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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 child on a swing sweeps out a distance of 45 ft on the first pass. If she is allowed to continue swinging until she stops, an

earnstyle [38]

Answer:

d = 90 ft

Explanation:

Here in each swing the distance sweeps by the swing is half of the initial distance that it will move

So here we can say that total distance in whole motion is given as

$d = 45 + \frac{45}{2} + \frac{45}{4} + \frac{45}{8}..........$

since it is half of the distance that it will move in each swing so it would be a geometric progression with common ratio of 1/2

so sum of such GP is given by the formula

$S = \frac{a}{1 - r}$

$d = \frac{45}{1 - \frac{1}{2}}$

$d = 90 ft$

6 0

3 years ago

A solid conducting sphere is given a positive charge q. How is the charge q distributed in or on the sphere?

leva [86]

Answer:

Explanation:

the sphere is solid and conducting, so the charge is uniformly distributed over its volume.

7 0

3 years ago

Read 2 more answers

A horizontal object-spring system oscillates with an amplitude of 2.8 cm. If the spring constant is 275 N/m and object has a mas

Lisa [10]

Answer:

(a) the mechanical energy of the system, U = 0.1078 J

(b) the maximum speed of the object, Vmax = 0.657 m/s

Explanation:

Given;

Amplitude of the spring, A = 2.8 cm = 0.028 m

Spring constant, K = 275 N/m

Mass of object, m = 0.5 kg

(a) the mechanical energy of the system

This is the potential energy of the system, U = ¹/₂KA²

U = ¹/₂ (275)(0.028)²

U = 0.1078 J

(b) the maximum speed of the object

$V_{max} =\omega*A= \sqrt{\frac{K}{M} } *A\\\\V_{max} = \sqrt{\frac{275}{0.5} } *0.028\\\\V_{max} = 0.657 \ m/s$

$a_{max} = \frac{KA}{M} \\\\a_{max} = \frac{275*0.028}{0.5}\\\\a_{max} = 15.4 \ m/s^2$

6 0

3 years ago

A 5.00 kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the en

Brut [27]

Answer:

75 N

Explanation:

In this problem, the position of the crate at time t is given by

$y(t)=2.80t+0.61t^3$

The velocity of the crate vs time is given by the derivative of the position, so it is:

$v(t)=y'(t)=\frac{d}{dt}(2.80t+0.61t^3)=2.80+1.83t^2$

Similarly, the acceleration of the crate vs time is given by the derivative of the velocity, so it is:

$a(t)=v'(t)=\frac{d}{dt}(2.80+1.83t^2)=3.66t$ [m/s^2]

According to Newton's second law of motion, the force acting on the crate is equal to the product between mass and acceleration, so:

$F(t)=ma(t)$

where

m = 5.00 kg is the mass of the crate

At t = 4.10 s, the acceleration of the crate is

$a(4.10)=3.66\cdot 4.10 =15.0 m/s^2$

And therefore, the force on the crate is:

$F=ma=(5.00)(15.0)=75 N$

7 0

3 years ago

What is the mass of an object that has an acceleration of 2.63m/s2 when a unbalanced force of 112N is applied to it?

exis [7]

Answer:

42.58kg

Explanation:

By Newton's second law, F = ma.

F is the force being applied, in this case 112N. a is the acceleration, in this case 2.63 m/s^2.

Thus, with some simple algebraic manipulation, we get the mass to equal:

m = F/a = 112N / 2.63 m/s^2 = 42.58kg

4 0

3 years ago