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wolverine [178]
2 years ago
13

When I drop a bouncy ball it bounces back up to me. What force causes it to bounce?

Physics
2 answers:
Darya [45]2 years ago
6 0
I do not think that there is a "force" that causes it to bounce. it's the rubber (or whatever material) makes up the ball.
SVEN [57.7K]2 years ago
6 0
Gravity pulls the ball down to the centre of the Earth but the ground resists the ball in the opposite direction. If there was no resistance the ball would phase through the ground. This is Newton's Second Law. However since the ball is made of rubber (just an assumption), the force of the floor resisting the ball is higher than the force of gravity, hence the ball then flies back up.
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A motorist, traveling east on an open highway, sets his cruise control at 90.0 km/h. How far will he travel in 0.75 hours?
dimaraw [331]

Answer:

<h2>The answer is 67.5 km</h2>

Explanation:

The distance covered by an object given it's velocity and time taken can be found by using the formula

distance = velocity × time

From the question we have

distance = 90 × 0.75

We have the final answer as

<h3>67.5 km</h3>

Hope this helps you

3 0
2 years ago
Read 2 more answers
An airplane flies in a horizontal circle of radius 500 m at a speed of 150 m/s. If the radius were changed to 1000 m, but the sp
laila [671]

Answer:

The centripetal acceleration changed by a factor of 0.5

Explanation:

Given;

first radius of the horizontal circle, r₁ = 500 m

speed of the airplane, v = 150 m/s

second radius of the airplane, r₂ = 1000 m

Centripetal acceleration is given as;

a = \frac{v^2}{r}

At constant speed, we will have;

v^2 =ar\\\\v = \sqrt{ar}\\\\at \ constant\ v;\\\sqrt{a_1r_1} = \sqrt{a_2r_2}\\\\a_1r_1 = a_2r_2\\\\a_2 = \frac{a_1r_1}{r_2} \\\\a_2 = \frac{a_1*500}{1000}\\\\a_2 = \frac{a_1}{2} \\\\a_2 = \frac{1}{2} a_1

a₂ = 0.5a₁

Therefore, the centripetal acceleration changed by a factor of 0.5

7 0
3 years ago
In a crash test, a truck with mass 2100 kg traveling at 22 m/s smashes head-on into a concrete wall without rebounding. The fron
-BARSIC- [3]

Answer:

a) 11 m/s

b) 0.0564 s

Explanation:

Given:

m = 2100 kg

vi = 22 ..... m/s before collision

vf = 0 ......after collision to stop

Δs = 0.62   distance traveled after collision .. crumpling of truck

Part a

V_{avg} = \frac{vi- vf}{2}\\\\V_{avg} = \frac{22- 0}{2}\\\\ V_{avg} = 11 m/s

Part b

vf = vi + a*t\\vf^2 = vi^2 + 2*a*s\\\\0 = 22 + a*t\\t = -22 /a\\\\0 = 484 +2*a*(0.62)\\a = - 390.3232 m/s^2\\\\t = -22/(-390.3232)\\\\t = 0.0564 s

7 0
3 years ago
Read 2 more answers
a conical pendulum is formed by attaching a 50g mass to a 1.2m string. The mass swings around the circle of radius 25cm (a) calc
kipiarov [429]
A. 0.5 miles an hour 
B. 0.2 miles an hour
4 0
3 years ago
A rocket is fired straight up. It contains two stages (Stage 1 and Stage 2) of solid rocket fuel that are designed to burn for 1
Archy [21]

Answer:

a)  y= 3.5 10³ m, b)   t = 64 s

Explanation:

a) For this exercise we use the vertical launch kinematics equation

Stage 1

          y₁ = y₀ + v₀ t + ½ a t²

          y₁ = 0 + 0 + ½ a₁ t²

Let's calculate

         y₁ = ½ 16 10²

         y₁ = 800 m

At the end of this stage it has a speed

        v₁ = vo + a₁ t₁

        v₁ = 0 + 16 10

        v₁ = 160 m / s

Stage 2

        y₂ = y₁ + v₁ (t-t₀) + ½ a₂ (t-t₀)²

        y₂ = 800 + 150 5 + ½ 11 5²

        y₂ = 1092.5 m

Speed ​​is

        v₂ = v₁ + a₂ t

        v₂ = 160 + 11 5

        v₂ = 215 m / s

The rocket continues to follow until the speed reaches zero (v₃ = 0)

         v₃² = v₂² - 2 g y₃

         0 = v₂² - 2g y₃

         y₃ = v₂² / 2g

         y₃ = 215²/2 9.8

         y₃ = 2358.4 m

The total height is

          y = y₃ + y₂

          y = 2358.4 + 1092.5

          y = 3450.9 m

           y= 3.5 10³ m

b) Flight time is the time to go up plus the time to go down

Let's look for the time of stage 3

          v₃ = v₂ - g t₃

          v₃ = 0

          t₃ = v₂ / g

          t₃ = 215 / 9.8

          t₃ = 21.94 s

The time to climb is

          t_{s} = t₁ + t₂ + t₃

          t_{s} = 10+ 5+ 21.94

          t_{s} = 36.94 s

The time to descend from the maximum height is

          y = v₀ t - ½ g t²

When it starts to slow down it's zero

         y = - ½ g t_{b}²

         t_{b}  = √-2y / g

         

        t_{b} = √(- 2 (-3450.9) /9.8)

        t_{b} = 26.54 s

Flight time is the rise time plus the descent date

        t = t_{s} + t_{b}

        t = 36.94 + 26.54

        t =63.84 s

        t = 64 s

3 0
2 years ago
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