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Ivan
2 years ago
15

When a force of 15 newtons is applied to a stationary chair, it starts moving. What can you say about the frictional force betwe

en the chair and the floor of the room?
Physics
1 answer:
harkovskaia [24]2 years ago
3 0
First you need to make a difference between friction while object is stationary and the friction while object is moving. Force required to start moving some object is slightly greater than force required to maintain objects movement. That means that to move a chair you need some force F1 but you can than slightly reduce force and chair will still be moving.

Now to the problem in this question: It can be said that "stationary friction force" is equal to 15 Newtons. Its also good to know that friction force between chair and floor while you are increasing your push is also increasing and is equal to force of your push. Once it reaches 15N which is it "critical value" for that chair, chair starts moving and friction force drops a little bit and now it is called friction force of moving chair.
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A 150 N boy rides a 60 N bicycle a total of 200 m at a constant speed. The frictional force against the forward motion of the bi
soldier1979 [14.2K]

Answer:

W = 7000 J

Explanation:

To solve this problem we use that the speed of the bicycle is constant, therefore its acceleration is zero

            F -fr = 0

            F = fr

where F is the force applied by the child

Work is defined by

           W = F. x

           W = F x cos θ

in this case the child's force is parallel to the movement, therefore the angle is zero and cos 0 = 1

           

let's calculate

           W = 35 200

           W = 7000 J

8 0
3 years ago
A 5 meter long ladder leans against a wall. The bottom of the ladder slides away from the wall at the constant rate of 1 3 m/s.
Oksana_A [137]

Answer:9.75 m/s

Explanation:

Given

Length of ladder (L)=5 m

Foot the ladder is moving away with speed of \frac{\mathrm{d} x}{\mathrm{d} t}=13 m/s

From diagram

x^2+y^2=L^2------1

at x=3

y^2=25-9=16

y=4 m

Now differentiating equation 1 w.r.t time

2x\frac{\mathrm{d} x}{\mathrm{d} t}+2y\frac{\mathrm{d} y}{\mathrm{d} t}=0

x\frac{\mathrm{d} x}{\mathrm{d} t}=-y\frac{\mathrm{d} y}{\mathrm{d} t}

3\times 13=-4\times \frac{\mathrm{d} y}{\mathrm{d} t}

\frac{\mathrm{d} y}{\mathrm{d} t}=-\frac{3\times 13}{4}=-9.75 m/s

negative indicates distance is decreasing with time

5 0
2 years ago
Force of 10N down,10N to the right,and 5N to the left are acting on a ball .it acceleration horizontally to the right.what other
denpristay [2]

Answer:

A 10 N force pointing up

Explanation:

If the net acceleration of the object is horizontal pointing to the right, that means that all vertical forces must have canceled out, and the only ones "unbalanced" are the horizontal ones (10 N to the right minus 5 N to the left giving a net force of 5 N to the right).

Since they mentioned only one vertical force pointing down (10 N), there must be another one of same magnitude but pointing in opposite direction (up).

Then there must also be a 10 N force pointing up acting on the object.

3 0
3 years ago
A circuit has a 9.0 V power supply and two 15 Ω resistors connected in parallel. what is the voltage across each resistor?
Flauer [41]

Answer:

B. 9.0 V

Explanation:

In parallel circuits, the voltage across each circuit is the same across each component, which is also equal to the total voltage of the power supplied. So in this case, the voltage across each resistor is still 9.0V.

The voltage only changes when the resistors can connected in series.

3 0
3 years ago
two long parrallel wires that are 0.30 m apart carry currents of 5.0 A and 8.0 A in the opposite direction.Find the magnitude of
Vera_Pavlovna [14]

Answer: The magnitude of force per length that each wire exert on the other wire is 2.67×10^-5 N/m.

The force is repulsive.

Explanation: Please see the attachments below

3 0
2 years ago
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