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Flura [38]
2 years ago
8

What direction can Earth's plates move

Physics
1 answer:
Burka [1]2 years ago
3 0

The movement of the plates creates three types of tectonic boundaries: convergent, where plates move into one another; divergent, where plates move apart; and transform, where plates move sideways in relation to each other. They move at a rate of one to two inches (three to five centimeters) per year.

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Which of the following is an example of a vector quantity
Tomtit [17]

Answer:

A

Explanation:

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A pendulum on a planet, where gravitational acceleration is unknown, oscillates with a time period 5 sec. If the mass is increas
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Explanation:

We have, a pendulum on a planet, oscillates with a time period 5 sec. The formula used to find the time period is given by :

T=2\pi \sqrt{\dfrac{l}{g}}

l is length of the pendulum

g is acceleration due to gravity on which it is placed

It is clear that, the time period of pendulum is independent of the mass. Hence, if the mass is increased six times, its time period remains the same.

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3 years ago
The majority of neritic sediments are of which type?
lesantik [10]
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8 0
3 years ago
WILL GIVE BRAINLIEST AND 5 STARS.
Zolol [24]
The right answer is b) copper
8 0
3 years ago
Read 2 more answers
Suppose you have two solid bars, both with square cross-sections of 1 cm2. They are both 24.6 cm long, but one is made of copper
vodka [1.7K]

Explanation:

Expression to calculate thermal resistance for iron (R_{I}) is as follows.

             R_{I} = \frac{L_{I}}{k_{I} \times A_{I}}  

where,   L_{I} = length of the iron bar

             k_{I} = thermal conductivity of iron

             A_{I} = Area of cross-section for the iron bar

Thermal resistance for copper (R_{c}) = \frac{L_{c}}{k_{c} \times A_{c}}[/tex]

where,  L_{c} = length of copper bar

             k_{c} = thermal conductivity of copper

            A_{c} = Area of cross-section for the copper bar

Now, expression for the transfer of heat per unit cell is as follows.

           Q = \frac{(100^{o} - 0^{o}}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}

 Putting the given values into the above formula as follows.

       Q = \frac{(100^{o} - 0^{o})}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}

  = \frac{(100^{o} - 0^{o})}{21 \times 10^{-2} m[\frac{1}{73 \times 10^{-4}m^{2}} + \frac{1}{386 \times 10^{-4}m^{2}}}

           = 2.92 Joule

It is known that heat transfer per unit time is equal to the power conducted through the rod. Hence,

                 P = \frac{Q}{T}

Here, T is 1 second so, power conducted is equal to heat transferred.

So,           P = 2.92 watt

Thus, we can conclude that 2.92 watt power will be conducted through the rod when it reaches steady state.

7 0
3 years ago
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