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dusya [7]
3 years ago
14

A heavy copper ball of mass 2 kg is dropped from a fiftieth-floor apartment window. Another one with mass 1 kg is dropped immedi

ately after 1 second. Air resistance is negligible. The difference between the speeds of the two balls:__________.
a. increases over time at first, but then stays constant.
b. decreases over time.
c. remains constant over time.
d. increases over time.
Physics
1 answer:
STALIN [3.7K]3 years ago
5 0

Answer:

C

Explanation:

Because everything on Earth falls at the same speed, the masses of the balls do not matter. Since the acceleration due to gravity is constant, their speeds will both be increasing at the same rate, and therefore the difference in speeds would remain constant until they hit the ground. Hope this helps!

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I believe the answer is C. Positive, since it is a sustainable plan to use a renewable source hope this helps :)
4 0
3 years ago
Please help me with this Physics question?
Leno4ka [110]

C 82.4 N sorry man if i am wrong but don't even think about reporting my answer

6 0
3 years ago
A ball is dropped from a height of 20 meters. At what height does the ball have a velocity of 10 meters/second?
borishaifa [10]

Answer:B

Explanation:

Initial velocity, u=0m/s

Distance,s=20m

a=+g=9.8m/s*s

Using v*v=u*u+2gs

v*v=0+2*9.8*20

v*v=392

v=19.8

When s=20m, v = 19.8m/s

Therefore when v = 10m/s, s= 10*20/19.8

s =10.1m

6 0
3 years ago
A cat falls from a tree (with zero initial velocity) at time t = 0. How far does the cat fall between t = 1 2 and t = 1 s? Use G
Romashka-Z-Leto [24]

There is one mistake in the question.The Correct question is here

A cat falls from a tree (with zero initial velocity) at time t = 0. How far does the cat fall between t = 1/2 and t = 1 s? Use Galileo's formula v(t) = −9.8t m/s.

Answer:

y(1s) - y(1/2s) =  - 3.675 m  

The cat falls 3.675 m between time 1/2 s and 1 s.

Explanation:

Given data

time=1/2 sec to 1 sec

v(t)=-9.8t m/s

To find

Distance

Solution

As the acceleration as first derivative of velocity with respect to time  

So

acceleration(-g)=  dv/dt

Solve it

dv  =  a dt

dv =  -g dt

v - v₀  =  -gt

v=  dy/dt

dy  =  v dt

dy =  ( v₀ - gt ) dt

y(1s) - y(1/2s)  =  ( v₀ ) ( 1 - 1/2 ) - ( g/2 )[ ( t1)² -( t1/2s )² ]

y(1s) - y(1/2s)  = ( - 9.8/2 ) [ ( 1 )² - ( 1/2 )² ]

y1s - y1/2s  = ( - 4.9 m/s² ) ( 3/4 s² )

y(1s) - y(1/2s) =  - 3.675 m  

The cat falls 3.675 m between time 1/2 s and 1 s.

6 0
3 years ago
a bal is launched upward with a velocity of v0 from the edge of a cliff of height D. it reaches a maximum height of H above its
lilavasa [31]

Answer:

D/H =15

Explanation:

  • We can find first the peak height H, taking into consideration, that at the maximum height, the ball will reach momentarily to a stop.
  • At this point, we can find the value of H, applying the following kinematic equation:

       v_{f} ^{2} -v_{0} ^{2} = 2* g* H (1)

  • If vf=0, if we assume that the positive direction is upwards, we can find the value of H as follows:

       H = \frac{v_{0} ^{2} }{2*g} (2)

  • We can use the same equation, to find the value of D, as follows:

        v_{f} ^{2} -v_{1} ^{2} = 2* g* D (3)

  • In order to find v₁, we can use the same kinematic equation that we used to get H, but now, we know that v₀ = 0.
  • When we replace these values in (1), we find that  v₁ = -v₀.
  • Replacing in (3), we have:

        (4*v_{0})^{2} - (-v_{0}) ^{2}  = 2* g* D\\ \\ 15*v_{0}^{2}  = 2*g*D

  • Solving for  D:

       D = \frac{15*v_{0} ^{2} }{2*g}

  • From (2) we know that H can be expressed as follows:

       H = \frac{v_{0} ^{2} }{2*g}

  • ⇒ D = 15 * H

        \frac{D}{H} = 15

3 0
3 years ago
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