Answer:
31.55 m/s
Explanation:
Let the initial velocity of the arrow is u metre per second.
Angle of projection, θ = 40 degree
range = 100 m
Use the formula for the range.
100 = u^2 Sin(2 x 40) / 9.8
100 x 9.8 = u^2 Sin 80
u^2 = 995.11
u = 31.55 m/s
A toy cart is pulled a distance of 6.00 m in a straight line across the floor. The force pulling the cart has a magnitude of 20.
0 N and is directed at 37° above the horizontal. What is the work done by this force?
2 answers:
Given Information:
Horizontal distance = d = 6.0 m
Force = F = 20 N
Angle = θ = 37°
Required Information:
Work done = W = ?
Answer:
Work done = 95.84 Joules
Explanation:
The work done on toy cart is given by
W = Fdcos(θ)
Where F is the force applied, d is the distance by which the toy cart has been displaced, and θ is the angle between F and horizontal floor.
As you can see in the attached diagram, The force pulling the cart makes an angle of 37 with respect to the horizontal surface.
W = 20*6*cos(37°)
W = 95.84 Joules
Therefore, 95.84 Joules of work has been by a force of 20 N to displace the toy cart by 6 meters.
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(b) The initial speed of the skater is
while the final speed is
and the time taken to accelerate to this velocity is t=2 s, so the acceleration of the skater is given by
(c) The initial speed of the skater is
while the final speed is
since she comes to a stop. The distance covered is S=8 m, so we can use the following relationship to find the acceleration of the skater:
from which we find
where the negative sign means it is a deceleration.
(b) The initial speed of the skater is
while the final speed is
and the time taken to accelerate to this velocity is t=2 s, so the acceleration of the skater is given by
(c) The initial speed of the skater is
while the final speed is
since she comes to a stop. The distance covered is S=8 m, so we can use the following relationship to find the acceleration of the skater:
from which we find
where the negative sign means it is a deceleration.
Answer:
Hz
Explanation:
We know that
1 cm = 0.01 m
= Length of the human ear canal = 2.5 cm = 0.025 m
= Speed of sound = 340 ms⁻¹
= First resonant frequency
The human ear canal behaves as a closed pipe and for a closed pipe, nth resonant frequency is given as
for first resonant frequency, we have n = 1
Inserting the values
Hz