Ask question

Ask question

All categories

3 years ago

12

a bucket of water of mass 20kg is pulled at a constant velocity up to a platform 40 meters above the ground. this takes 10 minut

es , during which time 5kg of water drips out at a steady rate through a hole in the bottom . find the work required to raise the bucket to the platform.

1 answer:

erma4kov [3.2K]3 years ago

5 0

<span>Taking into account the information above, we know the average mass of the bucket of water may be m=20-5/2=17.5kg. As the bucket of water is pulled at a "constant velocity" the work required to raise the bucket to the platform transformed into the potential energy of the bucket of water. That is why it should be W=mgh=17.5*9.8*40=6860J</span>

You might be interested in

PLEASE ANSWER ASAP!!!!!!!

Viefleur [7K]

Answer:

B. Mechanical energy= 50J+30J=80J

4 0

3 years ago

A rocket in deep space has an empty mass of 150 kg and exhausts the hot gases of burned fuel at 2500 m/s. It is loaded with 600

3241004551 [841]

Answer:

$v(10\,s) \approx 775.387\,\frac{m}{s}$

$v(20\,s)\approx 1905.350\,\frac{m}{s}$

$v(30\,s) \approx 4023.595\,\frac{m}{s}$

Explanation:

The speed of the rocket is given the Tsiolkovsky's differential equation, whose solution is:

$v (t) = v_{o} - v_{ex}\cdot \ln \frac{m}{m_{o}}$

Where:

$v_{o}$ - Initial speed of the rocket, in m/s.

$v_{ex}$ - Exhaust gas speed, in m/s.

$m_{o}$ - Initial total mass of the rocket, in kg.

$m$ - Current total mass of the rocket, in kg.

Let assume that fuel is burned linearly. So that,

$m(t) = m_{o} + r\cdot t$

The initial total mass of the rocket is:

$m_{o} = 750\,kg$

The fuel consumption rate is:

$r = -\frac{600\,kg}{30\,s}$

$r = -20\,\frac{kg}{s}$

The function for the current total mass of the rocket is:

$m(t) = 750\,kg - (20\,\frac{kg}{s} )\cdot t$

The speed function of the rocket is:

$v(t) = - 2500\,\frac{m}{s}\cdot \ln \frac{750\,kg -(20\,\frac{kg}{s} )\cdot t}{750\,kg}$

The speed of the rocket at given instants are:

$v(10\,s) \approx 775.387\,\frac{m}{s}$

$v(20\,s)\approx 1905.350\,\frac{m}{s}$

$v(30\,s) \approx 4023.595\,\frac{m}{s}$

7 0

3 years ago

A box is held at rest by two ropes that form 30° angles with the vertical. The tension T in either rope is 42 N. What is the wei

iragen [17]

<h3><u>Answer;</u></h3>

= 73 N

<h3><u>Explanation</u>;</h3>

Using the formula

2 T cos(30°) = w

Where; T is the tension on each string, while w is the weight of the box given by mg

Therefore;

W = 2Tcos 30°

= 2 × 42 cos 30°

= 84 cos 30°

= 72.74

<u> ≈ 73 N</u>

7 0

3 years ago

Is it better to wire a house using a series circuit or a parallel circuit?

Svetach [21]

Answer: its better to use parallel because, in parallel connection there will be more advantages than a series connection. and also the electronic devices are wired in series so thats why you should use parralel in house wiring

so its c.

because the parallel is wired through the whole house so if one of the circuits fail youŕe not screwed

Explanation:

5 0

3 years ago

A square coil (length of side = 24 cm) of wire consisting of two turns is placed in a uniform magnetic field that makes an angle

dalvyx [7]

Explanation:

It is given that,

Length of side of a square, l = 24 cm = 0.24 m

The uniform magnetic field makes an angle of 60° with the plane of the coil.

The magnetic field increases by 6.0 mT every 10 ms. We need to find the magnitude of the emf induced in the coil. The induced emf is given by :

$\epsilon=N\dfrac{d\phi}{dt}$

$\dfrac{d\phi}{dt}$ is the rate of change if magnetic flux.

$\phi=BA\ cos\theta$

$\theta$ is the angle between the magnetic field and the normal to area vector.

$\theta=90-60=30$

$\epsilon=NA\dfrac{dB}{dt}\times cos30$

$\epsilon=2\times (0.24\ m)^2\times \dfrac{6\ mT}{10\ mT}\times cos(30)$

$\epsilon=0.0598\ T$

$\epsilon=59.8\ mT$

or

EMF = 60 mT

So, the magnitude of emf induced in the coil is 60 mT. Hence, this is the required solution.

6 0

3 years ago