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3 years ago

Use the given data to calculate the total mass of hydrogen available for fusion over the lifetime of the sun.

1 answer:

8 0

Total mass of the Sun = 2x10^30kg

<span>So 76% of that = (2x10^30kg)*(0.76) = 1.52x10^30kg ----> total amount of Hydrogen i</span><span>f only 12% of that is used for fusion, then (1.52x10^30kg)*(0.12) = 1.82x10^9kg</span>

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Inside a freely falling runaway elevator, your

nasty-shy [4]

The answer is apparent weight is zero.

You are still accelerating downwards at 9.8m/s^2 (if you are on Earth).

You still are being affected by the Earth's gravity.

Not all because of the previous two statements.

Not none because apparent weight is zero as you are falling.

7 0

3 years ago

Read 2 more answers

Cheetahs, the fastest of the great cats, can reach 50.0 miles/hour in 2.22 s starting from rest. Assuming that they have constan

elena55 [62]

Answer:

$10.07m/s^2$

Explanation:

Information we have:

velocities:

initial velocity: $v_{i}=0mph$ (starts from rest)

final velocity: $v_{f}=50mph$

time: $t=2.22s$

Since we need the answer in $m/s^2$ , we nees to convert the speed to meters per second:

$v_{f}=\frac{50miles}{1hour}(\frac{1hour}{3,600s} )(\frac{1609.34meters}{1mile} ) \\\\v_{f}=\frac{50*1609.34}{3600} m/s\\\\v_{f}=22.35m/s$

We find the acceleration with the following formula:

$a=\frac{v_{f}-v_{i}}{t}$

substituting the known values:

$a=\frac{22.35m/s-0m/s}{2.22s}\\ \\a=10.07m/s^2$

the acceleration is 10.07 $m/s^2$

8 0

2 years ago

Read 2 more answers

What is the correct order of the structures of the motor neuron?

Ulleksa [173]

Dendrites, the cell body, axon, terminal branches of the axon

7 0

3 years ago

Which statement about polymers is true?

Mamont248 [21]

Where are the statements? You forgot to attach them lol

6 0

3 years ago

What is the acceleration of a proton moving with a speed of 6.5 m/s at right angles to a magnetic field of 1.5 T?

Brilliant_brown [7]

Answer:

The acceleration of the proton is 9.353 x 10⁸ m/s²

Explanation:

Given;

speed of the proton, u = 6.5 m/s

magnetic field strength, B = 1.5 T

The force of the proton is given by;

F = ma = qvB(sin90°)

ma = qvB

where;

m is mass of the proton, = 1.67 x 10⁻²⁷ kg

charge of the proton, q = 1.602 x 10⁻¹⁹ C

The acceleration of the proton is given by;

$a = \frac{qvB}{m}\\\\a = \frac{(1.602*10^{-19})(6.5)(1.5)}{1.67*10^{-27}}\\\\a = 9.353*10^8 \ m/s^2$

Therefore, the acceleration of the proton is 9.353 x 10⁸ m/s²

4 0

3 years ago