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MissTica
4 years ago
6

A 565 N rightward force pulls a large box across the floor with a constant velocity of 0.75 m/s. If the coefficient of friction

is 0.8, what is the weight of the box?
Physics
1 answer:
alukav5142 [94]4 years ago
6 0
The first thing you should know is that the friction force is equal to the coefficient of friction due to normal force.
 Therefore, clearing the normal force we have:
 The friction is 565N.
 (565 / 0.8) = 706.25N. weight.
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5 0
4 years ago
A ball of radius R and mass m is magically put inside a thin shell of the same mass and radius 2R. The system is at rest on a ho
dlinn [17]

Answer:

x =\frac{-R}{2}

Explanation:

From the question we are told that mass

Thin layer radius = 2R

Generally the expression for ths solution is given as

Xcm =(m*0 =m(-2R))/2m =-mR/(2m)=-R/2

the center of mass will not move at initial state  

Considering the center of mass of both bodies

xcm=\frac{m*x+m*x)}{2m} =x

x =\frac{-R}{2}

Therefore the enclosing layer moves x =\frac{-R}{2}                          

7 0
3 years ago
When I bump the table, the coffee in my cup spilled out. Newton's _____ law explains this reaction.
Ivan
When I bump the table, the coffee in my cup spilled out. Newton's 1st law explains this reaction.

Answer: A) or the first option.
7 0
3 years ago
Read 2 more answers
Find the net charge of a system consisting of 6.25×10^6 electrons and 7.75×10^6 protons. Express your answer using three signifi
jok3333 [9.3K]

Answer:

2.40 x 10⁻¹³ C

Explanation:

n_{e} = number of electrons = 6.25 x 10⁶

q_{e} = charge on electron = - 1.6 x 10⁻¹⁹ C

n_{p} = number of protons = 7.75 x 10⁶

q_{p} = charge on proton =  1.6 x 10⁻¹⁹ C

Net charge is given as

Q = n_{e} q_{e} + n_{p} q_{p}

Q = (- 1.6 x 10⁻¹⁹) (6.25 x 10⁶) + (1.6 x 10⁻¹⁹) (7.75 x 10⁶)

Q = 2.40 x 10⁻¹³ C

6 0
3 years ago
A 1150 kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 7.69 m before contacting the beam,
Natasha_Volkova [10]

Answer:

the average force 11226 N  

Explanation:

Let's analyze the problem we are asked for the average force, during the crash, we can find this from the impulse-momentum equation, but this equation needs the speeds and times of the crash that we could look for by kinematics.

Let's start looking for the stack speeds, it has a free fall, from rest  (Vo=0)

             

           Vf² = Vo² - 2gY

            Vf² = 0 - 2 9.8 7.69 = 150.7

            Vf = 12.3 m / s

This is the speed that the battery likes when it touches the beam.  They also give us the distance it travels before stopping, let's calculate the time

         

            Vf = Vo - g t

             0 = Vo - g t

             t = Vo / g

             t = 12.3 / 9.8

             t = 1.26 s

This is the time to stop

Now let's use the equation that relates the impulse to the amount of movement

                 I = Δp

                F t = pf-po

The amount of final movement is zero because the system stops

                F = - po / t

                F = - mv / t

                F = - 1150 12.3 / 1.26

                F = -11226 N

This is the average force exerted by the stack on the vean

7 0
3 years ago
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