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3 years ago

1) the strength of an electromagnet can be increased by

Physics

1 answer:

miss Akunina [59]3 years ago

5 0

Using coils of fewer turns on the electromagnet

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A 100-kg astronaut is floating in outer space. If the astronaut throws a 2-kilogram wrench at a speed of 10 meters per second, w

sergey [27]

Since Astronaut and wrench system is isolated in the space and there is no external force on it

So here momentum of the system will remain conserved

so here we can say

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

initially both are at rest

so here plug in all values

$0 = 100 v_{1f} + 2\times 10$

$v_{1f} = -0.20 m/s$

so here the astronaut will move in opposite direction and its speed will be equal to 0.20 m/s

3 0

3 years ago

The field-line representation of the e-field in a certain region in space is shown below. The dashed lines represent equipotenti

lorasvet [3.4K]

You need to go on google an dlook this up

3 0

3 years ago

A clarinetist, setting out for a performance, grabs his 3.010 kg clarinet case (including the clarinet) from the top of the pian

KATRIN_1 [288]

Answer:

-0.481 m/s^2

Explanation:

The force equation of this problem is given as:

F - W = ma

where F = upward force holding the clarinet bag

W = downward force (weight of the clarinet)

The mass of the clarinet bag is 3.010 kg, therefore, its weight is:

W = mg

W = 3.010 * 9.8 = 29.498

F = 28.05 N

Therefore:

28.05 - 29.498 = 3.010 * a

-1.448 = 3.010a

=> a = -1.448 / 3.010

a = -0.481 m/s^2

The acceleration of the bag is downward.

8 0

3 years ago

What type of energy is thermal energy?

BigorU [14]

Thermal energy is an example of kinetic energy , due to motion of particles .

5 0

3 years ago

A planet of mass m 6.75 x 1024 kg is orbiting in a circular path a star of mass M 2.75 x 1029 kg. The radius of the orbit is R 8

Umnica [9.8K]

Answer:

The orbital period of the planet is 387.62 days.

Explanation:

Given that,

Mass of planet $m =6.75\times10^{24}\ kg$

Mass of star $m'=2.75\times10^{29}\ kg$

Radius of the orbit $r =8.05\times10^{7}\ km$

Using centripetal and gravitational force

The centripetal force is given by

$F = \dfrac{mv^2}{r}$

$F=m\omega^2r$

We know that,

$\omega=\dfrac{2\pi}{T}$

$F=m(\dfrac{2\pi}{T})^2r$ ....(I)

The gravitational force is given by

$F = \dfrac{mm'G}{r^2}$ ....(II)

From equation (I) and (II)

$m(\dfrac{2\pi}{T})^2r=\dfrac{mm'G}{r^2}$

Where, m = mass of planet

m' = mass of star

G = gravitational constant

r = radius of the orbit

T = time period

Put the value into the formula

$T^2=\dfrac{4\pi^2R^3}{m'G}$

$T^2=\dfrac{4\times(3.14)^2\times(8.05\times10^{7})^3}{2.75\times10^{29}\times6.67\times10^{-11}}$

$T=2\times3.14\times\sqrt{\dfrac{(8.05\times10^{10})^3}{2.75\times10^{29}\times6.67\times10^{-11}}}$

$T =3.34\times10^{7}\ s$

$T= 387.62\days$

Hence, The orbital period of the planet is 387.62 days.

4 0

3 years ago

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