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Zinaida [17]
3 years ago
8

Ball A is thrown in a horizontal x-direction from a certain height. Ball B is dropped at the same time and from the same height.

Which of the following statements is true about Balls A and B? (Neglect Air Resistance}
Physics
1 answer:
Vedmedyk [2.9K]3 years ago
5 0

Answer:

Both balls hit the ground simultaneously

Explanation:

Let´s call b₁  ( the ball dropped ) the equation for a free fall movement is, tacking as the origin of coordinates  the level from which it was dropped

V = g*t  ( since V₀ initial velocity is 0 )

In the case of projectile movement the equation for Vx and Vy (these speeds are independent)

are:

Vx = V₀x      (constant)    and Vy = Voy + g*t  

Vy = V₀ * sin 0⁰ + g*t      Vy = 0 + g*t

As we can see V is the same in both cases then the balls hit the ground simultaneously

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Answer: Probably at-least 30 mins to an hour

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A stationary police car emits a sound of frequency of 1240 HZ that bounces off a car on the highway and returns with a frequency
Dmitrij [34]

Answer:

Explanation:

Since the sound bouncing back from car has greater frequency , the car must be moving towards the police car . If v be the velocity of car

f = f_0\times\frac{V+v}{V-v}

f is apparent frequency , f₀ is original frequency , V is velocity of sound and v is velocity of car

1270 = 1240\times\frac{340+v}{340-v}

1.02419 = \frac{340+v}{340-v}

348.22-1.02419v =340 +v

2.02419v  = 8.22

v = 4.06 m /s

B)

In this case , we shall take relative velocity in place of velocity of car .So

v = 20+4.06

= 24.06 m /s

f = 1240\times\frac{340+v}{340-v}

= 1240\times\frac{340+24}{340-24}

1240 x \frac{364}{316}

= 1428 Hz .

7 0
4 years ago
The drop in potential across a resistor in a circuit is 50 V. What is the current
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Explanation:

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3 years ago
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By means of a rope whose mass is negligible, two blocks are suspended over a pulley, as the drawing shows, with m1 = 12.1 kg and
puteri [66]

Answer:

14.8 kg

Explanation:

We are given that

m_1=43.7 kg

m_2=12.1 kg

g=9.8 m/s^2

a=\frac{1}{2}(9.8)=4.9 m/s^2

We have to find the mass of the pulley.

According to question

T_2-m_2 g=m_2 a

T_2=m_2a+m_2g=m_2(a+g)=12.1(9.8+4.9)=177.87 N

T_1=m_1(g-a)=43.7(9.8-4.9)=214.13 N

Moment of inertia of pulley=I=\frac{1}{2}Mr^2

(T_2-T_1)r=I(-\alpha)=\frac{1}{2}Mr^2(\frac{-a}{r})=\frac{1}{2}Mr(-4.9)

Where \alpha=\frac{a}{r}

(177.87-214.13)=-\frac{1}{2}(4.9)M

-36.26=-\frac{1}{2}(4.9)M

M=\frac{36.26\times 2}{4.9}=14.8 kg

Hence, the mass of the pulley=14.8 kg

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A 2 000-kg car is slowed down uniformly from 20.0 m/s to 5.00 m/s in 4.00 s. (a) What average force acted on the car during that
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Answer:

the answer is c

Explanation:

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