Answer:
<em>The force required is 3,104 N</em>
Explanation:
<u>Force</u>
According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:
F = ma
Where a is the acceleration of the object.
On the other hand, the equations of the Kinematics describe the motion of the object by the equation:

Where:
vf is the final speed
vo is the initial speed
a is the acceleration
t is the time
Solving for a:

We are given the initial speed as vo=20.4 m/s, the final speed as vf=0 (at rest), and the time taken to stop the car as t=7.4 s. The acceleration is:


The acceleration is negative because the car is braking (losing speed). Now compute the force exerted on the car of mass m=1,126 kg:

F= 3,104 N
The force required is 3,104 N
<span>First question: The type of energy involved when a river moves sediment and erodes its banks is: option d. Kinetic energy. Kinetic energy is the energy associated with motion. A body (in this case the water) that moves has an energy associated with its motion that is proportional to the speed (exactly to the square of the speed). When the water collides with the banks it is the kinetic energy of the river that erodes it Second question: the answer is the option d. As gravity pulls water down a slope potential energy changes to knietic energy. This is the, water loses altitude and gains velocity. The potential energy. which is proportional to the height, decreases and the kinetic energy, which is proportional to the square of the speed, increases.</span>
Pressure increases with increasing depth. h2=2hh
Sum of all forces = mass * acceleration
Ft= tension force
Fw= force of gravity (Fw= mass* acceleration of gravity which is 9.8 this only applies to force of gravity)
Ft- Fw = 0 (there is no acceleration)
Ft = Fw
Ft= m*g
Ft= 0.250kg*9.8m/s
Ft= 2.45N
On a similar problem wherein instead of 480 g, a 650 gram of bar is used:
Angular momentum L = Iω, where
<span>I = the moment of inertia about the axis of rotation, which for a long thin uniform rod rotating about its center as depicted in the diagram would be 1/12mℓ², where m is the mass of the rod and ℓ is its length. The mass of this particular rod is not given but the length of 2 meters is. The moment of inertia is therefore </span>
<span>I = 1/12m*2² = 1/3m kg*m² </span>
<span>The angular momentum ω = 2πf, where f is the frequency of rotation. If the angular momentum is to be in SI units, this frequency must be in revolutions per second. 120 rpm is 2 rev/s, so </span>
<span>ω = 2π * 2 rev/s = 4π s^(-1) </span>
<span>The angular momentum would therefore be </span>
<span>L = Iω </span>
<span>= 1/3m * 4π </span>
<span>= 4/3πm kg*m²/s, where m is the rod's mass in kg. </span>
<span>The direction of the angular momentum vector - pseudovector, actually - would be straight out of the diagram toward the viewer. </span>
<span>Edit: 650 g = 0.650 kg, so </span>
<span>L = 4/3π(0.650) kg*m²/s </span>
<span>≈ 2.72 kg*m²/s</span>