Answer:
So the force of attraction between the two objects is 3.3365*10^-6
Explanation:
m1=10kg
m2=50kg
d=10cm=0.1m
G=6.673*10^-11Nm^2kg^2
We have to find the force of attraction between them
F=Gm1m2/d^2
F=6.673*10^-11*10*50/0.1^2
F=3.3365*10^-8/0.01
F=3.3365*10^-6
5m/s
Explanation:
Given parameters:
Mass of ball = 0.1kg
Force on the ball = 5N
time taken = 0.1s
Unknown:
final speed of the ball = ?
Solution:
According to newton's second law "the net force on a body is the product of its mass and acceleration".
Force = mass x acceleration equation 1
Acceleration =
V is the final velocity
U is the initial velocity
T is the time taken
U = O since it is a stationary body;
a = 
Input "a" into equation 1
F = m x 
5 = 0.1 x 
V = 5m/s
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Newton's laws brainly.com/question/11411375
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The change in potential energy when the block falls to ground is -480J.
The maximum change in kinetic energy of the ball is 480 J.
The initial kinetic energy of the ball is 0 J.
The final kinetic energy of the ball is 0.148J.
The initial potential energy of the ball is 0.187 J.
The final potential energy of the ball is 0 J.
The work done by the air resistance is 0.039 J.
<h3>Change in potential energy when the block falls to ground</h3>
ΔP.E = -mgh
ΔP.E = -Wh
ΔP.E = - 40 x 12
ΔP.E = -480 J
<h3>Maximum change in kinetic energy of the ball</h3>
ΔK.E = - ΔP.E
ΔK.E = - (-480 J)
ΔK.E = 480 J
<h3>Initial kinetic energy of the ball</h3>
K.Ei = 0.5mv²
where;
- v is zero since it is initially at rest
K.Ei = 0.5m(0) = 0
<h3>Final kinetic energy</h3>
K.Ef = 0.5mv²
K.Ef = 0.5(0.0091)(5.7)²
K.Ef = 0.148 J
<h3>Initial potential energy of the ball</h3>
P.Ei = mghi
P.Ei = 0.0091 x 9.8 x 2.1
P.Ei = 0.187 J
<h3>Final potential energy</h3>
P.Ef = mghf
P.Ef = 0.0091 x 9.8 x 0
P.Ef = 0
<h3>Work done by the air resistance</h3>
W = ΔE
W = P.E - K.E
W = 0.187 J - 0.148 J
W = 0.039 J
Learn more about potential energy here: brainly.com/question/1242059
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<h3 />
Work is calculated as the product of Force, Distance, and
angular motion. In this case, the work done by gravity is perpendicular to the motion
of the cart, so θ = 90°
and W=Fdcosθ
W=35.0 N x 20.0 m x cos90
W=0 J
This means that work done perpendicular to the direction of
the motion is always zero.
To be honest I’m not really sure cause I’m in 5th grade but the red planet is the sun