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GalinKa [24]
3 years ago
11

Adam drops a ball from rest from the top floor of a building at the same time Bob throws a ball horizontally from the same locat

ion. Which ball hits the ground first? (Neglect any effects due to air resistance.) They both hit the ground at the same time. Adam's ball Bob's ball It depends on how fast Bob throws the ball. It depends on how fast the ball falls when Adam drops it.
Physics
1 answer:
guapka [62]3 years ago
3 0

Answer:

Both balls hit the ground at the same time

Explanation:

Adam drops the ball from rest, so the ball just "<em>falls</em>" in vertical direction, being gravity its only acceleration, for cinematic movements we use that:

y(t)=y_{0}+v_{0y}t+\frac{1}{2}gt^{2}

In this case we have that gravity is negative, and as Adam drops the ball, v_{0y}=0

Bob throws the ball horizontally, so the movement will be a <em>parabola</em>, we can divide into horizontal direction, and vertical direction.

But we only need to analize the vertical movement, in wich again the only acceleration is gravity, and compare it with Adam's ball. Again we have that gravity is negative, and as the initial throw is horizontal, v_{0y}=0

Finally, we have that

y(t)=h-\frac{1}{2}gt^{2}

where

h=y_{0}

both for Adam's vertical drop, and for Bob's vertical component of the parabolic throw.

Now, if we put y(t)=0 (the origin of the vertical coordinate), we get for both cases that

h=\frac{1}{2}gt^{2}

where we can clear the value for the time t, of the fall, wich will be the same in both cases.

Hence, both balls hit the ground at the same time.

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leva [86]

Answer:

c . slower and started moving in place.

Explanation:

Matter can exist generally in three phases, as a solid, liquid or gas. But it can be transformed from one phase to another by the removal or application of heat energy.

The water was initially in a liquid form in the sealed tank until energy was transferred out of the substance. Thus, this causes a change of state in which the water turns to a solid. Whereby during the process, the molecules of the water moved slowly until they are fixed at a point, and vibrates individually at their individual point.

Therefore the molecules of water moved slower and stated moving in place (i.e vibrating at a point). The water turns to an ice.

3 0
3 years ago
A stone is catapulted at time t = 0, with an initial velocity of magnitude 19.9 m/s and at an angle of 39.9° above the horizonta
larisa [96]

Answer:

Part a)

x = 15.76 m

Part b)

y = 7.94 m

Part c)

x = 26.16 m

Part d)

y = 7.49 m

Part e)

x = 83.23 m

Part f)

y = -75.6 m

Explanation:

As we know that catapult is projected with speed 19.9 m/s

so here we have

v_x = 19.9 cos39.9

v_x = 15.3 m/s

similarly we have

v_y = 19.9 sin39.9

v_y = 12.76 m/s

Part a)

Horizontal displacement in 1.03 s

x = v_x t

x = (15.3)(1.03)

x = 15.76 m

Part b)

Vertical direction we have

y = v_y t - \frac{1]{2}gt^2

y = (12.76)(1.03) - 4.9(1.03)^2

y = 7.94 m

Part c)

Horizontal displacement in 1.71 s

x = v_x t

x = (15.3)(1.71)

x = 26.16 m

Part d)

Vertical direction we have

y = v_y t - \frac{1]{2}gt^2

y = (12.76)(1.71) - 4.9(1.71)^2

y = 7.49 m

Part e)

Horizontal displacement in 5.44 s

x = v_x t

x = (15.3)(5.44)

x = 83.23 m

Part f)

Vertical direction we have

y = v_y t - \frac{1]{2}gt^2

y = (12.76)(5.44) - 4.9(5.44)^2

y = -75.6 m

6 0
3 years ago
A 100-kg running back runs at 5 m/s into a stationary linebacker. It takes 0.5 s for the running back to be completely stopped.
Elza [17]

Answer:

1000 N

Explanation:

First, we need to find the deceleration of the running back, which is given by:

a=\frac{v-u}{t}

where

v = 0 is his final velocity

u = 5 m/s is his initial velocity

t = 0.5 s is the time taken

Substituting, we have

a=\frac{0-5 m/s}{0.5 s}=-10 m/s^2

And now we can calculate the force exerted on the running back, by using Newton's second law:

F=ma=(100 kg)(-10 m/s^2)=-1000 N

so, the magnitude of the force is 1000 N.

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