Your cat "Ms." (mass 7.00 {\rm kg}) is trying to make it to the top of a frictionless ramp 2.00 {\rm m} long and inclined upward
at 30.0 ^\circ above the horizontal. Since the poor cat can't get any traction on the ramp, you push her up the entire length of the ramp by exerting a constant 100 {\rm N} force parallel to the ramp.
If Ms. takes a running start so that she is moving at 2.40 {\rm m/s} at the bottom of the ramp, what is her speed when she reaches the top of the incline? Use the work-energy theorem.
If Ms. takes a running start so that she is moving at 2.40 {\rm m/s} at the bottom of the ramp, what is her speed when she reaches the top of the incline? Use the work-energy theorem.
1 answer:
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Answer:
1.
109.6 cm , - 1.74 , real
2.
1.5
Explanation:
1.
d₀ = object distance = 63 cm
f = focal length of the lens = 40 cm
d = image distance = ?
using the lens equation
d = 109.6 cm
magnification is given as
m = - 1.74
The image is real
2
d₀ = object distance = a
d = image distance = - (a + 5)
f = focal length of lens = 30 cm
using the lens equation
a = 10
magnification is given as
m = 1.5
Explanation:
Given that,
Mass if the rock, m = 1 kg
It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.
We need to find the mass of the meter stick. The force acting by the stone is
F = 1 × 9.8 = 9.8 N
Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.
W = 3.266 N
The mass of the meters stick is :
So, the mass of the meter stick is 0.333 kg.