Answer:
The answer is Temperature.
Answer:
The magnitude of the force of friction equals the magnitude of my push
Explanation:
Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.
Let F = push and f = frictional force and f' = net force
F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0
So, F - f = 0
Thus, F = f
So, the magnitude of the force of friction equals the magnitude of my push.
T=distance over speed
T=40m over 9.8ms
T=answer
Answer:
Increase in temperature
Increase in pressure
Explanation:
For the rate of collisions to increase, It can happen in various ways
1. The gasses can be made to move faster: This can be done by increasing their temperature. An increase in the temperature of the gases directly relates to a corresponding increase in their kinetic energy. This means that the molecules of the gases can move with a greater velocity, which will increase collision rates.
2. The volume of the container can be reduced: Consider this; If the container is made smaller, the molecules of the gases will have less space to move about. This means that they will be bumping into the walls of the container more frequently. This reduction in volume leads to an increase in the pressure of the gas. This is exactly what happens in a piston-cylinder assembly
Answer:
I = 1.06886 N s
Explanation:
The expression for momentum is
I = F t = Δp
therefore the momentum is a vector quantity, for which we define a reference system parallel to the floor
Let's find the components of the initial velocity
sin 28.2 = v_y / v
cos 28.2= vₓ / v
v_y = v sin 282
vₓ = v cos 28.2
v_y = 42.8 sin 28.2 = 20.225 m / s
vₓ = 42.8 cos 28.2 = 37.72 m / s
since the ball is heading to the ground, the vertical velocity is negative and the horizontal velocity is positive, it can also be calculated by making
θ = -28.2
v_y = -20.55 m / s
v_x = 37.72 m / s
X axis
Iₓ = Δpₓ = 
since the ball moves in the x-axis without changing the velocity, the change in moment must be zero
Δpₓ = m 
- m v₀ₓ = 0
v_{fx} = v₀ₓ
therefore
Iₓ = 0
Y axis
I_y = Δp_y = p_{fy} -p_{oy}
when the ball reaches the floor its vertical speed is downwards and when it leaves the floor its speed has the same modulus but the direction is upwards
v_{fy} = - v_{oy}
Δp_y = 2 m v_{oy}
Δp_y = 2 0.0260 (20.55)
= 1.0686 N s
the total impulse is
I = Iₓ i ^ + I_y j ^
I = 1.06886 j^ N s
