Impulse = Integral of F(t) dt from 0.012s to 0.062 s
Given that you do not know the function F(t) you have to make an approximation.
The integral is the area under the curve.
The problem suggest you to approximate the area to a triangle.
In this triangle the base is the time: 0.062 s - 0.012 s = 0.050 s
The height is the peak force: 35 N.
Then, the area is [1/2] (0.05s) (35N) = 0.875 N*s
Answer> 0.875 N*s
1 watt = 1 joule/second
1 horsepower = 746 watts = 746 joule/second
(150 horsepower) x (746 watt/HP) x (1 joule/sec / watt) x (10 sec)
= (150 x 746 x 1 x 10) joule = 1,119,000 joules .
if correct plz mark brainly
Answer:
The velocity of the man is 0.144 m/s
Explanation:
This is a case of conservation of momentum.
The momentum of the moving ball before it was caught must equal the momentum of the man and the ball after he catches the ball.
Mass of ball = 0.65 kg
Mass of the man = 54 kg
Velocity of the ball = 12.1 m/s
Before collision, momentum of the ball = mass x velocity
= 0.65 x 12.1 = 7.865 kg-m/s
After collision the momentum of the man and ball system is
(0.65 + 54)Vf = 54.65Vf
Where Vf is their final common velocity.
Equating the initial and final momentum,
7.865 = 54.65Vf
Vf = 7.865/54.65 = 0.144 m/s
Answer:
x = 11.23 m
Explanation:
For this interesting exercise, we must use angular kinematics, linear kinematics and the relationship between angular and linear quantities.
Let's reduce to SI system units
θ = 155 rev (2pi rad / rev) = 310π rad
α = 2.00rev / s2 (2pi rad / 1 rev) = 4π rad / s²
Let's look for the angular velocity at the time the piece is released, with starting from rest the initial angular velocity is zero (wo = 0)
w² = w₀² + 2 α θ
w =√ 2 α θ
w = √(2 4pi 310pi)
w = 156.45 rad / s
The relationship between angular and linear velocity
v = w r
v = 156.45 0.175
v = 27.38 m / s
In this part we have the linear speed and the height that it travels to reach the floor, so with the projectile launch equations we can find the time it takes to arrive
y =
t - ½ g t²
As it leaves the highest point its speed is horizontal
y = 0 - ½ g t²
t = √ (-2y / g)
t = √ (-2 (-0.820) /9.8)
t = 0.41 s
With this time we calculate the horizontal distance, because the constant horizontal speed
x = vox t
x = 27.38 0.41
x = 11.23 m