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3 years ago

13

What happens when a penny is dropped from a height of 20 meters? A.it falls at a constant velocity B.its mass doubles for every

second it falls C.the force of gravity causes it to accelerate D. it changes direction every 9.8 meters

2 answers:

Alex_Xolod [135]3 years ago

7 0

I think it's either A or C because on B an object's mass can't change unless it is on a different planet, and on D it doesn't make sense unless the wind is blowing.

Travka [436]3 years ago

4 0

Answer: Option (c) is the correct answer.

Explanation:

When a penny is dropped from a height of 20 meters then it will achieve an acceleration.

As acceleration is the rate of change in velocity of an object with respect to time. Therefore, the velocity does not remain constant.

Whereas mass of the penny will remain the same as it will not get affected when it falls. Also, there will be no change in direction of the penny as it is falling only in one direction.

The acceleration of penny is due to the force of gravity.

Thus, we can conclude that the force of gravity causes it to accelerate.

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3 years ago

If two waves pass a point every second what is the frequency of the waves

marishachu [46]

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3 years ago

you have been called to testify as a as an expert witness in a trial involving a head-on collision Car A weighs 1515 pounds and

GrogVix [38]

Answer:

70.6 mph

Explanation:

Car A mass= 1515 lb

Car B mass=1125 lb

Speed of car B is 46 miles/h

Distance before locking, d=19.5 ft

Coefficient of kinetic friction is 0.75

Initial momentum of car B=mv where m is mass and v is velocity in ft/s

46 mph*1.46667=67.4666668 ft/s

$Momentum_B=1125*67.4666668 ft/s$

Initial momentum of car A is given by

$Momentum_A=1515v_a$ where $v_a$ is velocity of A

Taking East as positive and west as negative then the sum of initial momentum is

$1515v_a-(1125*67.4666668 ft/s)$

The common velocity is represented as $v_c$ hence after collision, the final momentum is

$Momentum_final=(m_a+m_b)v_c=(1515+1125)v_c=2640v_c$

From the law of conservation of linear momentum, sum of initial and final momentum equals each other hence

$1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

The acceleration of two cars $a=-\mug=-0.75*32.17=-24.1275 ft/s^{2}$

From kinematic equation

$v^{2}=u^{2}+2as$ hence

$v^{2}-u^{2}=2as$

$0^{2}-(v_c)^{2}=2*-24.1275*19.5$

$v_c=\sqrt{2*24.1275*19.5}=30.67 ft/s$

Substituting the value of $v_c$ in equation $1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

$1515v_a-(1125*67.4666668 ft/s)= 2640*30.67$

$1515v_a=(1125*67.4666668 ft/s)+2640*30.67$

$v_a=\frac {156868.8}{1515}=103.5438 ft/s$

$\frac {103.5438}{1.46667}=70.59787 mph\approx 70.60 mph$

3 0

3 years ago

A cyclist rides 16.0 km east, then 8.0 km west, then 8.0 km east, then 32.0 km west, and finally 11.2 km east.

Anna71 [15]

Answer:

Time, t = 12 minutes

Explanation:

It is given that,

A cyclist rides 16.0 km east, then 8.0 km west, then 8.0 km east, then 32.0 km west, and finally 11.2 km east. Let west direction is negative and east direction is positive. The displacement of the cyclist is :

$d=16-8+8-32+11.2=-4.8\ km$

d = 4800 m

Let us assumed that the average speed of the cyclist is, v = 24 km/h = 6.66667 m/s

Let t is the time taken by the cyclist to complete the trip. The velocity of an object is given by :

$v=\dfrac{d}{t}$

$t=\dfrac{d}{v}$

$t=\dfrac{4800\ m}{6.66667\ m/s}$

t = 719.99 seconds

t = 720 seconds

or

t = 12 minutes

So, the time taken by the cyclist to complete the trip is 12 minutes. Yes, the time taken by the cyclist to complete the trip is reasonable. Hence, this is the required solution.

7 0

4 years ago