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3 years ago

7

Do we "spend" anything in a resistance? Like for example: Current, mass, energy, field, potential, momentum? If the answer is ye

s, what do we spend? Explain your answer briefly.

2 answers:

masha68 [24]3 years ago

5 0

Resistance, like the word suggests, is the objects opposition to flow of current through it. I would think that as the object is causing resistance, some of the energy from the electricity is dissipated as heat.

tatuchka [14]3 years ago

3 0

Resistance dissipates electrical ENERGY in the form of heat.

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In an experiment, James studied the relationship between kinetic energy, mass, and velocity. Four masses were rolled across a sm

Alex777 [14]

Answer:

20

Explanation:

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4 years ago

Choose the correct statement about the acceleration of the car.

brilliants [131]

Answer:

Whats the question/word problem or where is the graph (if included) representing this problem?

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3 years ago

Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject materia

kramer

Answer:

The height reached by the material on Earth is 91 km.

Explanation:

Given that,

Mass $M_{Io}=8.93\times10^{22}\ kg$

Radius = 1821 km

Height $h_{Io}=500\ km$

Suppose we need to find that how high would this material go on earth if it were ejected with the same speed as on Io?

We need to calculate the acceleration due to gravity on Io

Using formula of gravity

$g =\dfrac{GM_{Io}}{(R_{Io})^2}$

Put the value into the formula

$g=\dfrac{6.67\times10^{-11}\times8.93\times10^{22}}{(1821\times10^{3})^2}$

$g=1.79\ m/s^2$

Let v be the speed at which the material is ejected.

We need to calculate the height

Using the formula of height

$H=\dfrac{v^2}{2g}$

Using ratio of height of earth and height of Io

$\dfrac{H_{e}}{H_{Io}}=\dfrac{\dfrac{v^2}{2g_{e}}}{\dfrac{v^2}{2g_{Io}}}$

$\dfrac{H_{e}}{H_{Io}}=\dfrac{g_{Io}}{g_{e}}$

Put the value into the formula

$\dfrac{H_{e}}{H_{Io}}=\dfrac{1.79}{9.8}$

$\dfrac{H_{e}}{H_{Io}}=0.182$

$H_{e}=0.182\times H_{Io}$

$H_{e}=0.182\times500$

$H_{e}=91\ km$

Hence, The height reached by the material on Earth is 91 km.

3 0

3 years ago

You throw a 3.00 N rock vertically into the air from ground level. You observe that when it is 16.0 m above the ground, it is tr

Morgarella [4.7K]

Answer:

The final speed of the stone as it lift the ground is 23.86 m/s.

Explanation:

Given that,

Force acting on the rock, F = 3 N

Distance, d = 16 m

Initial speed of the stone, u = 22 m/s

We need to find the rock's speed just as it left the ground. It can be calculated using work energy theorem as :

$W=\Delta E\\\\W=\dfrac{1}{2}m(v^2-u^2)\\\\Fd=\dfrac{1}{2}m(v^2-u^2)\\\\v^2=\dfrac{2Fd}{m}+u^2\\\\v^2=\dfrac{2mgd}{m}+u^2\\\\v^2=2\times 9.8\times 16+(16)^2\\\\v=23.86\ m/s$

So, the final speed of the stone as it lift the ground is 23.86 m/s.

4 0

3 years ago

Earth's neighboring galaxy, the Andromeda Galaxy, is a distance of 2.54×107 light-years from Earth. If the lifetime of a human i

Varvara68 [4.7K]

Answer:

$0.0018833\ \text{m/s}$

Explanation:

$d$ = Distance of Andromeda Galaxy from Earth = $2.54\times 10^7\ \text{ly}$

$t$ = Time taken = $90\ \text{years}$

$c$ = Speed of light = $3\times 10^8\ \text{m/s}$

We have the relation

$t=t_o\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow 90=2.54\times 10^7\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow \dfrac{90^2}{(2.54\times 10^7)^2}=1-\dfrac{v^2}{c^2}\\\Rightarrow 1-\dfrac{90^2}{(2.54\times 10^7)^2}=\dfrac{v^2}{c^2}\\\Rightarrow v=c\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}}$

$c-v=c(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=3\times 10^8(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=0.0018833\ \text{m/s}$

The required answer is $0.0018833\ \text{m/s}$ .

7 0

3 years ago