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3 years ago

You need to move a 132 kg sofa to a different location in the room. It takes a force

1 answer:

nataly862011 [7]3 years ago

6 0

Let w denote the weight of the sofa, n the magnitude of the normal force, and f the magnitude of the friction force. The sofa is in equilibrium, so by Newton's second law,

n - w = 0

125 N - f = 0

The sofa has a weight of

w = (132 kg) g = 1293.6 N

where g = 9.80 m/s² is the magnitude of the acceleration due to gravity. Since n = w, the normal force has the same magnitude.

The friction force is proportional to the normal force by a factor of the coefficient of static friction µ, such that

f = µ n

Our second equation tells us f = 125 N, so solve for µ :

125 N = µ (1293.6 N)

µ ≈ 0.0966

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A wire carrying a 29.0 A current passes between the poles of a strong magnet such that the wire is perpendicular to the magnet's

Dmitrij [34]

Answer:

2.59 T

Explanation:

Parameters given:

Current flowing through the wire, I = 29 A

Angle between the magnetic field and wire, θ = 90°

Magnetic force, F = 2.25 N

Length of wire, L = 3 cm = 0.03 m

The magnetic force, F, is related to the magnetic field, B, by the equation below:

F = I * L * B * sinθ

Inputting the given parameters:

2.25 = 29 * 0.03 * B * sin90

2.25 = 0.87 * B

=> B = 2.25/0.87

B = 2.59 T

The magnetic field strength between the poles is 2.59 T

4 0

3 years ago

Calculate net force 532N 215N

lorasvet [3.4K]

Answer:

FN is the forces acting on a body. When the body is at rest, the net force formula is given by, FNet = Fa + Fg.

Im in 7th and thats all I know so I hope it's enough

8 0

3 years ago

you give a shopping cart a job down the aisle the cart is full of groceries and has a mass of 18 kilograms the cart accelerates

mezya [45]

Force = (mass) x (acceleration)

Force = (18 kg) x (3 m/s²) = 54 newtons

As long as you continue pushing the cart with 54 newtons of force,
it will accelerate at 3 m/s².

At the instant you release it, or keep your hands on it but stop pushing,
it will stop accelerating. It'll continue forward at the speed it had when
the 54 newtons of force stopped.

6 0

2 years ago

Read 2 more answers

How does the electric force between two charged participles change if one particle’s charge is reduced by a factor of 3?

Eva8 [605]

Answer:

ick

Explanation:

Me down daddy

3 0

3 years ago

A 175-kg roller coaster car starts from rest at the top of an 18.0-m hill and rolls down the hill, then up a second hill that ha

Anni [7]

Answer:

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

From top of the first hill to the bottom

$m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss}$ (1)

From the bottom to the top of the second hill

$\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss}$ (2)

Where:

$m$ - Mass of the roller coaster car, in kilograms.

$v_{1}$ - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

$g$ - Gravitational acceleration, in meters per square second.

$h_{1}$ - Height of the first top of the hill with respect to the bottom, in meters.

$W_{1, loss}$ - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

$v_{2}$ - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

$h_{2}$ - Height of the second top of the hill with respect to the bottom, in meters.

$W_{2, loss}$ - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

$m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss}$ (3)

Where $W_{loss}$ is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

If we know that $m = 175\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $h_{1} = 18\,m$ , $h_{2} = 8\,m$ and $v_{2} = 11\,\frac{m}{s}$ , then the work done by non-conservative force is:

$W_{loss} = m\cdot\left[ g\cdot \left(h_{1}-h_{2}\right)-\frac{1}{2}\cdot v_{2}^{2} \right]$

$W_{loss} = 6574.75\,J$

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

8 0

2 years ago