All categories

2 years ago

Someone pls help fast

Physics

2 answers:

Helen [10]2 years ago

5 0

I think u forgot to add the question please add the question

Kruka [31]2 years ago

4 0

<h3>You forgot to add question...Add questions before asking so we can help</h3>

You might be interested in

Methods to reduce the frictional force between the an object and surface which it is in contact.

telo118 [61]

Use of lubricant
Use of ball bearers
Use of streamlined body
Use of graphite

7 0

3 years ago

enot [183]

Answer: 3 square feet

Explanation: I took the test

6 0

3 years ago

Use the drop-down menus to complete the statement about Earth’s magnetism.

Nataly [62]

Answer:

The Earth's magnetism is generated in the core, which is composed of iron that is constantly churning

Explanation:

Magnetic fields are produced by charges in motion, therefore by currents.

The outer core of the Earth consists mainly of melted iron that is in constant motion. This iron in motion actually acts as a giant current, and therefore it is responsible for the creation of the Earth's magnetic field.

The magnetic field of the Earth is very weak, in fact its magnitude is on average between 25 and 65 microtesla (for comparison, normal magnets can even produce magnetic fields of a few millitesla).

However, its role is very important for the Earth: in fact, it provides a shield that blocks most of the harmful radiation coming from the Sun.

5 0

3 years ago

Read 2 more answers

A 1-kilogram mass is attached to a spring whose constant is 14 N/m, and the entire system is then submerged in a liquid that imp

ch4aika [34]

Answer:

Part(a): The equation of motion is $\bf{x(t) = \dfrac{7}{5}~e^{-2t} - \dfrac{2}{5}~e^{-7t}}$ .

Part(b): The equation of motion is $\bf{x(t) = -e^{-2t} + \dfrac{11}{5}~e^{-7t}}$ .

Explanation:

If ' $m$ ' be the mass of the object, ' $k$ ' be the force constant and ' $\beta$ ' be the damping constant, then the equation of motion of the particle can be written as

$\dfrac{d^{2}x}{dt^{2}} + \dfrac{\beta}{m} \dfrac{dx}{dt} + \dfrac{k}{m}x= 0.........................................(I)$

Given $m$ = 1 Kg, $k$ = 14 $N~m^{-1}$ , $\beta$ = 9. Substituting these values in equation (I),

$\dfrac{d^{2}x}{dt^{2}} + 9~\dfrac{dx}{dt} + 14~x= 0$

Taking a trial solution $x(t) = e^{mt}$ , the auxiliary equation can be written as

$m^{2} + 9m + 14 = 0............................................................(II)$

and its solutions are $m_{1} = -2~and~m_{2} = -7$ , resulting the general solution

$x(t) = C_{1}~e^{-2t} + C_{2}~e^{-7t}....................................................................(III)$

The velocity at any instant of time of the mass is

$v(t) = -2C_{1}~e^{-2t} _7~C_{2}~e^{-7t}..............................................................(IV)$

Part(a):

Given $x(t=0) = 1 m,~and~v(t=0) = 0~m~s^{-1}$ . Substituting these values in equation (III) and (IV),

$&& 1 = C_{1} + C_{2}......................(V)\\&and,& 0 = -2C_{1} - 7C_{2}.......................(VI)$

Solving equations (V) and (VI), we have

$C_{1} = \dfrac{7}{5}~and~C_{2} = \dfrac{-2}{5}$

So the equation of motion is

$x(t) = \dfrac{7}{5}~e^{-2t} - \dfrac{2}{5}~e^{-7t}$

Part(b):

Given $x(t=0) = 1 m,~and~v(t=0) = - 12~m~s^{-1}$ . Substituting these values in equation (III) and (IV),

$&& 1 = C_{1} + C_{2}......................(VII)\\&and,& -12 = -2C_{1} -7C_{2}.......................(VIII)$

Solving equations (V) and (VI), we have

$C_{1} = -1~and~C_{2} = \dfrac{11}{5}$

So the equation of motion is

$x(t) = -e^{-2t} + \dfrac{11}{5}~e^{-7t}$

3 0

3 years ago

If we decrease the amount of force, and keep all other factors the same, what will happen to the amount of work?

kozerog [31]

The answer is A. I am sure of it.

8 0

3 years ago

Read 2 more answers