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3 years ago

A 3.0 kg box is at rest on a table. The static friction coefficient us between the box and table is 0.40, and

Physics

1 answer:

Oduvanchick [21]3 years ago

7 0

While the box is at rest, the net vertical force acting on it is

∑ F = n - w = 0

where n = magnitude of the normal force (table pushing up on the box) and w = weight of the box. So we have

n = w = mg = (3.0 kg) (9.8 m/s²) = 29.4 N

The coefficient of static friction between the table and box is 0.40, so the maximum magnitude of static friction is

f = 0.40 n = 0.40 (29.4 N) = 11.76 N

which is to say, the box will not move unless a force larger than this is applied to the box.

10 N is of course smaller than 11.76 N, so the box would stay at rest, and its acceleration would remain 0 m/s².

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A 2.1 ✕ 103-kg car starts from rest at the top of a 5.9-m-long driveway that is inclined at 19° with the horizontal. If an avera

Lina20 [59]

Answer:

3.9 m/s

Explanation:

We are given that

Mass of car,m= $2.1\times 10^3 kg$

Initial velocity,u=0

Distance,s=5.9 m

$\theta=19^{\circ}$

Average friction force,f= $4.0\times 10^3 N$

We have to find the speed of the car at the bottom of the driveway.

Net force, $F_{net}=mgsin\theta-f=2.1\times 10^3\times 9.8sin19-4.0\times 10^3$

Where $g=9.8 m/s^2$

Acceleration, $a=\frac{F_{net}}{m}=\frac{2.1\times 10^3\times 9.8sin19-4.0\times 10^3}{2.1\times 10^3}$

$v=\sqrt{2as}$

$v=\sqrt{2\times \frac{2.1\times 10^3\times 9.8sin19-4.0\times 10^3}{2.1\times 10^3}\times 5.9}$

v=3.9 m/s

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3 years ago

What are the two factors that affect the period of a pendulum?

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The two factors that affect the period of a pendulum are the length of the string and the distance in which the pendulum falls.

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A circular coil of wire of radius 5.0 cm has 20 turns and carries a current of 2.0 A. The coil lies in a magnetic field of magni

Korvikt [17]

Answer:a. Magnetic dipole moment is 0.3412Am²

b. Torque is zero(0)N.m

Explanation: The magnetic dipole moment U is given as the product of the number of turns n times the current I times the area A

That is,

U = n*I*A

But Area A is given as pi*radius² since it is a circular coil

Radius given is 5cm converting to meter we divide by 100 so we have our radius to be 0.05m. So area A is

A = 3.142*(0.05)² =7.86*EXP {-3} m²

Current I is 2 A

Number of turns is 20

So magnetic dipole moment U is

U = 20*2*7.86*EXP {-3}=0.3142A.m²

b. Torque is given as the cross product of the magnetic field B and magnetic dipole moment U

Torque = B x U =B*U*Sine(theta)

But since the magnetic field is directed parallel to the plane of the coil from the question, it means that the angle between them is zero and sine zero is equals 0(zero) if you substitute that into the formula for torque you will find out that your torque would equals zero(0)N.m

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3 years ago

The mass of 150cm³ of stone is 400g. its density is

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