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Mekhanik [1.2K]
3 years ago
15

A child’s toy arrow is launched horizontally from a bow at a speed of 8 m/s from a second story window 4 meters above the ground

, and the sticks in a pile of sand.
A. How far away is the pile of sand?

B. What were the horizontal and vertical speeds of the arrow when it hits the sand?

C. What angle does the arrow make with the ground when it sticks in the sand?
Physics
1 answer:
Westkost [7]3 years ago
7 0
First find the time it takes to hit the sand t
using s=1/2 g t^2
t^2= 8/9.8= 0.82 seconds
This means the sand was s = 8 * 0.82 = 6.5m away
The horizontal speed stays constant at 8 m/s
The vertical speed =.82 * 9.8 = 8.0 m/s
As the two velocities are equal the angle must be 45 degrees to the horizontal
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ivann1987 [24]

Explanation:

It is given that, the height of a certain tower is 862 feet i.e to reach on the ground the object should travel, s = 862 feet

The distance traveled by a freely falling object is given by :

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16t^2=862

t=\sqrt{53.875}

t = 7.34 seconds

So, the object will take 7.34 seconds to fall to the ground from the top of the​ building. Hence, this is the required solution.

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Match the layers of the atmosphere with the description that best matches it according to the graph titled Temperature Profile.
Finger [1]
The correct matches are as follows:

<span>Troposphere
A) Layer closest to the Earth where all weather occurs

Mesosphere
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Mesopause
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Stratosphere
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Hope this answers the question. Have a nice day.
3 0
3 years ago
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The mcb of rupa's room is tripped and keeps on tripping again and again . if it is a domestic circuit, what could be the reason
kap26 [50]

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Explanation:

The reason reason why the mcb of rupa's room keeps tripping is due to the fact that excessive current has being supplied to his room.

MCB stands for a miniature circuit breaker.

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4 0
3 years ago
A man works on a striaght road from his home to market 2.5km away dwith a speed of 5km/h.Finding the market closed, he instantly
umka21 [38]

Answer:

5.625km/h

Explanation:

We are given that

Distance between home and market, d=2.5 km

Speed, v1=5km/h

Speed, v2=7.5 km/h

We have to find the average speed of the man over the interval of time 0 to 40min.

Time,t=\frac{distance}{speed}

Using the formula

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Therefore,

Total distance covered=2.5+2.5/2=3.75 km

Time=40 min=40/60=2/3 hour

Average speed=\frac{total\;distance}{total\;time}

Average speed=\frac{3.75}{2/3}=5.625km/h

7 0
2 years ago
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