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Kazeer [188]
3 years ago
8

A 1.0 kg ball falling vertically hits a floor with a velocity of 3.0 m/s and bounces vertically up with a velocity of 2.0 m/s .

If the ball is in contact with the floor for 0.10 s, the average force on the floor by the ball is:
Physics
1 answer:
ale4655 [162]3 years ago
5 0

Answer:

50 N

Explanation:

given,

mass of ball = 1 Kg

initial velocity = 3 m/s

final velocity = 2 m/s

time = 0.10 s

impulse = change in momentum

I = m ( v - u)            

I = 1( 2 -(-3))                

I = 5 kg.m/s          

Force is equal to impulse per unit time

average force = \dfrac{I}{t}

                       = \dfrac{5}{0.1}

                       = 50 N

Average force on the floor will be equal to 50 N

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You serve a volleyball with a mass of 2.1 kg. The ball leaves your hand with a speed of 30 m/s. The ball has—— energy. Calculate
Slav-nsk [51]

Answer:

945 j

Explanation:

You have just given the ball kinetic energy, which is given by the following equation:

KE= 1⁄2 m v2 = 1⁄2 (2.1 kg)(30 m/s)2 = 945 Joules

3 0
4 years ago
Jim is driving a 2268-kg pickup truck at 22 m/s and releases his foot from the accelerator pedal. The car eventually stops due t
shutvik [7]

Answer:

610 meters.

Explanation:

Because Jim released the accelerator, the truck started to slow down, so the friction force will eventually stop the truck.

the kinetic energy of the truck just after Jim released the pedal is:

E_k=\frac{1}{2}*m*v^2\\E_k=\frac{1}{2}*2268*(22)^2=548856J

The work done by the friction force is given by:

W_f=F_s*d\\\\d=\frac{548856J}{900N}\\\\d=610m

6 0
3 years ago
I dont know how to do this at all please help
worty [1.4K]
Wow !  I understand your shock.  I shook and vibrated a little
when I looked at this one too.

The reason for our shock is all the extra junk in the question,
put there just to shock and distract us.

"Neutron star", "5.5 solar masses", "condensed burned-out star".
That's all very picturesque, and it excites cosmic fantasies in
out brains when we read it, but it's just malicious decoration.
It only gets in the way, and doesn't help a bit.

The real question is:

What is the acceleration of gravity 2000 m from
the center of a mass of 1.1 x 10³¹ kg ?

Acceleration of gravity is

                           G  ·  M / R²

      =  (6.67 x 10⁻¹¹ N·m²/kg²) · (1.1 x 10³¹ kg) / (2000 m)²

      =  (6.67 x 10⁻¹¹  ·  1.1 x 10³¹ / 4 x 10⁶)      (N) · m² · kg / kg² · m²

      =             1.83 x 10¹⁴           (kg · m / s²) · m² · kg / kg² · m²

      =             1.83 x 10¹⁴            m / s²      

That's about  1.87 x 10¹³  times the acceleration of gravity on
Earth's surface.

In other words, if I  were standing on the surface of that neutron star,
I would weigh  1.82 x 10¹² tons, give or take.     
3 0
3 years ago
Given the two sets, which statement is true?<br><br> A = {1, 2}<br> B = {1, 2, 3, 4}
Mnenie [13.5K]
What are the two sets? I only see the two answers?
3 0
3 years ago
To make ice, a freezer that is a reverse Carnot engine extracts 37 kJ as heat at -17°C during each cycle, with coefficient of pe
Dvinal [7]

Answer:

a) 6.4 kJ

b) 43.4 kJ

Explanation:

a)

Q_{a} = Heat absorbed = 37 kJ

β  = Coefficient of performance = 5.8

W = Work done

Heat absorbed is given as

Q_{a} = β W

37 = (5.8) W

W = 6.4 kJ

b)

Q  = work per cycle required

Q  = Q_{a} + W

Q  = 37 + 6.4

Q  = 43.4 kJ

5 0
3 years ago
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