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Norma-Jean [14]
3 years ago
5

A horizontal force of 150 N is used to push a 38.0 kg packing crate a distance of 6.85 m on a rough horizontal surface. If the c

rate moves at constant speed, find each of the following.
a) find the work done by the 150N force:
b) find the coefficient of kinetic energy between the crate and rough surface
Physics
1 answer:
SIZIF [17.4K]3 years ago
3 0

Answer:

0.40

Explanation:

a) Work done is expressed as

W = Force × distance

Given.

Force = 150N

Distance = 6.85m

Workdone = 150×6.85

Work done = 1027.5Joules

b) According to Newton's second law

\sumFx = ma

Fm - Ff = ma

Since speed is constant, acceleration us zero

Fm - Ff = 0

Fm = Ff = nR

Fm is the moving force

Ff is the frictional force

n is the coefficient of kinetic energy between the crate and rough surface

R is the reaction

From the formula;

Fm = nR

n = Fm/R

n = Fm/mg

n = 150/38(9.8)

n = 150/372.4

n = 0.40

Hence the coefficient of kinetic energy between the crate and rough surface is 0.40

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A stretched string has a mass per unit length of 5.40 g/cm and a tension of 17.5 N. A sinusoidal wave on this string has an ampl
kondaur [170]

Answer:

Part a)

y_m = 0.157 mm

part b)

k = 101.8 rad/m

Part c)

\omega = 579.3 rad/s

Part d)

here since wave is moving in negative direction so the sign of \omega must be positive

Explanation:

As we know that the speed of wave in string is given by

v = \sqrt{\frac{T}{m/L}}

so we have

T = 17.5 N

m/L = 5.4 g/cm = 0.54 kg/m

now we have

v = \sqrt{\frac{17.5}{0.54}}

v = 5.69 m/s

now we have

Part a)

y_m = amplitude of wave

y_m = 0.157 mm

part b)

k = \frac{\omega}{v}

here we know that

\omega = 2\pi f

\omega = 2\pi(92.2) = 579.3 rad/s

so we  have

k = \frac{579.3}{5.69}

k = 101.8 rad/m

Part c)

\omega = 579.3 rad/s

Part d)

here since wave is moving in negative direction so the sign of \omega must be positive

4 0
3 years ago
A weight lifter does 700J of work on a weight that he lifts in 3.1s. What is the power with
IgorLugansk [536]

Power = Work done/Time taken

So, keeping this in mind,we can solve it as follows:

= 700/3.1

= 7000/31

= 225.80 W

8 0
2 years ago
A 500 N force is applied to an object accelerating at 25 m/s^2.<br> what is the mass of the object?
krok68 [10]

Answer:

there are go fella hope u understood

4 0
2 years ago
A park ranger driving on a back country road suddenly sees a deer in his headlights 20
olya-2409 [2.1K]

Answer:

17.1

Explanation:

The distance ahead, of the deer when it is sighted by the park ranger, d = 20 m

The initial speed with which the ranger was driving, u = 11.4 m/s

The acceleration rate with which the ranger slows down, a = (-)3.80 m/s² (For a vehicle slowing down, the acceleration is negative)

The distance required for the ranger to come to rest, s = Required

The kinematic equation of motion that can be used to find the distance the ranger's vehicle travels before coming to rest (the distance 's'), is given as follows;

v² = u² + 2·a·s

∴ s = (v² - u²)/(2·a)

Where;

v = The final velocity = 0 m/s (the vehicle comes to rest (stops))

Plugging in the values for 'v', 'u', and 'a', gives;

s = (0² - 11.4²)/(2 × -3.8) = 17.1

The distance the required for the ranger's vehicle to com to rest, s = 17.1 (meters).

6 0
3 years ago
Que fuerza será necesaria aplicar a un cuerpo de 20kg de masa para imprimirle una aceleración a=4m/s²
WARRIOR [948]

Answer:

La fuerza que será necesaria aplicar a un cuerpo de 20kg de masa para imprimirle una aceleración a=4m/s² es 80 N.

Explanation:

La segunda ley de Newton, llamada ley fundamental o principio fundamental de la dinámica, plantea que un cuerpo se acelera si se le aplica una fuerza.

De esta manera, esta ley establece que las aceleraciones que experimenta un cuerpo son proporcionales a las fuerzas que recibe. Dicho de otra forma, la aceleración de un cuerpo es proporcional a la fuerza neta que se le aplica. Cuanto mayor es la fuerza que se le aplica a  un objeto con una masa dada, mayor será su aceleración.

La segunda Ley de Newton se expresa matemáticamente como:

F = m*a

Donde:

  • F es la fuerza neta. Se expresa en Newton (N)
  • m es la masa del cuerpo. Se expresa en kilogramos (Kg.).
  • a es la aceleración que adquiere el cuerpo. Se expresa en metros sobre segundo al cuadrado (m/s²).

En este caso:

  • m= 20 kg
  • a= 4 m/s²

Reemplazando:

F= 20 kg* 4 m/s²

Resolviendo:

F= 80 N

<u><em>La fuerza que será necesaria aplicar a un cuerpo de 20kg de masa para imprimirle una aceleración a=4m/s² es 80 N.</em></u>

4 0
3 years ago
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