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velikii [3]
3 years ago
11

Rest and motion are relative terms why?​

Physics
1 answer:
krok68 [10]3 years ago
3 0

Answer:

Rest and motion are the relative terms because they depend on the observer's frame of reference. So if two different observers are not at rest with respect to each other, then they too get different results when they observe the motion or rest of a body.

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How do surface currents affect climate
Bumek [7]
Ocean currents<span> act much like a conveyer belt, transporting warm water and precipitation from the equator toward the poles and cold water from the poles back to the tropics. Therefore, </span>currents<span> regulate global </span>climate<span>, helping to counteract the uneven distribution of solar radiation reaching Earth's </span>surface<span>.</span>
8 0
4 years ago
What mass of ice (in g) can be melted if 27.2 kJ of thermal energy are added at the freezing point? Use molar mass = 18.02 g/mol
san4es73 [151]

Answer : The mass of ice melted can be, 3.98 grams.

Explanation :

First we have to calculate the moles of ice.

Q=\frac{\Delta H}{n}

where,

Q = energy absorbed = 27.2 kJ

\Delta H = enthalpy of fusion of ice = 6.01 kJ/mol

n = moles = ?

Now put all the given values in the above expression, we get:

27.2kJ=\frac{6.01kJ/mol}{n}

n=0.221mol

Now we have to calculate the mass of ice.

\text{Mass of ice}=\text{Moles of ice}\times \text{Molar mass of ice}

Molar mass of ice = 18.02 g/mol

\text{Mass of ice}=0.221mol\times 18.02g/mol=3.98g

Thus, the mass of ice melted can be, 3.98 grams.

3 0
4 years ago
What is the weight of a feather (mass = 0.0001 kg) that floats through earth's and the moon's atmospheres?
olya-2409 [2.1K]

Weight = (mass) x (acceleration of gravity)

Acceleration of gravity = 9.81 m/s² on Earth, 1.62 m/s² on the Moon.

The feather's weight is . . .

On Earth:  (0.0001 kg) x (9.81 m/s²) = <em>0.000981 Newton </em>

On the Moon:  (0.0001 kg) x (1.62 m/s²) = <em>0.000162 N</em>

The presence or absence of atmosphere makes no difference.  In fact, the numbers would be the same if the feather were sealed in a jar, or spinning wildly in a tornado, or hanging by a thread, or floating in a bowl of water or chicken soup.  Weight is just the force of gravity between the feather and the Earth.  It's not affected by what's around the feather, or what's happening to it.

6 0
3 years ago
Differences between weightlessness in space and weightlessness in earth​
zhuklara [117]

Answer:

it depends on a person's own weight

6 0
2 years ago
Read 2 more answers
A bug is 12 cm from the center of a turntable that is rotating with a frequency of 45 rev/min . What minimum coefficient frictio
Agata [3.3K]

Answer:

The minimum coefficient of friction is 0.27.

Explanation:

To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.

Centripetal force is written as

F_c = m\frac{v^2}{r}

with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as

v = r\omega

with ω denoting the angular velocity, which we are given. With that, the above becomes:

F_c = m\frac{v^2}{r}=m\omega^2 r

Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?

F_r=\mu mg \geq m\omega^2 r = F_c\implies\\\mu \geq \frac{\omega^2 r}{g}

Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:

\frac{1 rev}{1 min}\cdot\frac{\frac{2\pi rad}{rev}}{\frac{60s}{1 min}}=\frac{2\pi}{60}\frac{rad}{s}

and so 45 rev/min = 4.71 rad/s.

\mu \geq \frac{\omega^2 r}{g}=\frac{4.71^2\frac{1}{s^2}\cdot 0.12m}{9.8\frac{m}{s^2}}=0.27

A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.



3 0
3 years ago
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