What are the two conditions for work done

A 25-foot ladder is leaning against a house. The base of the ladder is pulled away from the house at a rate of 2 feet per second

grandymaker [24]

A ladder 25 feet long is leaning against a house. The base of the ladder is pulled away at a rate of 2 ft/sec.

a.) How fast is the top of the ladder moving down the wall when the base of the ladder is 12 feet from the wall?

Answer:

dy/dt = -1.094ft/sec

Explanation:

Given that:

dz/dt = 0,

dx/dt = 2,

dy/dt = ?

Hence, we have the following

Using Pythagoras theorem

We have 25ft as the hypotenuse, y as the opposite or height of wall, and x as the base of the triangle

X² + y² = z²,

12² + y² = 25²,

y² = 25² - 12²

y = √481

Therefore, we have the following:

2x dx/dt + 2y dy/dt,

= 2z dz/dt,

= 12 (2) √481 dy/dt,

= √481 dy/dt = -24,

= dy/dt = -1.094ft/sec

Therefore, final answer is -1.094ft/sec

4 0

3 years ago

If an electron moves in a circle of radius 21 cm perpendicular to a B field of 0.4 T, what are the speed of the electron and the

kodGreya [7K]

Answer:

a)

$v = 4.048 *10^6$ m/s

b)

Angular frequency = $1.92 * 10^7$

Explanation:

As we know

$v = \frac{qBr}{m}$

q is the charge on the electron = $3.2 * 10^{-19}$ C

B is the magnetic field in Tesla = $0.4$ T

r is the radius of the circle = $0.21$ m

mass of the electrons = $6.64 * 10^{-27}$ Kg

a)

Substituting the given values in above equation, we get -

$v = \frac{3.2 * 10^{-19}*0.4*0.21}{6.64 * 10^{-27}} \\v = 4.048 *10^6$ m/s

b)

Angular frequency =

$\frac{4.048 * 10^6 }{0.21} \\1.92 * 10^7$

8 0

3 years ago

A 62.0-kg skier is moving at 6.30 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.90

Klio2033 [76]

Here are the missing questions:
(a) How fast is the skier moving when she gets to the bottom of the hill?
(b) How much internal energy was generated in crossing the rough patch?
Part A
The initial kinetic energy of the skier is:
$E_{k0}=m\frac{v_0^2}{2}$
Part of this energy is then used to do work against the force of friction. Force of friction on the horizontal surface can be calculated using following formula:
$F_f=mg\mu$
The work is simply the force times the length:
$W_f=F_f\cdot L=mg\mu L$
So when the skier passes over the rough patch its energy is:
$E=E_{k0}-W_f$
When the skier is going down the skill gravitational potential energy is transformed into the kinetic energy:
$E_p=E_{k1}\\ mgh=E_{k1}$
So the final energy of the skier is:
$E_f=E_{k0}-W_f+E_{k1}\\ E_f=m\frac{v_0^2}{2}-mg\mu L+mgh=1856.86$J$
This energy is the kinetic energy of the skier:
$E_f=m\frac{v_f^2}{2}\\ v_f=\sqrt{\frac{2E_f}{m}}=7.74\frac{m}{s}$
Part B
We know that skier lost some of its kinetic energy when crossing over the rough patch. This energy is equal to the work done by the skier against the force of friction.
$E_{int}=W_f\\
E_{int}=mg\mu L=894.1$J$

4 0

4 years ago

Why is the reflection of a wave at a free boundary different from reflection at a fixed boundary??

Gnoma [55]

Here is the main thing is medium. In the free boundary wave face the air medium, but in a fixed boundary wave face different medium. The change of medium effect on the change of the reflection of wave.

6 0

3 years ago

What is the atomic mass of Jupiter?

lana [24]

The mass of Jupiter is 1.9 x 1027 kg.

4 0

3 years ago

What are the two conditions for work done​

What are the two conditions for work done