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Natali5045456 [20]
3 years ago
11

A girl throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of 25 m/s, and it bounc

es back with this same speed. The ball is in contact with the wall 0.050 s. What is the magnitude of the average force exerted on the wall by the ball
Physics
1 answer:
erma4kov [3.2K]3 years ago
4 0

Answer:

F = 800N

the magnitude of the average force exerted on the wall by the ball is 800N

Explanation:

Applying the impulse-momentum equation;

Impulse = change in momentum

Ft = m∆v

F = (m∆v)/t

Where;

F = force

t = time

m = mass

∆v = v2 - v1 = change in velocity

Given;

m = 0.80 kg

t = 0.050 s

The ball strikes the wall horizontally with a speed of 25 m/s, and it bounces back with this same speed.

v2 = 25 m/s

v1 = -25 m/s

∆v = v2 - v1 = 25 - (-25) m/s = 25 +25 = 50 m/s

Substituting the values;

F = (m∆v)/t

F = (0.80×50)/0.05

F = 800N

the magnitude of the average force exerted on the wall by the ball is 800N

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Answer:

0.186 N-m        

Explanation:

mass of the grindstone, m=1.7 kg

radius, r=8 cm

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time, t=9s

final angular velocity, \omega=0

Initial angular velocity,

\omega_o=2\pi f\\=2\pi (12.16) rad/s\\= 76.36 rad/s

Angular acceleration of the grind stone is:

\alpha=\frac{\omega-\omega_o}{t}\\\Rightarrow \alpha =\frac{0-76.36}{9} = -8.48 rad/s^2

Moment of inertia:

I=mr^2+mr^2=2mr^2

I=2\times 1.7 kg\times (0.08m)^2= 0.022kg-m^2

Torque exerted by the ax on the grind stone is:

\tau=I\alpha\\\tau=0.022\times (-8.48) \\\tau=0.186N-m

8 0
4 years ago
An astronaut landed on a far away planet that has a sea of water. To determine the gravitational acceleration on the planet's su
ra1l [238]

Answer:

Acceleration due to gravity will be g=17.3m/sec^2

Explanation:

We have given gauge pressure P = 3.8 atm = 3.8×101325 = 385035 Pa

Depth h = 24.3 m

Density \rho =1000kg/m^3

We have find the acceleration due to gravity at the surface of planet

We know that pressure is given by

P=\rho gh

So 385035=1000\times g\times 24.3

g=17.3m/sec^2

Acceleration due to gravity will be g=17.3m/sec^2

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4 years ago
A cold glass of iced tea warms quickly on a hot day through the process of A. conversion. B. insulation. C. conduction. D. expan
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C) conduction. let me know if I'm wrong
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A 22-turn circular coil of radius 3.00 cm and resistance 1.00 Ω is placed in a magnetic field directed perpendicular to the plan
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Answer:

23.5 mV

Explanation:

number of turn coil  'N' =22

radius 'r' =3.00 cm=> 0.03m

resistance = 1.00 Ω

B= 0.0100t + 0.0400t²

Time 't'= 4.60s

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The magnitude of induced EMF is given by,

lƩl =ΔφB/Δt = N (dB/dt)A

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3 years ago
answer A vertical spring stretches 3.4 cm when a 8-g object is hung from it. The object is replaced with a block of mass 26 g th
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Answer:

0.695s

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From Hooke's law, the restoring force is given has

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Also recall,

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Where m is the mass of object, g is the acceleration due to gravity.

Equating 1 and 2

Ky = mg

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K= ( 0.008kg × 9.8m/s2 ) ÷ 0.034

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T=2×3.142√0.025/2.24

T=6.284√0.0111

T=0.659seconds

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