1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
SVETLANKA909090 [29]
3 years ago
15

A force of 42N is needed to start a box sliding across the floor. The weight of the box is 55N.

Physics
2 answers:
Bezzdna [24]3 years ago
4 0
I think the right answer is c
GenaCL600 [577]3 years ago
3 0

Explanation:

It is given that,

Force applied to the box, F = 42 N

Weight of the box, W = 55 N

(a) Let f is the force of the friction acting on the box. The force of friction always opposes the motion of an object. It always equal to the applied force as :

f = -F = -42 N

(b) The frictional force is kinetic in nature because when the force is applied to the box it starts to move.

(c)  Let \mu is the coefficient of friction. It is given by :

\mu=\dfrac{f}{N}

\mu=\dfrac{42}{55}

\mu=0.76

Hence, this is the required solution.

You might be interested in
Allison is looking through a telescope at an object in space. The object looks like a very small planet, and it does not have a
Verizon [17]

Answer:

Allison is probably looking at the asteroid.

Explanation:

  • Asteroids are the giant gas balls of inner solar system that are present in the space.
  • Asteroids orbits around the sun in very irregular ans strange path unlike the other planets.
  • This is also the reason why it's called as the minor planets.
  • It could have been comet but if it had been a comets he must have seen the tail but asteroids do not have any tail like structure.
8 0
3 years ago
What is the maximum acceleration the belt can have without the crate slipping? express your answer using two significant figures
Montano1993 [528]

To prevent the crate from slipping, the maximum force that the belt can exert on the crate must be equal to the static friction force.


Ff = 0.5 * 16 * 9.8 = 78.4 N

a = 4.9 m/s^2


If acceleration of the belt exceeds the value determined in the previous question, what is the acceleration of the crate?


In this situation, the kinetic friction force is causing the crate to decelerate. So the net force on the crate is 78.4 N minus the kinetic friction force.


Ff = 0.28 * 16 * 9.8 = 43.904 N

Net force = 78.4 – 43.904 = 34.496 N

To determine the acceleration, divide by the mass of the crate.

a = 34.496 ÷ 16 = 2.156 m/s^2



8 0
3 years ago
During a baseball game, a batter hits a high .
Harman [31]

Answer: Due that we don't know the initial speed after hitting the ball, we are going to accept that the ball goes up for half of the time and then falls during other half part, that is 3.0 seconds each. Then we know that ball's movement is ruled by the acceleration of gravity formula, as follows: H = Vi * T + 1/2 * g * T^2 V = Vi + g * T where: H is height, Vi initial speed, g gravity acceleration and T time When we only consider the second half of the trajectory, we have that initial speed at the top of that movement is zero, because ball goes up till top, where stops and starts to go down, so : H = 0 * 3 + 1/2 * 32 * 3^2 = 144 ft. So the height of the pop-up is 144 feet.

7 0
3 years ago
Busco<br>n o<br><br><br><br><br>Vi<br><br><br><br><br>oookuflufurlgxbmhxmdhkhdkydkdul​
stira [4]

Answer:

Explanation:

Need help? With something

4 0
3 years ago
A rod extending between x = 0 and x = 13.0 cm has uniform cross-sectional area A = 8.00 cm2. Its density increases steadily betw
mezya [45]

Answer:

The mass is  m  = 3.45408 kg

Explanation:

From the question we are told that

    The extension  of the rod is from , \   x_1 = 0 \to x_2 = 13.0

     The area is  A =  8.0 cm^2

      The density increase as follows from  \  \rho_1 =2.5 g/cm^2 \to  \rho_2= 19.0 g/cm^3

    The equation  \rho =  B + Cx

at  x_1= 0  \rho_1 =2.5 g/cm^2

So

      2.5  =  B  + 0

=>  B =2.5

    So at x_2 = 13.0 ,  \rho_2= 19.0 g/cm^3

So

            19.0 = 2.5 + C(13)

       =>   C = 1.27

Now  

       m  =  8   \int\limits^{13}_{0} {2.5 + 1.27x} \, dx

      m  =  8   [{2.5 +\frac{ 1.27x^2}{2} } ]\left  | 13} \atop {0}} \right.

      m  =  8   [{2.5 +\frac{ 1.27(13)^2}{2} } ]

      m  = 3454.08 g

        m  = 3.45408 kg

         

8 0
3 years ago
Other questions:
  • What is flens, the focal length of the lens? if the lens is converging flens is positive. it the lens is diverging, flens is neg
    15·1 answer
  • PHYSICS *30 POINTS + BRAINLIEST ANSWER*
    14·1 answer
  • 26. A 40 kg boy jumps from a height of 4m onto a plate-form mounted on springs. As the
    14·1 answer
  • Rosie is reading an article about living cells. The article states that all living cells are composed of one or more cells, that
    14·1 answer
  • How much work is needed to lift a 3kg create a vertical displacement of 22m
    10·1 answer
  • Which statement(s) accurately describe the conditions immediately before and after the firecracker explodes:
    14·1 answer
  • How long will it take for an object to hit the ground when dropped from a height of 7m?
    10·1 answer
  • Light travels in a straight line at a constant speed of 3,0 x 108 m/s for 4,1
    13·1 answer
  • A 1200 kg car is travelling west at 45 m/s. What is the momentum of the car?
    5·1 answer
  • The range of a projectile is found by multiplying which two factors?
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!