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Viktor [21]
3 years ago
8

The ball is dropped from a certain height and is falling to the ground. If its acceleration is a constant 9.8 meters per second

squared, how many seconds would it have dropped when the velocity is 49.0 m/s ????
A. 4.9 secs
B. 4 secs
C. 5 secs
D. 9.8 secs

Physics
1 answer:
lord [1]3 years ago
7 0
5 secs I did 49 divided by 9.8 and got 5
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Answer:

the answer is A.) -1 * 10^3[N]

Explanation:

The solution consists of two steps, the first step is using the following kinematic equation:

v=v_{i} +a*t\\where:\\v=final velocity [m/s]\\v_{i}=initial velocity [m/s]\\a=acceleration[m/^2]\\t=time[s]\\

The initial velocity is 10 [m/s], and the final velocity is zero because the car stops in 0.5[s].

Replacing:

0=10+a*(0.5)\\a=-20[m/s^2]

Now in the second part, we need to use the second law of Newton, this law relates the forces with the acceleration of a body.

In the moment when the car stops suddenly the driver will feel the force of the seatbelt acting in the opposite direction of the movement.

F=m*a\\F=50[kg]*(-20[m/s^2])\\units\[kg]*[m/s^2]=[N]\\F=-1000[N] or -1*10^{3} [N]

The minus sign means that the force is acting in the opposite direction of the movement.

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3 years ago
How to understand physics
NeTakaya

Answer:

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3 years ago
A stationary police car emits a sound of frequency 1240 HzHz that bounces off of a car on the highway and returns with a frequen
lara [203]

Answer:

frequency =  1475.45 Hz

Explanation:

given data

frequency f1 = 1215 Hz,

frequency f2 = 1265 Hz

police car moving vp = 25.0 m/s

solution

speed of sound u = 343 m / s

speed of the other car = v

when the police car is stationary

the frequency the other car receives is

f2 =  f1  ×  \dfrac{u+v}{u}      ................1

and

the frequency the police car receives is

 f2 =  f1  ×  \dfrac{u}{u-v}      ..................2

now from equation 1 and 2

\frac{f2}{f1} = \dfrac{u+v}{u-v}

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v =\frac{1275-1240}{1275+1240}\times 343  

v = 4.77 m/s

and

frequency the other car receives is  

f2 = f1 ×   \dfrac{u+v}{u-vp}       ......................3

and

the frequency the police car receives is

f2 = f1 ×  \dfrac{u+vp}{u - v}       .......................4

now we get

f2 = f1 ×  \dfrac{(u+v)(u + vp)}{(u-v)(u-vp)}      

f2 =    1240\times \frac{(343+4.77)(343+25)}{(343-4.77)(343-25)}        

f2 =  1475.45 Hz

 

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3 years ago
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And yeah it's done :)

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Answer:

Correct, is there another part to the question?

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