PHYSICS COLLEGE

Name the four forces in physics?

Answers

Answer 1

Answer:

Answer:

Gravitational

Electrostatic

magnetic

Frictional

Answer 2

Answer:

gravitational

electrostatic

magnetic

frictional

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Related Questions

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A 1 225.0 kg car traveling initially with a speed of 25.000 m/s in an easterly direction crashes into the back of a 9 700.0 kg truck moving in the same direction at 20.000 m/s. The velocity of the car right after the collision is 18.000 m/s to the east.(a) What is the velocity of the truck right after the collision? (Give your answer to five significant figures.) m/s east (b) What is the change in mechanical energy of the cartruck system in the collision? J (c) Account for this change in mechanical energy.
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An object of mass m swings in a horizontal circle on a string of length L that tilts downward at angle θ.Find an expression for the angular velocity ω in terms of g, L and angle θ.

Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 400 N while the other pulls in the same direction with a force of 290 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor

Answers

Answer:

The kinetic coefficient of friction of the crate is 0.235.

Explanation:

As a first step, we need to construct a free body diagram for the crate, which is included below as attachment. Let supposed that forces exerted on the crate by both workers are in the positive direction. According to the Newton's First Law, a body is unable to change its state of motion when it is at rest or moves uniformly (at constant velocity). In consequence, magnitud of friction force must be equal to the sum of the two external forces. The equations of equilibrium of the crate are:

$\Sigma F_(x) = P+T-\mu_(k)\cdot N = 0$ (Ec. 1)

$\Sigma F_(y) = N - W = 0$ (Ec. 2)

Where:

$P$ - Pushing force, measured in newtons.

$T$ - Tension, measured in newtons.

$\mu_(k)$ - Coefficient of kinetic friction, dimensionless.

$N$ - Normal force, measured in newtons.

$W$ - Weight of the crate, measured in newtons.

The system of equations is now reduced by algebraic means:

$P+T -\mu_(k)\cdot W = 0$

And we finally clear the coefficient of kinetic friction and apply the definition of weight:

$\mu_(k) =(P+T)/(m\cdot g)$

If we know that $P = 400\,N$ , $T = 290\,N$ , $m = 300\,kg$ and $g = 9.807\,(m)/(s^(2))$ , then:

$\mu_(k) = (400\,N+290\,N)/((300\,kg)\cdot \left(9.807\,(m)/(s^(2)) \right))$

$\mu_(k) = 0.235$

The kinetic coefficient of friction of the crate is 0.235.

Final answer:

The calculation of the coefficient of kinetic friction involves setting the total force exerted by the workers equal to the force of friction, as the crate moves at a constant speed. The coefficient of kinetic friction is then calculated by dividing the force of friction by the normal force, which is the weight of the crate. The coefficient of kinetic friction for the crate on the floor is approximately 0.235.

Explanation:

To calculate the coefficient of kinetic friction, we first must understand that the crate moves at a constant velocity, indicating that the net force acting on it is zero. Thus, the total force exerted by the workers (400 N + 290 N = 690 N) is equal to the force of friction acting in the opposite direction.

Since the frictional force (F) equals the normal force (N) times the coefficient of kinetic friction (μk), we can write the equation as F = μkN. Here, the normal force is the weight of the crate, determined by multiplying the mass (m) of the crate by gravity (g), i.e., N = mg = 300 kg * 9.8 m/s² = 2940 N.

Next, we rearrange the equation to solve for the coefficient of kinetic friction: μk = F / N. Substituting the known values (F=690 N, N=2940 N), we find: μk = 690 N / 2940 N = 0.2347. Thus, the coefficient of kinetic friction for the crate on the floor is approximately 0.235.

Learn more about the Coefficient of kinetic friction here:

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Assuming the same initial conditions as described in FNT 2.2.1-1, use the energy-interaction model in two different ways (parts (a) and (b) below) to determine the speed of the ball when it is 4 meters above the floor headed down: a) Construct a particular model of the entire physical process, with the initial time when the ball leaves Christine’s hand, and the final time when the ball is 4 meters above the floor headed down.
b) Divide the overall process into two physical processes by constructing two energy-system diagrams and applying energy conservation for each, one diagram for the interval corresponding to the ball traveling from Christine’s hand to the maximum height, and then one diagram corresponding to the interval for the ball traveling from the maximum height to 4 meters above the floor headed down.
c) Did you get different answers (in parts (a) and (b)) for the speed of the ball when it is 4 meters above the floor headed down?

Answers

Answer:

(a). Vf = 7.14 m/s

(b). Vf = 7.14 m/s

(c). same answer

Explanation:

for question (a), we would be applying conservation of energy principle.

but the initial height is h = 1.5 m

and the initial upward velocity of the ball is Vi = 10 m/s

Therefore

(a). using conservation law

Ef = Ei

where Ef = 1/2mVf² + mghf ........................(1)

also Ei = 1/2mVi² + mghi ........................(2)

equating both we have

1/2mVf² + mghf = 1/2mVi² + mghi

eliminating same terms gives,

Vf = √(Vi² + 2g (hi -hf))

Vf = √(10² + -2*9.8*2.5) = 7.14 m/s

Vf = 7.14 m/s

(b). Same process as done in previous;

Ef = Ei

but here the Ef = mghf ...........(3)

and Ei = 1/2mVi² + mghi ...........(4)

solving for the final height (hf) we relate both equation 3 and 4 to give

mghf = 1/2mVi² + mghi ..............(5)

canceling out same terms

hf = hi + Vi²/2g

hf = 1.5 + 10²/2*9.8 = 6.60204m ............(6)

recalling conservation energy,

Ef = Ei

1/2mVf² + mghf = mghi

inputting values of hf and hi we have

Vf = √(2g(hi -hf)) = 7.14 m/s

Vf = 7.14 m/s

(c). From answer in option a and c, we can see there were no changes in the answers.

The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2.00 x 103N with an effective perpendicular arm of 3.0 cm, producing an angular acceleration of the forearm of 130 rad/s2

What is the moment of inertia of the boxer's forearm?

Answers

To solve this problem it is necessary to apply the concepts related to Torque as a function of Force and distance. Basically the torque is located in the forearm and would be determined by the effective perpendicular lever arm and force, that is

$\tau = F * r$

Where,

F = Force

r = Distance

Replacing,

$\tau = 2*10^3*0.03$

$\tau = 60N\cdot m$

The moment of inertia of the boxer's forearm can be calculated from the relation between torque and moment of inertia and angular acceleration

$\tau = I \alpha$

I = Moment of inertia

$\alpha$ = Angular acceleration

Replacing with our values we have that

$I = (\tau)/(\alpha)$

$I = (60)/(120)$

$I = 0.5kg\cdot m^2$

Therefore the value of moment of inertia is $0.5kg\cdot m^2$

Which is true about the radiation force of light shining on a surface? The force is greater if the light reflects back along its incident path than in some other direction. The force is greater if the light is absorbed instead of being reflected. The force is greater if the light is reflected in some direction other than back along the incident path.

Answers

Answer:

The force is greater if the light is absorbed instead of being reflected

Explanation:

Light could either be reflected or absorbed. Reflection takes place where light is concentrated back to another surface whereas absorption takes place when light is incorporated into a surface thereby providing kinetic energy. The kinetic energy produced by absorption provides more force than reflection which just involves concentrating back to another surface.

If the population of penguins increased, then this would have a direct effect on the populations of?

Answers

Answer:

Globel warming

Explanation:

hope this helpex

A proton moves perpendicular to a uniform magnetic field B with arrow at a speed of 2.20 107 m/s and experiences an acceleration of 1.90 1013 m/s2 in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field.

Answers

Answer:

The magnitude and direction of the magnetic field is 0.009014 T in the negative y direction.

Explanation:

Given that,

Speed $v = 2.20*10^7\ m/s$

Acceleration $a=1.90*10^(13)\ m/s^2$

We need to calculate the magnetic field

Using formula of magnetic field

$F=qvB$ ....(I)

Using newton's second law

$F= ma$ ....(II)

From equation (I) and (II)

$ma=qvB$

Put the value into the formula

$1.90*10^(13)*1.67*10^(-27)=1.6*10^(-19)*2.20*10^(7)*B$

$3.173*10^(-14)=1.6*10^(-19)*2.20*10^(7)*B$

$B=(3.173*10^(-14))/(1.6*10^(-19)*2.20*10^(7))$

$B=0.009014\ T$

We need to calculate the direction of the field

Using the right hand rule, point the right hand fingers along the velocity which is in the positive z direction.

Now, if we curl the fingers along the direction of magnetic field that is in the negative y direction, then the thumb will point in the positive x direction.

Hence, The magnitude and direction of the magnetic field is 0.009014 T in the negative y direction.

Other Questions

Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 360°C. For the water, determine the heat transfer, in kJ per kg of water. Kinetic and potential energy effects can be ignored.

A 1850 kg car traveling at 13.8 m/s collides with a 3100 kg car that is initally at rest at a stoplight. The cars stick together and move 1.91 m before friction causes them to stop. Determine the coefficient of kinetic friction between the cars and the road, assuming that the negative acceleration is constant and all wheels on both cars lock at the time of impact.

Leaving the distance between the 181 kg and the 712 kg masses fixed, at what distance from the 712 kg mass (other than infinitely remote ones) does the 72.6 kg mass experience a net force of zero? Answer in units of m

G Water enters a house through a pipe 2.40 cm in diameter, at an absolute pressure of 4.10 atm. The pipe leading to the second-floor bathroom, 5.20 m above, is 1.20 cm in diameter. The flow speed at the inlet pipe is 4.75 m/s a) What is the algebraic expression for flow speed in the bathroom? b) Calculate the flow speed in the bathroom. c) What is algebraic expression for the pressure in the bathroom? d) Calculate the water pressure in the bathroom. Report your answer in the (atm) unit.