6. And what's up with ChiN?

In your own words..Describe Friction Force..<br><br> In your own words..describe Applied force...

BARSIC [14]

I think force is something to do with a force being applied to a movement or is just to pull but not sure sorry if this aint right

5 0

4 years ago

Read 2 more answers

Question 2 of 10

mamaluj [8]

Answer:

2.00 A

Explanation:

Total resistance = 10 + 20 + 30 = 60 ohms

Potential difference (V) = 120volts

Current (I) =?

from Ohms law V = IR

==> I = V/R = 120/60 = 2 A

Note: for resistors in series equal amount of electric current flows through the circuit

7 0

3 years ago

In longitudinal waves the places where the coils are bunched together are called *

Firdavs [7]

In longitudinal waves the places where the coils are bunched together are called *
Compressions

3 0

3 years ago

Jake is helping Fin push a box at a constant velocity up an incline that makes an angle of 30.0° above the horizontal by applyin

andre [41]

Given data

The angle of inclination of the plane is theta = 30 degree

The applied force in the inclined plane is F = 94 N

The distance moved in the inclined plane is d = 2.30 m

The coefficient of kinetic friction is u_k = 0.280

The free-body diagram of the above configuration is shown below:

Here, the normal reaction force on the box is N, the acceleration due to gravity is denoted as g, the friction force on the box is F_f, and the mass of the box is denoted as m.

(a)

The expression for the work done by the pushing force is given as:

$W=Fd$

Substitute the value in the above equation.

$\begin{gathered} W=94\text{ N}\times2.30\text{ m} \\ W=216.2\text{ J} \end{gathered}$

Thus, the work done by the pushing force is 216.2 J.

(b)

The box is moving at the constant velocity, therefore, the pushing force will be equal to the frictional force and the component of the gravitational force in the inclined plane.

$\begin{gathered} F=F_f+mg\sin \theta \\ F=\mu_kN+mg\sin \theta \end{gathered}$

The expression for the normal reaction force is given as:

$N=mg\cos \theta$

The expression for the mass of the box is given as:

$\begin{gathered} F=\mu_k\times mg\cos \theta+mg\sin \theta \\ m=\frac{F}{\mu_kg\cos \theta+g\sin \theta} \end{gathered}$

Substitute the value in the above equation.

$\begin{gathered} m=\frac{94\text{ N}}{0.28\times9.8m/s^2\times\cos 30^o+9.8m/s^2\times\sin 30^0} \\ m=12.9\text{ kg} \end{gathered}$

Thus, the mass of the box is 12.9 kg.

7 0

1 year ago

1. Which mathematical representation correctly identifies impulse?

horsena [70]

Answer:

1. B. Impulse = Force × Time

2. A. The momentum of each ball changes, and the total momentum stays the same

3. -55 kg·m/s

4. B. 3.5 kg

5. C. 6.3 m/s

Explanation:

1. The impulse is the momentum change of an object due to a force applied for a given period

2. Given that the objects collide, and the force of the 3 kg mass moving with 24 kg·m/s acts on the 1 kg mass while the total momentum is conserved;

The stationary ball of mass 1 kg begins to moves at certain velocity after collision and therefore changes momentum, while the velocity of the ball of mass 3.0 kg reduces and the total combined momentum of the two balls in the closed system remains the same

3. By the principle of conservation of linear momentum, we have;

The sum of the momentum before the collision = The sum of the momentum after collision

Given that the objects move together after the collision, the total momentum is therefore;

Total momentum = 110 kg·m/s + -65 kg·m/s + -100 kg·m/s = 110 kg·m/s - 65 kg·m/s - 100 kg·m/s = -55kg·m/s

4. Given that the final velocity of the two objects (m₁ + m₂) combined = 50 m/s

Where;

m₁ = The mass of the first object

m₂ = The mass of the second object

The total momentum of the system = 250 kg·m/s

From momentum = Mass × Velocity, we have;

Mass = Momentum/Velocity = 250 kg·m/s/(50 m/s) = 5.0 kg

The mass (m₁ + m₂) = 5.0 kg

Given that m₁ = 1.5 kg, we have;

m₂ = 5.0 kg - m₁ = 5.0 kg - 1.5 kg = 3.5 kg

The mass of the second object = 3.5 kg

5. The mass of the cue stick = 0.5 kg

The velocity of the cue stick = 2.5 m/s

The mass of the ball = 0.2 kg

The initial velocity of the ball = 0 m/s

Given that total initial momentum = Total final momentum, we have;

0.5 kg × 2.5 m/s + 0.2 kg × 0 = 0.2 kg × v + 0.5 kg × 0

0.5 kg × 2.5 m/s = 0.2 kg × v

v = (0.5 kg × 2.5 m/s)/(0.2 kg) = 6.25 m/s ≈ 6.3 m/s

3 0

4 years ago