Answer:
A. 69.9m
Explanation:
Given parameters:
Initial velocity = 10.5m/s
Final velocity = 21.7m/s
Time = 4.34s
Unknown:
Distance traveled = ?
Solution:
Let us first find the acceleration of the car;
Acceleration =
v is final velocity
u is initial velocity
t is the time
Acceleration =
= 2.58m/s²
Distance traveled;
V² = U² + 2aS
21.7² = 10.5² + 2 x 2.58 x S
360.64 = 2 x 2.58 x S
S = 69.9m
Answer:
a) P = 807.85 N, b) P = 392.15 N, c) P = 444.12 N
Explanation:
For this exercise, let's use Newton's second law, let's set a reference frame with the x-axis parallel to the plane and the direction rising as positive, and the y-axis perpendicular to the plane.
Let's use trigonometry to break down the weight
sin θ = Wₓ / W
cos θ = W_y / W
Wₓ = W sin θ
W_y = W cos θ
Wₓ = 1200 sin 30 = 600 N
W_y = 1200 cos 30 = 1039.23 N
Y axis
N- W_y = 0
N = W_y = 1039.23 N
Remember that the friction force always opposes the movement
a) in this case, the system will begin to move upwards, which is why friction is static
P -Wₓ -fr = 0
P = Wₓ + fr
as the system is moving the friction coefficient is dynamic
fr = μ N
fr = 0.20 1039.23
fr = 207.85 N
we substitute
P = 600+ 207.85
P = 807.85 N
b) to avoid downward movement implies that the system is stopped, therefore the friction coefficient is static
P + fr -Wx = 0
fr = μ N
fr = 0.20 1039.23
fr = 207.85 N
we substitute
P = Wₓ -fr
P = 600 - 207,846
P = 392.15 N
c) as the movement is continuous, the friction coefficient is dynamic
P - Wₓ + fr = 0
P = Wₓ - fr
fr = 0.15 1039.23
fr = 155.88 N
P = 600 - 155.88
P = 444.12 N
Answer:
The focal length of the lens is 34.047 cm
The power of the needed corrective lens is 2.937 diopter.
Explanation:
Distance of the object from the lens,u = 26 cm
Distance of the image from the lens ,v= -110 cm
(Image is forming on the other side of the lens)
Since ,lens of the human eye is converging lens,convex lens.
Using a lens formula:


f = 34.047 cm = 0.3404 m
Power of the lens = P

Answer:
7.09683 m
1.20285 s
2.4057 s
11.8 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s² (negative up, positive down)
From equation of motion we have

The maximum height above the ground that the ball reaches is 7.09683 m

Time taken to go up is 1.20285 s it will take the same time to come down so total time taken to reach the ground after it is shot is 1.20285+1.20285 = 2.4057 s

The velocity just before it hits the ground is 11.8 m/s