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4 years ago

The static friction force is

Physics

1 answer:

blagie [28]4 years ago

3 0

Answer:

The answer is the third option, the minimum force required to overcome static friction and move an object.

Explanation:

The first choice is talking about an object moving, and anything moving relates to <em>kinetic </em>friction, not static friction. I don't think it's the fourth option because, even with the net force being 0, the object could still be moving (again relating to kinetic friction).

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Select the correct answer.

svet-max [94.6K]

Heya!!!

Answer to your question:
C.
Energy transferred to a machine equals energy transferred from it.

Law of conservation of energy states that total energy of an isolated system remains constant
Energy can neither be created nor be destroyed. Machine follows the same rule.

Hope it helps ^_^

6 0

3 years ago

Read 2 more answers

An 5-kg animal accelerates from rest to 10 m/s in 10 m and encounters an average drag force of 12 N. What is the average power o

MArishka [77]

Answer:

60 watts

Explanation:

Given that

Mass of the animal, m = 5 kg

Velocity of the animal, v = 10 m/s

Distance moved by the animal, d = 10 m

Drag force, F(d) = 12 N

Time taken, t = 2 seconds

To start with, we need to find the power output in itself before proceeding to find the average power output. And as such, we have

Power = Force * Distance/Time

But Distance/Time is velocity, so

Power = Force * Velocity

Power = 12 * 10

Power = 120 W.

We then use this power gotten to find the average power output.

Power(avg) = P/t

Power(avg) = 120 / 2

Power(avg) = 60 Watts.

Therefore, the average power output is found to be 60 Watts

5 0

3 years ago

A parallel plate capacitor is connected to a DC battery supplying a constant DC voltage V0= 600V via a resistor R=1845MΩ. The ba

tensa zangetsu [6.8K]

Answer:

See explanation

Explanation:

Given:-

- The DC power supply, Vo = 600 V

- The resistor, R = 1845 MΩ

- The plate area, A = 58.3 cm^2

- Left plate , ground, V = 0

- The right plate, positive potential.

- The distance between the two plates, D = 0.3 m

- The mass of the charge, m = 0.4 g

- The charge, q = 3*10^-5 C

- The point C = ( 0.25 , 12 )

- The point A = ( 0.05 , 12 )

Find:-

What is the speed, v, of that charge when it reaches point A(0.05,12)?

How long would it take the charge to reach point A?

Solution:-

- The Electric field strength ( E ) between the capacitor plates, can be evaluated by the potential difference ( Vo ) of the Dc power supply.

E = Vo / D

E = 600 / 0.3

E = 2,000 V / m

- The electrostatic force (Fe) experienced by the charge placed at point C, can be evaluated:

Fe = E*q

Fe = (2,000 V / m) * ( 3*10^-5 C)

Fe = 0.06 N

- Assuming the gravitational forces ( Weight of the particle ) to be insignificant. The motion of the particle is only in "x" direction under the influence of Electric force (Fe). Apply Newton's equation of motion:

Fnet = m*a

Where, a : The acceleration of the object/particle.

- The only unbalanced force acting on the particle is (Fe):

Fe = m*a

a = Fe / m

a = 0.06 / 0.0004

a = 150 m/s^2

- The particle has a constant acceleration ( a = 150 m/s^2 ). Now the distance between (s) between two points is:

s = C - A

s = ( 0.25 , 12 ) - ( 0.05 , 12 )

s = 0.2 m

- The particle was placed at point C; hence, velocity vi = 0 m/s. Then the velocity at point A would be vf. The particle accelerates under the influence of electric field. Using third equation of motion, evaluate (vf) at point A:

vf^2 = vi^2 + 2*a*s

vf^2 = 0 + 2*0.2*150

vf = √60

vf = 7.746 m/s

- Now, use the first equation of motion to determine the time taken (t) by particle to reach point A:

vf - vi = a*t

t = ( 7.746 - 0 ) / 150

t = 0.0516 s

- The charge placed at point C, the Dc power supply is connected across the capacitor plates. The capacitor starts to charge at a certain rate with respect to time (t). The charge (Q) at time t is given by:

$Q = c*Vo*[ 1 - e^(^-^t^/^R^C^)]$

- Where, The constant c : The capacitance of the capacitor.

- The Electric field strength (E) across the plates; hence, the electrostatic force ( Fe ) is also a function of time:

$E = \frac{Vo*[ 1 - e^(^-^t^/^R^C^)]}{D} \\\\Fe = \frac{Vo*[ 1 - e^(^-^t^/^R^C^)]}{D}*q\\\\$

- Again, apply the Newton's second law of motion and determine the acceleration (a):

Fe = m*a

a = Fe / m

$a = \frac{Vo*q*[ 1 - e^(^-^t^/^R^C^)]}{m*D}$

- Where the acceleration is rate of change of velocity "dv/dt":

$\frac{dv}{dt} = \frac{Vo*q}{m*D} - \frac{Vo*q*[ e^(^-^t^/^R^C^)]}{m*D}\\\\B = \frac{600*3*10^-^5}{0.0004*0.3} = 150, \\\\\frac{dv}{dt} = 150*( 1 - [ e^(^-^t^/^R^C^)])\\\\$

- Where the capacitance (c) for a parallel plate capacitor can be determined from the following equation:

$c = \frac{A*eo}{d}$

Where, eo = 8.854 * 10^-12 .... permittivity of free space.

$K = \frac{1}{RC} = \frac{D}{R*A*eo} = \frac{0.3}{1845*58.3*8.854*10^-^1^2*1000} = 315\\\\$

- The differential equation turns out ot be:

$\frac{dv}{dt} = 150*( 1 - [ e^(^-^K^t^)]) = 150*( 1 - [ e^(^-^3^1^5^t^)]) \\\\$

- Separate the variables the integrate over the interval :

t : ( 0 , t )

v : ( 0 , vf )

Therefore,

$\int\limits^v_0 {dv} \, = \int\limits^t_0 {150*( 1 - [ e^(^-^3^1^5^t^)])} .dt \\\\\\vf = 150*( t + \frac{e^(^-^3^1^5^t^)}{315} )^t_0\\\\vf = 150*( t + \frac{e^(^-^3^1^5^t^) - 1}{315} )$

- The final velocity at point A for the particle is given by the expression derived above. So for t = 0.0516 s, The final velocity would be:

$vf = 150*( 0.0516 + \frac{e^(^-^3^1^5^*^0^.^0^5^1^6^) - 1}{315} )\\\\vf = 7.264 m/s$

- The final velocity of particle while charging the capacitor would be:

vf = 7.264 m/s ... slightly less for the fully charged capacitor

7 0

3 years ago

What is the purpose of the rorschach test

krok68 [10]

Answer:

to examine a person's personality characteristics and emotional functioning

Explanation:

The Rorschach test is a psychological test in which subjects' perceptions of inkblots are recorded and then analyzed using psychological interpretation, complex algorithms, or both. Some psychologists use this test to examine a person's personality characteristics and emotional functioning.

4 0

3 years ago

What must be the orbital speed of a satellite in a circular orbit 740 km above the surface of the earth?

BigorU [14]

The formula for orbital speed v is v=(G*Me/r)^1/2
Wher G= 6.67E-11, Me= 6E24, r= Re+h= 6.4E6+740000
putting values in the formula we get

v= 7486.7 m/s or v= 7.4867 km/s

3 0

4 years ago