Answer:
303.29N and 1.44m/s^2
Explanation:
Make sure to label each vector with none, mg, fk, a, FN or T
Given
Mass m = 68.0 kg
Angle θ = 15.0°
g = 9.8m/s^2
Coefficient of static friction μs = 0.50
Coefficient of kinetic friction μk =0.35
Solution
Vertically
N = mg - Fsinθ
Horizontally
Fs = F cos θ
μsN = Fcos θ
μs( mg- Fsinθ) = Fcos θ
μsmg - μsFsinθ = Fcos θ
μsmg = Fcos θ + μsFsinθ
F = μsmg/ cos θ + μs sinθ
F = 0.5×68×9.8/cos 15×0.5×sin15
F = 332.2/0.9659+0.5×0.2588
F =332.2/1.0953
F = 303.29N
Fnet = F - Fk
ma = F - μkN
a = F - μk( mg - Fsinθ)
a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0
303.29-0.35( 666.4 - 303.29*0.2588)/68.0
303.29-0.35(666.4-78.491)/68.0
303.29-0.35(587.90)/68.0
(303.29-205.45)/68.0
97.83/68.0
a = 1.438m/s^2
a = 1.44m/s^2
Answer:
<h2> 0.041kg</h2>
Explanation:
Step one:
given data
initial velocity u= 0m/s
Force F= 443N
time t= 4.3 ms= 0.0043seconds
final velocity v= 45.7m/s
Step two:
Required
mass m
we know that the expression for impulse is given as
Ft=mv
make m subject of the formula
m=Ft/v
substitute we have
m=443*0.0043/45.7
m=1.90/45.7
m=0.04kg
The mass of the ball is 0.041kg
Answer:
Momentum is the product of a moving object's mass and velocity . ... When two objects collide the total momentum before the collision is equal to the total momentum after the collision (in the absence of external forces). This is the law of conservation of momentum. It is true for all collisions.
Explanation:
Answer: The density is greater at point C
Explanation: At that point
The convention current slowly cools off
Momentum = mass x velocity = 91x9 = 819 kg-m/s.
