Answer:as per as Newtons second law, The forces exerted on the rope create tension.
As such,The tension is equal to the applied force.The tension is trasmitted to the opposite side and of the rope delivering the applied force.
Hope this helps.. :)
Answer: 11.025 meters.
Explanation:
Ok, we know that the initial velocity is only horizontal, so it does not affect the vertical problem.
Looking only at the vertical problem (because we want to know how high is the bridge) we have that:
The acceleration is the gravitational acceleration, g = 9.8m/s^2
A(t) = -9.8m/s^2
Where the negative sign is because this acceleration is downwards.
For the vertical velocity we integrate over time, as we do not have an initial vertical velocity, there is no constant of integration.
V(t) = (-9.8m/s^2)*t
For the position we integrate over time again, here the constant of integration is the initial vertical position, H, that is the height of the bridge.
P(t) = 0.5* (-9.8m/s^2)*t^2 + H.
Now we know that the ball hits the ground 1.5s after it was kicked, then:
p(1.5s) = 0m
With that we can find the value of H.
0 = 0.5* (-9.8m/s^2)*(1.5s)^2 + H.
H = 0.5*(9.8m/s^2)*(1.5s)^2 = 11.025m
Answer:
A) - 9.4 cm
B) 1.3 cm
Explanation:
We shall use mirror formula
\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\\
v = ?, u = + 15 cm , f = - 25 cm.
\frac{1}{v}+\frac{1}{15}=\frac{1}{-25}\\
\frac{1}{v}=-\frac{1}{15}-\frac{1}{25}\\
v=\frac{-10-6}{150}\\\\
v=-9.375cm
magnification =\frac{v}{u}\\
=\frac{9.375}{15}\\
Size of image = magnification multiply size of object
=\frac{9.375}{15}\times 2\\
1.25cm
