A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s.

Answers

Answer 1

Answer:

The work ( $W_f$ ) done by the friction force between the ramp and the skateboarder is given by $-\mu_k \cdot m_s \cdot g \cdot h_y$ .

The workdone by the friction force ( $W_f$ ) can be calculated using the formula for work, which is the product of the force applied ( $F_f$ ) and the displacement (d) over which the force is applied:

$W_f = F_f \cdot d$

In this scenario, the frictionforce works against the skateboarder's momentum down the ramp, therefore it does no good.

Given:

Mass of skateboarder ( $m_s$ ) = 54 kg

Height of the ramp ( $h_y$ ) = 3.3 m

Final velocity ( $v_f$ ) = 6.2 m/s

Coefficient of kineticfriction ( $\mu_k$ ) between skateboarder and ramp

Acceleration due to gravity (g) = $9.81 \, \text{m/s}^2$

The normal force ( $F_{\text{normal}}$ ) is equal to the weight of the skateboarder:

$F_{\text{normal}} = m_s \cdot g$

The displacement (d) is the vertical distance ( $h_y$ ) that the skateboarder descends down the ramp.

Now we can write the expression for the work done by the friction force ( $W_f$ ):

$W_f = -\mu_k \cdot F_{\text{normal}} \cdot d$

Substitute the expression for the normal force:

$W_f = -\mu_k \cdot (m_s \cdot g) \cdot h_y$

Thus, this expression represents the work done by the friction force between the ramp and the skateboarder in terms of the given variables.

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Your question seems incomplete, the probable complete question is:

A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s.

Write an expression for the work, Wf, done by the friction force between the ramp and the skateboarder in terms of the variables given in the problem statement.

Answer 2

Answer:

Final answer:

The momentum of the box with respect to the floor can be found by multiplying its mass by its velocity. When the box is put down on the frictionless skating surface, its velocity becomes zero and its momentum with respect to the floor is also zero.

Explanation:

To find the momentum of the box, we can use the formula:

Momentum = mass x velocity

a. The momentum of the box with respect to the floor is: 5 kg x 5 m/s = 25 kg·m/s

b. When the box is put down on the frictionless skating surface, its velocity becomes zero. So, the momentum of the box with respect to the floor is also zero.

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If the speed of light in a medium is 2 x 10^8 m/s, the medium's index of refraction is?

Answers

speed of light in the air is 3 x 10^8
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Answer: n=1.5

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Which describes the motion of the box based on the resulting free-body diagram?1. It is moving up with a net force of 20 N.
2. It is moving to the right with a net force of 10 N.
3. It is in dynamic equilibrium with a net force of 0 N.
4. It is in static equilibrium with a net force of 0 N.

Answers

The statement "It is in dynamic equilibrium with a net force of 0 N" describes the motion of the box based on the resulting free-body diagram. (option 3)

What is a free-body diagram?

A free-body diagram is a diagram that shows all the forces acting on an object. If the net force on an object is zero, then the object is in equilibrium. This means that the object is not accelerating and is either at rest or moving with constant velocity.

In the case of the box in the free-body diagram, there are two forces acting on it: the force of gravity and the force of the table pushing up on the box. The force of gravity is pulling the box down, but the force of the table is pushing the box up.

These two forces are equal in magnitude and opposite in direction, so they cancel each other out. This means that the net force on the box is zero and the box is in dynamic equilibrium.

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Answer:

4. It is in static equilibrium with a net force of 0 N.

Explanation:

Just got it right :)

An object of mass m swings in a horizontal circle on a string of length L that tilts downward at angle θ.Find an expression for the angular velocity ω in terms of g, L and angle θ.

Answers

Answer:

The expression would be ω = $\sqrt{(g)/(L sin 0 ) }$

Explanation:

Given that ω is the angular velocity

g is the acceleration due to gravity

L is the length

θ is the angle of downward tilt

For an object we compare the horizontal and vertical component of the forces acting on the body;

For vertical component

T sinθ = mg............1

For the horizontal component

T cos θ = $(mv^(2) )/(R)$ .............2

R is our radius and is = L cos θ

v = ωR

substituting into equation 2 we have

T cos θ = m(ωR $)^(2)$ /R

T cos θ=m(ω $)^(2)$ R ..................3

Now comparing the vertical and the horizontal component we have;

equation 1 divided by equation 3 we have

T sin θ /T cos θ = mg / m(ω $)^(2)$ R

Tan θ = g / (ω $)^(2)$ R............4

Making ω the subject formula we have;

(ω $)^(2)$ = g/ R Tan θ

But R = L cos θ and Tan θ = sin θ/ cosθ

putting into equation 4 we have;

(ω $)^(2)$ = g /[( L cos θ) x( sin θ/ cosθ)]

(ω $)^(2)$ = g/ L sinθ

ω = $\sqrt{(g)/(L sin 0 ) }$

Therefor the expression for the angular velocity ω in terms of g, L and angle θ would be ω = $\sqrt{(g)/(L sin 0 ) }$

Let g, r, L and T are gravity, radius, length, and angle of string w/r/t vertical, respectively.
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ω²r = rg/Lcos T
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A certain lightning bolt moves 50.0 c of charge. how many fundamental units of charge (qe) is this?

Answers

The lightning bolt that moves 50.0 C of charge is converted to qe units. The conversion factor is 1 coulumb is equivalent to 6.24150975·10^18 e. Thus, converting 50 C to e is:

50.0 C * (6.24150975x10^18 e)/1C
equals 3.120754875x10^20 e

Answer: There are $3.125\tims 10^(20)$ number of electrons.

Explanation:

We are given 50 Coulombs of charge and we need to find the number of electrons that can hold this much amount of charge. So, to calculate that we will use the equation:

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n = number of electrons

$q_e$ Charge of one electron = $1.6* 10^(-19)C$

Q = Total charge = 50 C.

Putting values in above equation, we get:

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Hence, there are $3.125\tims 10^(20)$ number of electrons.

A 90 kg painter is standing on a horizontal wooden scaffolding of length 10 m, which is supported on each end by a rope. The painter is standing 2 m from the first rope and 8 m from the second rope. What is the tension in the first rope (in Newtons)

Answers

Explanation:

For equilibrium, $\sum M = 0$ .

So, $8 m * mg - (10 m) T_(1)$ = 0

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= $(8 * 90 * 9.8)/(10)$

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Also, for equilibrium $\sum F_(y)$ = 0

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or, $T_(2) = mg - T_(1)$

= $90 * 9.8 - 705.6$

= 176.4 N

Thus, we can conclude that the tension in the first rope is 176.4 N.

Feest. Fysics and motion11
Select the correct answer
You travel in a circle, whose circumference is 8 kilometers, at an average speed of 8 kilometers/hour. If you stop at the same point you started
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A
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B.
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4 kilometers/hour
D
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E.
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Rese

Answers

Velocity depends on the straight-line distance between your start-point and your end-point, regardless of what route you follow to get there.

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So the average velocity around any "CLOSED" path is zero. (A)

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