What do opposite poles in magnets do?

Which tool would be most useful to identify areas with frequent earthquakes to predict future earthquakes?

enyata [817]

It is a seismograph that would be most useful.

6 0

3 years ago

How much work is done on 10.0C of charge to move it through a potential difference of 9V in 10s?

Norma-Jean [14]

Here is the link with ans on it
https://moorsscience.wikispaces.com/file/view/chapter+12+answers.pdf
hope it helps

5 0

3 years ago

What is the mass of a 2 kg object on the Earth and on the moon?

zubka84 [21]

Answer:

Same

Explanation:

Mass is the quantity of matter in a certain object.

WHEREVER you take a 2kg object, the mass will remain 2kg. All that changes is the Weight ..Weight the force which the centre of a Planet uses to pull everything towards itself.

On earth, it is 9.81 whereas on the Moon it is 1.6

7 0

3 years ago

A farmer lifts his hay bales into the top loft of his barn by walking his horse forward with a constant velocity of 1 ft/s. Dete

Lesechka [4]

Answer:

The velocity of the hay bale is - 0.5 ft/s and the acceleration is $6.25\times 10^{- 3} ft/s^{2}$

Solution:

As per the question:

Constant velocity of the horse in the horizontal, $v_{x} = 1 ft/s$

Distance of the horse on the horizontal axis, x = 10 ft

Vertical distance, y = 20 ft

Now,

Apply Pythagoras theorem to find the length:

$20^{2} + 10^{2} = l^{2}$

$l^{2}= 500$

Now,

$x^{2} + y^{2} = 500$ (1)

Differentiating equation (1) w.r.t 't':

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

$x\frac{dx}{dt} = - y\frac{dy}{dt}$

where

$\frac{dx}{dt}$ = Rate of change of displacement along the horizontal

$\frac{dy}{dt}$ = Rate of change of displacement along the vertical

$v_{x}$ = velocity along the x-axis.

$v_{y}$ = velocity along the y-axis

$xv_{x} = -yv_{y}$

$v_{y} = - 10\times \frac{1}{20} = - 0.5 ft/s$

$|v_{y}| = 0.5\ ft/s$

Acceleration of the hay bale is given by the kinematic equation:

$v_{y}^{2} = u_{y} + 2ay$

$(-0.5)^{2} =0 + 2ay$

$0.25 = 2ay$

$\frac{0.25}{2y} = a$

$a = \frac{0.25}{2\times 20} = 6.25\times 10^{- 3} ft/s^{2}$

7 0

4 years ago

Two people, one with mass m1 and the other with mass m2, stand on a stationary sled with mass M on a frozen lake. Assume that th

lozanna [386]

Answer:

Part a)

Velocity of sled

$v = \frac{m_1 s}{m_1 + m_2 + M}$

velocity of first man who jump off

$v_1 = -\frac{(m_2 + M) s}{m_1 + m_2 + M}$

Part b)

Velocity of sled

$v_f = (\frac{m_1 s}{m_1 + m_2 + M}) + (\frac{m_2}{m_2 + M})s$

Also the speed of second person is given as

$v_2 = (\frac{m_1 s}{m_1 + m_2 + M}) - \frac{Ms}{m_2 + M}$

Part c)

change in kinetic energy of sled + two people is given as

$KE = \frac{1}{2}Mv_f^2 + \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$

Explanation:

As we know that here we we consider both people + sled as a system then there is no external force on it

So here we can use momentum conservation

since both people + sled is at rest initially so initial total momentum is zero

now when first people will jump with relative velocity "s" then let say the sled + other people will move off with speed v

so by momentum conservation we have

$0 = m_1(v - s) + (m_2 + M)v$

$v = \frac{m_1 s}{m_1 + m_2 + M}$

so velocity of the sled + other person is

$v = \frac{m_1 s}{m_1 + m_2 + M}$

velocity of first man who jump off

$v_1 = \frac{m_1 s}{m_1 + m_2 + M} - s$

$v_1 = -\frac{(m_2 + M) s}{m_1 + m_2 + M}$

Part b)

now when other man also jump off with same relative velocity

so let say the sled is now moving with speed vf

so by momentum conservation we have

$(m_2 + M)(\frac{m_1 s}{m_1 + m_2 + M}) = m_2(v_f - s) + Mv_f$

$(m_2 + M)(\frac{m_1 s}{m_1 + m_2 + M}) + m_2s = (m_2 + M)v_f$

Now we have

$v_f = (\frac{m_1 s}{m_1 + m_2 + M}) + (\frac{m_2}{m_2 + M})s$

Also the speed of second person is given as

$v_2 = (\frac{m_1 s}{m_1 + m_2 + M}) + (\frac{m_2}{m_2 + M})s - s$

$v_2 = (\frac{m_1 s}{m_1 + m_2 + M}) - \frac{Ms}{m_2 + M}$

Part c)

change in kinetic energy of sled + two people is given as

$KE = \frac{1}{2}Mv_f^2 + \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$

here we know all values of speed as we found it in part a) and part b)

4 0

3 years ago