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suter [353]
3 years ago
10

A ball flies through the air. Right before it hit the ground, it’s peed is 10 meters/second and it has 500 joules of kinetic ene

rgy. The mass of the ball is ___ kilograms
Physics
1 answer:
irakobra [83]3 years ago
5 0

Answer: 10 kg

Explanation:

Kinetic Energy = 1/2 * M (V square )

M is mass and V is velocity of the ball

500J = 1/2 *M *( 10)*(10)

M = 10 kg

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When the starter motor on a car is engaged, there is a 310 A current in the wires between the battery and the motor. Suppose the
Blababa [14]

Answer:

Diameter will be 351.42 mm

Explanation:

We have given current flowing in the copper wire i = 310 A

Voltage drop across the wire V = 0.55 volt

We know that resistance is given by R=\frac{V}{i}=\frac{0.55}{310}=0.00177\Omega

Length of the copper wire l = 1 m

Resisitivity of the copper wire \rho =1.72\times 10^{-8}\Omega m

We know that resistance R=\frac{\rho l}{A}

0.00177=\frac{1.72\times 10^{-8}\times  1}{A}

A=969.4545\times 10^{-8}m^2

As area A=\pi r^2

3.14\times r^2=969.4545\times 10^{-8}

r=17.57\times 10^{-4}m

So diameter d=17.57\times 10^{-4}\times 2=35.142\times \times 10^{-4}m = 351.42 mm

3 0
4 years ago
If a person weighs 600N on earth, how much does he weigh on the moon
Snezhnost [94]

Answer:

His weight would be 100 N

Explanation:

5 0
4 years ago
2. A stone is thrown horizontally at a speed of 6.0 m/s from the top of a cliff 78.4 m high.
Ostrovityanka [42]

Answer:

a) 8 seconds if you are using earth's gravity.

b) 48m if the velocity does not change

c) 9.8m/s

Explanation:

3 0
3 years ago
A child bounces a 60 g superball on the sidewalk. The velocity change of the superball is from 22 m/s downward to 15 m/s upward.
fgiga [73]

complete question:

A child bounces a 60 g superball on the sidewalk. The velocity change of the superball is from 22 m/s downward to 15 m/s upward. If the contact time with the sidewalk is 1/800 s, what is the magnitude of the average force exerted on the superball by the sidewalk

Answer:

F = 1776  N

Explanation:

mass of ball = 60 g = 0.06 kg

velocity of downward direction = 22 m/s = v1

velocity of upward direction = 15 m/s = v2

Δt = 1/800 = 0.00125 s

Linear momentum of a particle with mass and velocity is the product of the mass and it velocity.

p = mv

When a particle move freely and interact with another system within a period of time and again move freely like in this scenario it has a definite change in momentum. This change is defined as Impulse .

I = pf − pi = ∆p

F =  ∆p/∆t  =  I/∆t

let the upward velocity be the positive

Δp =  mv2 - m(-v1)

Δp =  mv2 - m(-v1)

Δp = m (v2 + v1)

Δp = 0.06( 15 + 22)

Δp = 0.06(37)

Δp = 2.22 kg m/s

∆t  = 0.00125

F =  ∆p/∆t

F =  2.22/0.00125

F = 1776  N

4 0
3 years ago
Assume that at sea-level the air pressure is 1.0 atm and the air density is 1.3 kg/m3.
scoundrel [369]

Answer

Pressure, P = 1 atm

air density, ρ = 1.3 kg/m³

a) height of the atmosphere when the density is constant

   Pressure at sea level = 1 atm = 101300 Pa

   we know

   P = ρ g h

   h = \dfrac{P}{\rho\ g}

   h = \dfrac{101300}{1.3\times 9.8}

          h = 7951.33 m

height of the atmosphere will be equal to 7951.33 m

b) when air density decreased linearly to zero.

  at x = 0  air density = 0

  at x= h   ρ_l = ρ_sl

 assuming density is zero at x - distance

 \rho_x = \dfrac{\rho_{sl}}{h}\times x

now, Pressure at depth x

dP = \rho_x g dx

dP = \dfrac{\rho_{sl}}{h}\times x g dx

integrating both side

P = g\dfrac{\rho_{sl}}{h}\times \int_0^h x dx

P =\dfrac{\rho_{sl}\times g h}{2}

 now,

h=\dfrac{2P}{\rho_{sl}\times g}

h=\dfrac{2\times 101300}{1.3\times 9.8}

  h = 15902.67 m

height of the atmosphere is equal to 15902.67 m.

6 0
3 years ago
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