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Tamiku [17]
3 years ago
8

What will happen when a ball rolls down a hill?

Physics
2 answers:
Solnce55 [7]3 years ago
7 0
It will roll .. down the hill .. what else
mojhsa [17]3 years ago
5 0
Gravity pushes on the ball so when a ball rolls down a hill it will move faster until it is stoped
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What is a science term for trapping heat?
expeople1 [14]
Insulation or insulating. :) Hope that helps!
4 0
4 years ago
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A solar sail allows a spacecraft to use radiation pressure for propulsion, similar to the way wind propels a sailboat. The sails
Neporo4naja [7]

Answer:

  F = 2 I A / c

Explanation:

The radiation pressure on a reflective surface is

        P = 2 S / c

Where S is the Poynting Vector and c the speed of light

Furthermore pressure is defined as the ratio of force to area

        P = F / A

Let's replace

        F / A = 2 S / c

        F = 2 S A / c

The poynting vector is the power per unit area that is equal to the intensity

      S = I

       F = 2 I A / c

7 0
3 years ago
An 88 kg person steps into a car of mass 2002 kg, causing it to sink 5.36 cm on itssprings. Assuming no damping, with what fre-q
Marina CMI [18]

Answer:

The required frequency = 0.442 Hz

Explanation:

Frequency f  = ( \dfrac{1}{2 \pi}) \omega

where;

\omega = \sqrt{\dfrac{k}{m} }

Then;

f = \Bigg ( \dfrac{1}{2 \pi}  \Bigg )   \Bigg( \sqrt{\dfrac{k}{m} }  \Bigg )

However;

k  = \dfrac{F}{x} and;

mass m = m_{car } + m_{person}

f = \Bigg ( \dfrac{1}{2 \pi}  \Bigg )   \Bigg( \sqrt{\dfrac{\dfrac{F}{x}}{m_{car}+m_{person}} }  \Bigg )

f = \Bigg ( \dfrac{1}{2 \pi}  \Bigg )   \Bigg( \sqrt{\dfrac{{F}}{x(m_{car}+m_{person})} }  \Bigg )

where;

F = m_{person}g

Then;

f = \Bigg ( \dfrac{1}{2 \pi}  \Bigg )   \Bigg( \sqrt{\dfrac{ {m_{person}g }}{x(m_{car}+m_{person})} }  \Bigg )

replacing the values;

f = \Bigg ( \dfrac{1}{2 \pi}  \Bigg )   \Bigg( \sqrt{\dfrac{ {(88 \ kg)* (9.81 \ m/s^2) }}{(5.36 \times 10^{-2} \ m) (2002 \ kg +88 \ kg)} }  \Bigg )

\mathbf{f = 0.442 \ Hz}

8 0
3 years ago
A 2-kg bowling ball rolls at a speed of 10 m/s on the ground.
liberstina [14]
It's kinetic energy as the ball the ball isn't raised above the ground it does not have any gravitational potential energy. 
To find the kinetic energy of the ball you will have to use the formula:
KE=0.5 x m x v squared
m being mass and v being velocity 
so the calculation is:
0.5 x 2 x 10 x 10= 100J
8 0
4 years ago
After t hours a freight train is s(t) = 18t2 − 2t3 miles due north of its starting point (for 0 ≤ t ≤ 9). (a) Find its velocity
BartSMP [9]

Answer:

Explanation:

Given the equation modelled by the height of the train given as:

s(t) = 18t²-2t³ for for 0 ≤ t ≤ 9

a) Velocity is the rate of change of displacement.

Velocity = dS(t)/dt

V = dS(t)/dt = 36t - 6t² miles

Velocity at t = 3hrs is determiner by substituting t = 3 into the velocity function.

V = 36(3) -6(3)²

V= 108 - 72

Velocity = 36mi/hr

b) for Velocity at time = 7hrs

V(7) = 36(7) - 6(7)²

V(7) = 252 - 294

V(7) = -42mi/hr

The velocity at t = 7hrs is -42mi/hr

c) Acceleration is the rate of change of velocity.

a(t) = dV(t)/dt

Given v(t) = 36t - 6t²

a(t) = 36 - 12t

Acceleration at t=1 is given as:

a(1) = 36 -12(1)

a(1) = 24mi/hr²

4 0
3 years ago
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