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3 years ago

Please what is the work done by a man who is pulling a box of 45kg of mass by means of rope which makes angle of 45 degrees ?

Physics

1 answer:

Tpy6a [65]3 years ago

5 0

Answer:

No work is done since no distance is given

Explanation:

Since no distance is given, the force is not doing any work

No work is done by the man since we do not know the distance or displacement.

Work is only said to be done when the force applied on an object moves it through a particular distance.

Work done = Force x distance.

Since no distance is given in this problem, we can as well assume that the force applied is doing no work on the object.

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A ball with a mass of 2000 g is floating on the surface of a pool of water. What is the minimum volume that the ball could have

Doss [256]

Answer:

$2000\; {\rm cm^{3}}$ .

Explanation:

When the ball is placed in this pool of water, part of the ball would be beneath the surface of the pool. The volume of the water that this ball displaced is equal to the volume of the ball that is beneath the water surface.

The buoyancy force on this ball would be equal in magnitude to the weight of water that this ball has displaced.

Let $m(\text{ball})$ denote the mass of this ball. Let $m(\text{water})$ denote the mass of water that this ball has displaced.

Let $g$ denote the gravitational field strength. The weight of this ball would be $m(\text{ball}) \, g$ . Likewise, the weight of water displaced would be $m(\text{water})\, g$ .

For this ball to stay afloat, the buoyancy force on this ball should be greater than or equal to the weight of this ball. In other words:

$\text{buoyancy} \ge m(\text{ball})\, g$ .

At the same time, buoyancy is equal in magnitude the the weight of water displaced. Thus:

$\text{buoyancy} = m(\text{water}) \, g$ .

Therefore:

$m(\text{water})\, g = \text{buoyancy} \ge m(\text{ball})\, g$ .

$m(\text{water}) \ge m(\text{ball})$ .

In other words, the mass of water that this ball displaced should be greater than or equal to the mass of of the ball. Let $\rho(\text{water})$ denote the density of water. The volume of water that this ball should displace would be:

$\begin{aligned}V(\text{water}) &= \frac{m(\text{water})}{\rho(\text{water})} \\ &\ge \frac{m(\text{ball}))}{\rho(\text{water})} \end{aligned}$ .

Given that $m(\text{ball}) = 2000\; {\rm g}$ while $\rho = 1.00\; {\rm g\cdot cm^{-3}}$ :

$\begin{aligned}V(\text{water}) &\ge \frac{m(\text{ball}))}{\rho(\text{water})} \\ &= \frac{2000\; {\rm g}}{1.00\; {\rm g\cdot cm^{-3}}} \\ &= 2000\; {\rm cm^{3}}\end{aligned}$ .

In other words, for this ball to stay afloat, at least $2000\; {\rm cm^{3}}$ of the volume of this ball should be under water. Therefore, the volume of this ball should be at least $2000\; {\rm cm^{3}}\!$ .

3 0

2 years ago

What does the energy of an electromagnetic wave depend on

lora16 [44]

<span>Energy of an electromagnetic wave depends on it's "Frequency"

Hope this helps!</span>

4 0

4 years ago

Read 2 more answers

As the pendulum swings through the origin, it has a speed of v0 = 1.1 m/s. If the mass on the end of the pendulum is 15 g, and t

serg [7]

Answer:

1: 6.18 cm

2: 52.5609 degrees

Explanation:

We have the pendulum speed at the origin, and in that moment, all energy is kinetic, so we can calculate the pendulum energy by:

Ec = 0.5*m*v^2 = 0.5*0.015*1.1^2 = 0.0091 J

Now with that energy, we can calculate the height the pendulum will reach, as in that moment, the kinetic energy is totally converted to gravitational potencial energy:

Eg = m*g*h = 0.0091

0.015 * 9.81 * h = 0.0091

h = 0.0091 / (0.015 * 9.81 ) = 0.0618 m = 6.18 cm

Looking at the image attached, we can see that the pendulum will form a triangle, and one of the cathetus will be the length of the pendulum minus the height it went up, and the hypotenusa will be the pendulum length.

So, we know that the sine of the angle will be the division between the opposite cathetus and the hypotenusa:

sin(angle) = (30-6.18)/30 = 23.82/30 = 0.794 -> angle = 52.5609 degrees

4 0

3 years ago

How fast and in what direction must galaxy A be moving 17 an absorption Line found at 500 nm for a stationary galaxy 1s shifted

emmainna [20.7K]

Answer:

v = 0.22 c and wavelength is decrease so galaxy A is move away from stationary galaxy

v = 0.095 c and wavelength is increase so galaxy B come toward stationary galaxy

Explanation:

Given data

found = 500 nm

shifted A = 400 nm

shifted B = 550 nm

to find out

How fast and in what direction is galaxy

solution

we use here formula that is

(λ) shifted = √(1-β / 1+β) (λ)found

1-β / 1+β = (4/5)² ..................1

1-β / 1+β = 16 / 25

β = 0.22

v/c = 0.22

v = 0.22 c

here wavelength is decrease so galaxy A is move away from stationary galaxy

and

here according to equation 1

and we use shifted 550 nm

1-β / 1+β = (5.5/5)²

1-β / 1+β = 30.25 / 25

β = 0.095

so v/c = 0.095

v = 0.095 c

here wavelength is increase so galaxy B come toward stationary galaxy

3 0

3 years ago

(m = 4.0 kg) the cockroach rides on the rim (at radius R) of a disk (M = 6.0 kg) that turns about its center like a merry-go-rou

kvasek [131]

Answer:

ω' = 2.5 rad/s

Explanation:

mass of cockroach, m = 4 kg

mass of disk, M = 6 kg

Radius of disc= R

initial angular velocity, ω = 2 rad/s

Let the final angular velocity is ω'

As no external torque is applied, so the angular momentum is constant.

Angular momentum = Moment of inertia x angular velocity

I ω = I' ω'

$\frac{1}{2}\left ( M+m \right )R^{2}\times {2} = \left (\frac{1}{2}MR^{2}+m(0.5R)^{2}) \right )\omega '$

$10R^{2} = 4R^{2}\omega '$

ω' = 2.5 rad/s

5 0

3 years ago