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GaryK [48]
3 years ago
9

It takes 77.4 N to push a 43.8 kg

Physics
1 answer:
Sladkaya [172]3 years ago
7 0

Answer:

0.180

Explanation:

(Applied Force) / (Weight) = Coefficient of kinetic friction

(77.4 N) / (43.8 kg * 9.81 m/s/s) = 0.180

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If a car used 260.000 W of power to complete a race in 15 s. how much work did the car do?
marta [7]

Work done by the car is 3900 J

Explanation:

  • Power and work are related by the equation, Power = Work Done/Time
  • Power is the rate at which work is done.
  • Here, the car uses power of 260 W and time taken is 15 s.

Work Done = Power × Time

                   = 260 × 15 = 3900 J

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4 years ago
¿Cuál es la intensidad del campo eléctrico a una distancia de 2 m de una carga de -12μC? *
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3 years ago
What do we mean by the observable universe?
mote1985 [20]
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On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 103 N/C. Compare the
Nina [5.8K]

Answer:

a) FE = 0.764FG

b) a = 2.30 m/s^2

Explanation:

a) To compare the gravitational and electric force over the particle you calculate the following ratio:

\frac{F_E}{F_G}=\frac{qE}{mg}              (1)

FE: electric force

FG: gravitational force

q: charge of the particle = 1.6*10^-19 C

g: gravitational acceleration = 9.8 m/s^2

E: electric field = 103N/C

m: mass of the particle = 2.2*10^-15 g = 2.2*10^-18 kg

You replace the values of all parameters in the equation (1):

\frac{F_E}{F_G}=\frac{(1.6*10^{-19}C)(103N/C)}{(2.2*10^{-18}kg)(9.8m/s^2)}\\\\\frac{F_E}{F_G}=0.764

Then, the gravitational force is 0.764 times the electric force on the particle

b)

The acceleration of the particle is obtained by using the second Newton law:

F_E-F_G=ma\\\\a=\frac{qE-mg}{m}

you replace the values of all variables:

a=\frac{(1.6*10^{-19}C)(103N/C)-(2.2*10^{-18}kg)(9.8m/s^2)}{2.2*10^{-18}kg}\\\\a=-2.30\frac{m}{s^2}

hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.

7 0
3 years ago
A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is
aniked [119]

1) At the moment of being at the top, the piston will not only tend to push the penny up but will also descend at a faster rate at which the penny can reach in 'free fall', in that short distance. Therefore, at the highest point, the penny will lose contact with the piston. Therefore the correct answer is C.

2) To solve this problem we will apply the equations related to the simple harmonic movement, hence we have that the acceleration can be defined as

a = -\omega^2 A

Where,

a = Acceleration

A = Amplitude

\omega= Angular velocity

From a reference system in which the downward acceleration is negative due to the force of gravity we will have to

a = -g

-\omega^2 A = -g

\omega = \sqrt{\frac{g}{A}}

From the definition of frequency and angular velocity we have to

\omega = 2\pi f

f = \frac{1}{2\pi} \sqrt{\frac{g}{A}}

f = \frac{1}{2\pi} \sqrt{\frac{9.8}{4*10^{-2}}}

f = 2.5Hz

Therefore the maximum frequency for which the penny just barely remains in place for the full cycle is 2.5Hz

6 0
3 years ago
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