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Harman [31]
3 years ago
14

What is the amount of work done when JoAnne throws a baseball 2 meters at a force of 40

Physics
1 answer:
sergiy2304 [10]3 years ago
5 0

Answer:

Amount of work done by Joanne = 80 joule

Explanation:

Given:

Displacement of ball = 2 meters

Force applied = 40 newtons

Find:

Amount of work done by Joanne

Computation;

Work done = Force applied x Displacement

Amount of work done by Joanne = Force applied x Displacement of ball

Amount of work done by Joanne = 40 x 2

Amount of work done by Joanne = 80 joule

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<span>V=distance/time </span>

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3 years ago
In the diagram, a force of 20 newtons is applied to a block. The block is in dynamic equilibrium. What is the magnitude and dire
sladkih [1.3K]

Answer:B 20 newtons opposite to the direction of the applied force

Explanation:

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3 years ago
Read 2 more answers
the pygmy shrew has an average mass of 2.0 g if 49 of these shrew are placed on a spring scale with a spring constant of 24 N/m
olga_2 [115]

Answer:

Spring's displacement, x = -0.04 meters.

Explanation:

Let the spring's displacement be x.

Given the following data;

Mass of each shrew, m = 2.0 g to kilograms = 2/1000 = 0.002 kg

Number of shrews, n = 49

Spring constant, k = 24 N/m

We know that acceleration due to gravity, g is equal to 9.8 m/s².

To find the spring's displacement;

At equilibrium position:

Fnet = Felastic + Fg = 0

But, Felastic = -kx

Total mass, Mt = nm

Fg = -Mt = -nmg

-kx -nmg = 0

Rearranging, we have;

kx = -nmg

Making x the subject of formula, we have;

x = \frac {-nmg}{k}

Substituting into the formula, we have;

x = \frac {-49*0.002*9.8}{24}

x = \frac {-0.9604}{24}

x = -0.04 m

Therefore, the spring's displacement is -0.04 meters.

3 0
3 years ago
What energy is directly dependent upon velocity and mass?
erastova [34]

Answer:

Kinetic energy

Explanation:

3 0
3 years ago
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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