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Ierofanga [76]
3 years ago
14

Help please me please

Physics
1 answer:
Colt1911 [192]3 years ago
8 0
B

add them all by direction
13 East
10 West
subtract difference
3 E
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A well is being dug. A 4.5-kg bucket is filled with 28.0 kg of dirt and pulled vertically upward at a constant speed through a d
andreyandreev [35.5K]

The work done on the filled bucket in raising out of the hole is 2, 925 Joules

<h3>How to determine the work done</h3>

Using the formula:

Work done = force * distance

Note that force = mass * acceleration

F = mg + ma

F = 4. 5 * 10 + 28 * 10

F = 45 + 280

F = 325 Newton

Distance = 9m

Substitute into formula

Work done = 325 * 9

Work done = 2, 925 Joules

Therefore, the work done is 2, 925 Joules

Learn more about work done here:

brainly.com/question/25573309

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4 0
1 year ago
Last night Mookie Betts hit a baseball at 32.5 m/s at a 45° angle. Betts
podryga [215]

Answer:

a) Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.

b) The total distance traveled by the baseball was 108.7 m.

Explanation:

a) To know if the hit was a home run we need to calculate the height of the ball at 99 m:

y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}

Where:

y_{f}: is the final height =?

y_{0}: is the initial height = 1 m

v_{0_{y}: is the initial vertical velocity = v₀sin(45)

v₀: is the initial velocity = 32.5 m/s

g: is the gravity = 9.81 m/s²

t: is the time    

First, we need to find the time by using the following equation:

t = \frac{x}{v_{0_{x}}} = \frac{99 m}{32.5 m/s*cos(45)} = 4.31 s

Now, the height is:

y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2} = 1m + 32.5 m/s*sin(45)*4.31 s - \frac{1}{2}9.81 m/s^{2}*(4.31 s)^{2} = 8.93 m      

Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.

b) To find the distance traveled by the baseball first we need to find the time of flight:

y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}

0 = 1 m + 32.5m/s*sin(45)t - \frac{1}{2}9.81 m/s^{2}t^{2}

1 m + 32.5m/s*sin(45)t - \frac{1}{2}9.81 m/s^{2}t^{2} = 0

By solving the above quadratic equation we have:

t = 4.73 s

Finally, with that time we can find the distance traveled by the baseball:

x = v_{0_{x}}*t = 32.5 m/s*cos(45)*4.73 s = 108.7 m

Hence, the total distance traveled by the baseball was 108.7 m.

I hope it helps you!                                                                                  

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When a high-mass star explodes, which of the following is left over in the center?
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