/s

3.) t = 1.43 seconds

Explanation:

1.) A soccer ball is dropped from the top of a building. It takes 5.8 seconds to fall to the ground. The height of the building is...? 

Since the soccer ball is dropped from the building, the initial velocity U will be equal to zero

Using second equation of motion

h = Ut + 1/2gt^2

Substitutes the time into the formula

h = 1/2 × 9.8 × 5.8^2

h = 164.8 m

2. The Falcon 9 launches to a height of 123 meters. What is its vertical initial velocity?

At maximum height final velocity = 0

Using the third law of motion

V^2 = U^2 - 2gH

0 = U^2 - 2 × 9.8 × 123

U^2 = 2410.8

U = 49.1 m/s

3. An apple falls from rest off a 10.m m tree. How long will it take before it hits the ground?

Since the apple fall from rest, the initial velocity U will be equal to zero

Using the second equation of motion,

h = Ut + 1/2gt^2

substitute all the parameters into the formula

10 = 1/2 × 9.8 × t^2

10 = 4.9t^2

t^2 = 10/4.9

t^2 = 2.04

t = 1.43 seconds

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3 years ago
Subtracting a negative number follows the same process as a positive number
Sav [38]
This would be true!

When you subtract a positive number, it goes to the left of the number line. When you subtract a negative number, it goes to the left of the number line.

The thing that confuses most people is that the first number (when negative) continues to increase as a negative. It follows the same process, just different signs.

I hope this helps!
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A railroad freight car, mass 15,000 kg, is allowed to coast along a level track at a speed of 2.0 m/s. It collides and couples w
gayaneshka [121]

Answer:

The speed of the two cars after coupling is 0.46 m/s.

Explanation:

It is given that,

Mass of car 1, m₁ = 15,000 kg

Mass of car 2, m₂ = 50,000 kg

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Initial speed of car 2, u₂ = 0

Let V is the speed of the two cars after coupling. It is the case of inelastic collision. Applying the conservation of momentum as :

m_1u_1+m_2u_2=(m_1+m_2)V

V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}

V=\dfrac{15000\ kg\times 2\ m/s+0}{(15000\ kg+50000\ kg)}  

V = 0.46 m/s

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Answer:

a. distance = s x t

Explanation:

The equation we know for speed is speed = distance/time, shortened to s = d/t

If you want to find the distance, you'd rearrange the formula to make d the subject. To isolate d, multiply by t on both sides (since it's being divided on the right. This leads to

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Therefore, to find the distance an object travelled you'd use a. distance = s x t

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galben [10]

Answer:

3.19*10^-24

Explanation:

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