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OLEGan [10]
2 years ago
10

A bullet is fired horizontally at a height of 2 meters at a velocity of 930 m/s. Assume no air resistance. How far did the bulle

t travel horizontally when it hit the ground?
595.2 m
478.4 m
364.0 m
247.2 m
Physics
2 answers:
Vesnalui [34]2 years ago
5 0
It’s most likely A , i’m not so sure though
USPshnik [31]2 years ago
4 0

Answer:

im pretty sure its a

Explanation:

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SCALCET8 3.9.018.MI. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the
Firlakuza [10]

Answer:

The length of his shadow is decreasing at a rate of 1.13 m/s

Explanation:

The ray of light hitting the ground forms a right angled triangle of height H, which is the height of the building and width, D which is the distance of the tip of the shadow from the building.

Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.

By similar triangles,

H/D = h/L

L = hD/H

Also, when the man is 4 m from the building, the length of his shadow is L = D - 4

So, D - 4 = hD/H

H(D - 4) = hD

H = hD/(D - 4)

Since h = 2 m and D = 12 m,

H = 2 m × 12 m/(12 m - 4 m)

H = 24 m²/8 m

H = 3 m

Since L = hD/H

and h and H are constant, differentiating L with respect to time, we have

dL/dt = d(hD/H)/dt

dL/dt = h(dD/dt)/H

Now dD/dt = velocity(speed) of man = -1.7 m/s ( negative since he is moving towards the building in the negative x - direction)

Since h = 2 m and H = 3 m,

dL/dt = h(dD/dt)/H

dL/dt = 2 m(-1.7 m/s)/3 m

dL/dt = -3.4/3 m/s

dL/dt = -1.13 m/s

So, the length of his shadow is decreasing at a rate of 1.13 m/s

5 0
2 years ago
A car travels up a hill at a constant speed of 38 km/h and returns down the hill at a constant speed of 66 km/h. Calculate the a
mojhsa [17]

Answer:

Average speed will be 48.23 km/h

Explanation:

Let the distance up to hill is = d km

Speed when car goes to hill = 38 km/h

So time required t=\frac{distance}{speed}=\frac{d}{38}hour

Speed when car return from hill = 66 km/h

So time required to return fro hill t=\frac{d}{66}h

Total time t_{total}=\frac{t}{38}+\frac{t}{66}

Total distance = d+d =2d

So average speed=\frac{total\ distance}{total\ time}=\frac{2d}{\frac{d}{38}+\frac{d}{66}}=48.23km/h

8 0
3 years ago
When an object moves in uniform circular motion, the direction of its acceleration is?
Harrizon [31]
Clock wise idk i think you should double check my answer

7 0
2 years ago
To test the quality of a tennis ball, you drop it onto a floor from a height of 4.00m . It rebounds to a height of 2.00 m. If th
Agata [3.3K]
In this problem, we apply the equation regarding  kinematics  expressed as  vf^2 = v0^2 + 2as  vf eventually becomes zero because the ball stops in the end. a = -9.8 m/s2s = 2 metres this time 
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change in velocity/change in time = average acceleration 15.11/(12/1000) = 1259.167 m/s^2
3 0
2 years ago
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What is an example of vaporization?
Alenkasestr [34]

Answer:

just search it up you'll get ur answer

7 0
3 years ago
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