All categories

3 years ago

What is released through nuclear fusion in stars? energy gas mass pressure

Physics

2 answers:

trasher [3.6K]3 years ago

7 0

Answer:

Energy is released through nuclear fusion

ohaa [14]3 years ago

6 0

Answer:

A) Energy

Explanation:

You might be interested in

A scientist is conducting research to determine how much pressure a dam can withstand. Why are good observations important to hi

Natalija [7]

The things that a scientist should consider while observing the force is the environmental conditions,the force that is expected to act on the dam, the means to contain that force, and compare different types of designs in accordance with the location of the dam

3 0

3 years ago

The Atoms Family

ioda

What question are you asking?

4 0

3 years ago

Convert 4 radians to degrees A. 242.6° B. 9.6° C. 229.2° D. 6.8°

muminat

The central angle of a circle is 360° or 2π radians.

Therefore
1 radian = (360 degrees)/(2π radians) = 180/π degrees/radian.
4 radians = (4 radians)*(180/π degrees/radian) = 229.18 degrees.

Answer: C. 229.2°

8 0

3 years ago

Read 2 more answers

We start with 5.00 moles of an ideal monatomic gas with an initial temperature of 128 ∘C. The gas expands and, in the process, a

o-na [289]

Answer:

The final temperature of the gas is <em>114.53°C</em>.

Explanation:

Firstly, we calculate the change in internal energy, ΔU from the first law of thermodynamics:

ΔU=Q - W

ΔU = 1180 J - 2020 J = -840 J

Secondly, from the ideal gas law, we calculate the final temperature of the gas, using the change in internal energy:

$ΔU=\frac{3}{2} nRΔT$

$ΔU=\frac{3}{2} nR(T_{2} -T_{1} )$

Then we make the final temperature, T₂, subject of the formula:

$T_{2} =\frac{2ΔU}{3nR} +T_{1}$

$T_{2} =\frac{2(-840J)}{(3)(5)(8.314J/mol.K)} +128 deg.C$

$T_{2} =114.53 deg.C$

Therefore the final temperature of the gas, T₂, is 114.53°C.

7 0

3 years ago

A football is kicked from the ground with a velocity of 38m/s at an angle of 40 degrees and eventually lands at the same height.

Anastasy [175]

Initially, the velocity vector is $\langle 38cos(40^{\circ}),38sin(40^{\circ}) \rangle=\langle 29.110, 24.426 \rangle$ . At the same height, the x-value of the vector will be the same, and the y-value will be opposite (assuming no air resistance). Assuming perfect reflection off the ground, the velocity vector is the same. After 0.2 seconds at 9.8 seconds, the y-value has decreased by $4.9(0.2)^2$ , so the velocity is $\langle 29.110, 24.426-0.196 \rangle = \langle 29.110, 24.23 \rangle$ .

Converting back to direction and magnitude, we get $\langle r,\theta \rangle=\langle \sqrt{29.11^2+24.23^2},tan^{-1}(\frac{29.11}{24.23}) \rangle = \langle 37.87,50.2^{\circ}\rangle$

4 0

3 years ago