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3 years ago

14

Which statement describes why energy is released in a nuclear fission reaction based on mass-energy equivalence?

1 answer:

lawyer [7]3 years ago

9 0

Answer:

A is the answer!

Explanation:

Edge 2021

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A baseball has an approximate mass of 0.15 kg. If a bat strikes the baseball with a force of 6 N, what is the acceleration of th

Oksanka [162]

Answer:

a = 40 [m/s²]

Explanation:

These kinds of problems can be solved using Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F = m*a

where:

F = force = 6 [N]

m = mass = 0.15 [kg]

a = acceleration [m/s²]

$a=F/m\\a=6/0.15\\a=40[m/s^{2} ]$

3 0

3 years ago

Newton's First Law of Motion states that an object will remain at rest or in uniform motion in a straight line unless acted upon

atroni [7]

This is sometimes called the Law of Inertia

3 0

3 years ago

When one bumper car is still and the other one hits it (both cars have the same mass) what happens?

statuscvo [17]

the first bump car will move

Explanation:

this is because as long as the other bumper car is coming with force, it mass changes and then , it is able to push the other one

5 0

3 years ago

What is not a example of potiential energy

grin007 [14]

Answer:

when an object is moving

Explanation:

it turns into kinetic energy

3 0

3 years ago

Read 2 more answers

Water is flowing into a factory in a horizontal pipe with a radius of 0.0183 m at ground level. This pipe is then connected to a

timama [110]

Answer:

$0.0168 m^3/s$

Explanation:

We are given that

$r_1=0.0183 m$

$h_1=0$

$r_2=0.0420 m$

$h_2=12.6 m$

Let $P_1=P_2=P$

By using Bernoulli theorem

$P+\frac{1}{2}\rho v^2_1+\rho gh_1=P+\frac{1}{2}\rho v^2_2+\rho gh_2$

$\frac{1}{2}\rho v^2_1+\rho gh_1=\frac{1}{2}\rho v^2_2+\rho gh_2$

$v^2_1+2gh_1=v^2_2+2gh_2$

$A_1v_1=A_2v_2$

$v_1=\frac{A_2v_2}{A_1}$

$(\frac{A_2}{A_1})^2v^2_2+2g\times 0=v^2_2+2\times 9.8\times 12.6$

$(\frac{\pi r^2_2}{\pi r^2_1})^2v^2_2-v^2_2=246.96$

$v^2_2((\frac{r^2_2}{r^2_1})^2-1)=246.96$

$v^2_2=246.96\frac{r^4_1}{r^2_4-r^4_1}$

$v_2=\sqrt{246.96\frac{r^4_1}{r^4_2-r^4_1}}$

$v_2=\sqrt{246.96\times \frac{(0.0183)^4}{(0.042)^4-(0.0183)^4}}$

$v_2=3.038 m/s$

Volume flow rate = $A_2v_2$

Volume flow rate = $\pi r^2_2v_2=\pi (0.042)^2\times 3.038=0.0168 m^3/s$

3 0

3 years ago