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Sauron [17]
3 years ago
8

On his way to class, a student on a skateboard is accelerating on a downhill stretch. Which of the following statements is true?

(Select all that apply.)
A.The skateboarder's acceleration is perpendicular to the net force.
B.The skateboarder is moving in the direction of the net force.
C.The skateboarder's acceleration is in the direction of the net force.
D.The skateboarder is subjected to a net external force in the direction of motion.
Physics
1 answer:
sasho [114]3 years ago
3 0

Answer:

B is correct

Explanation:

Because of gravity

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In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the sp
MAVERICK [17]

Answer:

The ball would have landed 3.31m farther if the downward angle were 6.0° instead.

Explanation:

In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).

We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.

So, first we need to determine the components of the velocity of the ball, like this:

V_{Ax}=V_{A}cos\theta

V_{Ax}=(21m/s)cos(-14^{o})

V_{Ax}=20.38 m/s

V_{Ay}=V_{A}sin\theta

V_{Ay}=(21m/s)sin(-14^{o})

V_{Ay}=-5.08 m/s

we pick the positive one, so it takes 0.317s for the ball to hit on point A.

so now we can find the distance from the net to point A with this time. We can find it like this:

x_{A}=V_{Ax}t

x_{A}=(20.38m/s)(0.317s)

x_{A}=6.46m

Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:

V_{Bx}=20.88 m/s

V_{By}=-2.195 m/s

y_{Bf}=y_{B0}+V_{0}t-\frac{1}{2}at^{2}

0=2.1m+(-2.195m/s)t-\frac{1}{2}(-9.8m/s^{2})t^{2}

-4.9t^{2}-2.195t+2.1=0

t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}

t=\frac{-(-2.195)\pm\sqrt{(-2.195)^{2}-4(-4.9)(2.1)}}{2(-4.9)}

t= -0.9159s    or   t=0.468s

we pick the positive one, so it takes 0.468s for the ball to hit on point B.

so now we can find the distance from the net to point B with this time. We can find it like this:

x_{B}=V_{Bx}t

x_{B}=(20.88m/s)(0.468s)

x_{B}=9.77m

So once we got the two distances we can now find the difference between them:

x_{B}-x_{A}=9.77m-6.46m=3.31m

so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.

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3 years ago
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You have a 1.7 μf and a 2.2 μf capacitor. what values of capacitance could you get by connecting them in parallel?
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Read 2 more answers
A 77.0 kg woman slides down a 42.6 m long waterslide inclined at 42.3. at the bottom, she is moving 20.3 m/s.how much work did f
zhuklara [117]

Answer:

5.791244495 KNm

Explanation:

The height h is given by, h=42.6sin42.3^{o}

Potential energy, PE is given by

PE=mgh where m is mass of the woman, g is acceleration due to gravity whose value is taken as 9.81 m/s^{2} and h is already given hence substituting 77 Kg for m we obtain

PE=77*9.81*42.6sin42.3^{o}=21656.7095 Nm

PE=21.6567095 KNm

We also know that Kinetic energy is given by0.5mv^{2} where v is the velocity and substituting v for 20.3 we obtain

KE=0.5*77*20.3^{2}=15865.465 Nm

KE=15.865465 KNm

Friction work is the difference between PE and KE hence

Friction work=21.6567095 KNm-15865.465 Nm=5.791244495 KNm

8 0
3 years ago
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