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zloy xaker [14]

2 years ago

10

List the names of two tectonic plates that are NOT part of the ring of fire

1 answer:

Ann [662]2 years ago

5 0

Nolur acil lütfen yalvarırım yalvarırım

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Explain how the global sea surface temperature is affected by solar activity.

Dahasolnce [82]

The higher the solar activity, and the hotter it is, the more the oceans heat up and water evaporates into the atmosphere. High amounts of water vapor causes things like hurricanes. However, oceans have a high absorption rate of heat so they can absorb a lot of it without heating up the atmosphere, which is pretty good or else the Earth would overheat.

8 0

3 years ago

Rebecca heated 50mL of water from 0 degrees Celsius to 60 degrees Celsius. How much energy did she use to heat the water? Rememb

kramer

Answer:

12552 J or 3000 calories

Explanation:

Q = m × c × ∆T

Where;

Q = amount of heat energy (J)

m = mass of water (g)

c = specific heat capacity (4.184 J/g°C)

∆T = change in temperature

For 50mL of water, there are 50g, hence, m = 50g, c = 4.184 J/g°C, initial temperature = 0°C, final temperature = 60°C.

Q = m × c × ∆T

Q = 50 × 4.184 × (60 - 0)

Q = 209.2 × 60

Q = 12552 J

Hence, the amount of heat energy used to heat the water is 12552 J or 3000 calories

3 0

3 years ago

What would a velocity of a rock dropped from a cliff be after falling for 6 seconds?

Ivenika [448]

Answer:

A. 58.8m/s

Explanation:

The acceleration due to gravity is 9.8 m/s², so the velocity after 6 seconds is ...

v = at

v = (9.8 m/s²)(6 s) = 58.8 m/s

5 0

3 years ago

What are 2 things that legal immigrants can NOT do.

Alexandra [31]

One thing they can’t do is go somewhere out of their country

7 0

3 years ago

A thin rod of length 0.83 m and mass 110 g is suspended freely from one end. It is pulled to one side and then allowed to swing

VARVARA [1.3K]

Explanation:

(a) The given data is as follows.

Length of the rod, L = 0.83 m

Mass of the rod, m = 110 g = 0.11 (as 1 kg = 1000 g)

At the lowest point, angular speed of the rod ( $\omega$ ) = 5.71 rad/s

First, we will calculate the rotational inertia of the rod about an axis passing through its fixed end is as follows.

I = $I_{CM} + mh^{2}$

= $\frac{1}{12}mL^{2} + m(\frac{L}{2})$

= $\frac{1}{12} \times 0.11 \times (0.83)^{2} + 0.11 \times \frac{0.83}{2}$

= 0.00631 + 0.415

= 0.42131 $kg m^{2}$

Therefore, kinetic energy of the rod at its lowest point will be calculated as follows.

K = $\frac{1}{2}I \omega^{2}$

= $\frac{1}{2} \times 0.42131 kg m^{2} \times (5.71)^{2}$

= 6.86 J

Hence, kinetic energy of the rod at its lowest point is 6.86 J.

(b) According to the conservation of total mechanical energy of the rod, we have

$K_{i} + U_{i} = K_{f} + U_{f}$

$K_{i} = U_{f} - U_{i}$

or, mgh = K = 6.86 J

Therefore, h = $\frac{6.86}{mg}$

= $\frac{0.63}{0.11 \times 9.8}$

= 0.584 m

Hence, the center of mass rises 0.584 m far above that position.

6 0

3 years ago

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