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3 years ago

The Sun produces 9.5 x 10^38 He atoms every second. Calculate the amount of energy produced by

Physics

1 answer:

Rama09 [41]3 years ago

6 0

Answer:

You need to look up the figure for the energy released in the formation of one He atom, and then multiply that by the number of He atoms formed each second, and the result will be the total energy release that you seek.

Explanation:

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A force of 20 N acts on a 4 kg cart for 10 s . How far will it go (using the second law) ?

Elden [556K]

This took me a short while to figure out, but I am still not entirely sure if this is correct, this is just from my basic understanding of Newtons Second Law of Motion.

You have a 4kg cart with a force of 20N acting on it.

The formula for working out the acceleration is.
a=Fnet÷mass

Substitute in the information.
a=20N÷4kg

Now you solve it to give you.
a=5m/s

So now what you should be able to do is figure out that after 10 seconds the cart travelling at 5m/s would have travelled 10 metres.

This is achieved by finding out how many 5's go into 10 which is 2.
So you do 5×2 which equals 10.

The 4kg cart has travelled 10 meters in 10 seconds with a force of 20N acting upon it.

I hope that this has helped you.

8 0

3 years ago

If a wave has a frequency of 5 Hz, what is its period?

tino4ka555 [31]

Answer: period = 0.2 s

Explanation:

5Hz=0.2s

6 0

4 years ago

An open pipe is 1.42 m long

lora16 [44]

Answer:

the fundamental frequency produced by the open pipe is 120.78 Hz

Explanation:

Given;

length of the open pipe, L = 1.42 m

speed of sound in air, v = 343 m/s

The length of the open pipe for the fundamental frequency is equivalent to half of wavelength;

$L = \frac{\lambda}{2} \\\\\lambda = 2L$

The fundamental frequency produced by the open pipe is calculated as;

$f_o = \frac{v}{\lambda} \\\\f_o = \frac{v}{2L} \\\\f_o = \frac{343}{2 \times 1.42} \\\\f_o = 120.78 \ Hz$

Therefore, the fundamental frequency produced by the open pipe is 120.78 Hz

6 0

3 years ago

An object has a mass of 50.0 g and a volume of 10.5 cm3. What is the object's density?

seraphim [82]

Volume=mass/density

volume=455.6/19.3

volume=23.6 mL

4 0

3 years ago

The length of a certain wire is kept same while its radius is doubled. what is the change in the resistance of this wire?

kaheart [24]

The resistance of a wire is given by
$R= \frac{\rho L}{A}$
where $\rho$ is the resistivity of the material, L the length of the wire and A its cross-sectional area.

In the problem, $\rho$ and L remain the same, while A changes because the radius changes. The area is given by:
$A=\pi r^2$
This means that if we double the radius (2r), the area becomes
$A_{new}= \pi (2r)^2 = 4 \pi r^2 =4A$
And therefore, the new value of the resistance is
$R_{new} = \frac{\rho L}{4 A}= \frac{1}{4}R$

So, when the radius is doubled, the resistance becomes $\frac{1}{4}$ of its original value.

6 0

4 years ago