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DerKrebs [107]
3 years ago
10

A horizontal force of 120 N is required to push a bookcase across the floor at a constant velocity how much is the friction forc

e that acts on the sliding bookcase
Physics
1 answer:
belka [17]3 years ago
8 0

The Friction Force that acts on a sliding bookcase is -120 N

<u>Explanation:</u>

  • Frictional force is the force developed by the two surfaces where they contact and slide against each other. They mainly affected by amount of force impelled them together and by the angle and position of the object.
  •             Frictional Force F = Coefficient of friction * normal force
  • At a constant velocity, there will be no acceleration and the net force will be zero.

               Net Force        = m * a = 0

               120 - Friction   = 0

               Friction Force  = -120 N

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27 degrees to kelvin, 20 dg to milligrams, and 3 to decimeters.
Gala2k [10]
K = C + 273, so 27°C = 27+273 = 300 K

1 dg = 100 mg, so 20 dg = 20×100 = 2,000 mg
7 0
3 years ago
A block of mass 2 kg slides down a frictionless ramp of length 1.3 m tilted at an angle 25o to the horizontal. At the bottom of
marin [14]

Answer:

Diagrams in pictures

Explanation:

Using energy I can get

m g h = 1/2 m v^2

So the velocity at the end of the ramp is the squareroot of two times the initial height of the box times the gravity constant.

(H= 1,3m sin25)

V=2,32m/s

V= a t

And

X= v t +1/2 a t^2

Knowing v=2,32 m/s and x= 1,3 m

I can get

a= 6,21m/s2

F= m a

I can get the force of the box when it collides with the spring

F= 12, 425 N

The force the spring makes on the box then is

F = -12,425N = -k d

Then the spring's constant is k= 51,75N/m

To make the two diagrams I need the functions of time when the box slows down

I use the same two equations

V= a t

And

X= v t + 1/2 a t^2

Being now 2,32 my initial velocity and 0 my final velocity, and my distance 0,24 m.

I get there the time t=0,0689 seconds and the acceleration a= -33,67 m/s2 (negative because it's slowing down).

Then,

V(t)= - 33,67 m/s2 t for time between 0 and 0,689 sec

X(t)= 2,32 m/s t + 1/2 33,67 m/s2 t^2.

for time between 0 and 0,689 sec

Diagrams and equations are in the pictures

7 0
3 years ago
A meteorologist plans to release a weather balloon from ground level, to be used for high-altitude atmospheric measurements. The
Slav-nsk [51]

Answer:

563.86 N

Explanation:

We know the buoyant force F = weight of air displaced by the balloon.

F = ρgV where ρ = density of air = 1.29 kg/m³, g = acceleration due to gravity = 9.8 m/s² and V = volume of balloon = 4πr/3 (since it is a sphere) where r = radius of balloon = 2.20 m

So, F = ρgV = ρg4πr³/3

substituting the values of the variables into the equation, we have

F =  1.29 kg/m³ × 9.8 m/s² × 4π × (2.20 m)³/3

= 1691.58 N/3

= 563.86 N

8 0
3 years ago
Calculate the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from =7 to =1.
Harrizon [31]

1.549×10-19lJ is the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from =7 to =1.

The equation E= hcE =hc, where h is Planck's constant and c is the speed of light, describes the inverse relationship between a photon's energy (E) and the wavelength of light ().

The Rydberg formula is used to determine the energy change.

Rydberg's original formula used wavelengths, but we may rewrite it using units of energy instead. The result is the following.

aaΔE=R(1n2f−1n2i) aa

were

2.17810-18lJ is the Rydberg constant.

The initial and ultimate energy levels are ni and nf.

As a change of pace from

n=5 to n=3 gives us

ΔE

=2.178×10-18lJ (132−152)

=2.178×10-18lJ (19−125)

=2.178×10-18lJ×25 - 9/25×9

=2.178×10-18lJ×16/225

=1.549×10-19lJ

Learn more about Rydberg formula here-

brainly.com/question/13185515

#SPJ4

8 0
2 years ago
If you lift the front wheel of a poorly maintained bicycle off the ground and then start it spinning at 0.69 rev/s , friction in
Sindrei [870]

Answer:

magnitude of the frictional torque is 0.11 Nm

Explanation:

Moment of inertia I = 0.33 kg⋅m2

Initial angular velocity w° = 0.69 rev/s = 2 x 3.142 x 0.69 = 4.34 rad/s

Final angular velocity w = 0 (since it stops)

Time t = 13 secs

Using w = w° + §t

Where § is angular acceleration

O = 4.34 + 13§

§ = -4.34/13 = -0.33 rad/s2

The negative sign implies it's a negative acceleration.

Frictional torque that brought it to rest must be equal to the original torque.

Torqu = I x §

T = 0.33 x 0.33 = 0.11 Nm

5 0
3 years ago
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