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Vadim26 [7]
3 years ago
8

If the ball that is thrown downward has an acceleration of magnitude aaa at the instant of its release (i.e., when there is no l

onger any force on the ball due to the woman's hand), what is the relationship between aaa and ggg, the magnitude of the acceleration of gravity
Physics
1 answer:
kakasveta [241]3 years ago
5 0

Explanation:

At the instant of release there is no force but an acceleration of a, this means the ball is falling freely under the force of gravity. Then the acceleration would be due to force of gravity and acceleration a = g =9.81 m/s^2.

g= acceleration due to gravity

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Two plane coaxial disks are separated by a distance 0.6 m. The lower disk is solid with a diameter 0.80 m and a temperature 300
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Answer:

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3 years ago
Consider a 100 g object dropped from a height of 1 m. Assuming no air friction (drag), when will the object hit the ground and a
Katyanochek1 [597]

Answer:

speed and time are Vf = 4.43 m/s and  t = 0.45 s

Explanation:

This is a problem of free fall, we have the equations of kinematics

      Vf² = Vo² + 2g x

As the object is released the initial velocity is zero, let's look at the final velocity with the equation

      Vf = √( 2 g X)

      Vf = √(2 9.8  1)

      Vf = 4.43 m/s

This is the speed with which it reaches the ground

 

Having the final speed we can find the time

      Vf = Vo + g t

       t = Vf / g

       t = 4.43 / 9.8

       t = 0.45 s

This is the time of fall of the body to touch the ground

3 0
4 years ago
Gggcdfubrdegubtcwftvf y day rx u e HHS’s forget h
aev [14]

Answer:

um . . . yes ?

8 0
3 years ago
Read 2 more answers
A motorcycle is following a car that is traveling at a constant speed on a straight highway. Initially, the car and the motorcyc
Artist 52 [7]

Answer:

(a) 3.807 s

(b) 145.581 m

Explanation:

Let Δt = t2 - t1 be the time it takes from the moment when the motorcycle starts to accelerate until it catches up with the car. We know that before the acceleration, both vehicles are travelling at a constant speed. So they would maintain a distance of 58 m prior to the acceleration.

The distance traveled by car after Δt (seconds) at v_c = 23m/s speed is

s_c = \Delta t v_c = 23\Delta t

The distance traveled by the motorcycle after Δt (seconds) at m_m = 23 m/s speed and acceleration of a = 8 m/s2 is

s_m = \Delta t v_m + a\Delta t^2/2

s_m = 23\Delta t + 8\Delta t^2/2 = 23 \Delta t + 4 \Delta t^2

We know that the motorcycle catches up to the car after Δt, so it must have covered the distance that the car travels, plus their initial distance:

s_m = s_c + 58

23 \Delta t + 4 \Delta t^2 = 23\Delta t + 58

4 \Delta t^2 = 58

\Delta t^2 = 14.5

\Delta t = \sqrt{14.5} = 3.807s

(b)

s_m = 23 \Delta t + 4 \Delta t^2

s_m = 23*3.807 + 58 = 145.581 m

5 0
3 years ago
The outer planets are similar to the planet Earth; they just happen to be farther away from the Sun.
Harrizon [31]

Answer:

False

Explanation:

The inner planets are called terrestrial planets due to the surfaces are solid (similar to Earth)-made up of heavy metals, either have no moons or few moons.

The outer planets are called Jovian planets or gas giants because they are encased in gas.  They all have rings with plenty of moons.

5 0
3 years ago
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