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Shkiper50 [21]
4 years ago
7

When many atoms are split in a chain reaction, a large explosion occurs. This is an example of what type of energy conversion? (

2 points)
nuclear to mechanical

thermal to nuclear

chemical to mechanical

nuclear to chemical
Physics
2 answers:
VMariaS [17]4 years ago
6 0
The second option is correct
wariber [46]4 years ago
3 0

The second answer isn't correct, the correct answer is c.

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If an object has two forces that are equal in magnitude, but acting on it in opposite directions, what are some possible things
Oksi-84 [34.3K]

Answer: The body will be in a position of rest

Explanation:

Since both forces are equal and acting in opposite direction on the body. It will make the body to be in a state of rest or equilibrium because the sum of the forces acting in the body will be zero.

5 0
3 years ago
The centripetal force depends on the mass of body, angular velocity and radius of circular path. Find the expression for the cen
vitfil [10]

Answer:

F = m ω² r

Explanation:

Centripetal force is mass times centripetal acceleration:

F = ma

F = m v² / r

In terms of angular velocity:

F = m (ωr)² / r

F = m ω² r

5 0
3 years ago
a chuck wagon with an initial velocity of 4 m/s and a mass of 35 kg gets a push with 350 joules of force. what is the wagon's fi
Kitty [74]

Answer:

the final velocity of the wagon is 6 m/s.

Explanation:

Given;

initial velocity of the wagon, u = 4 m/s

mass of the wagon, m = 35 kg

energy applied to the wagon, E = 350 J

The final velocity of the wagon is calculated as;

E = ¹/₂m(v² - u²)

m(v^2-u^2) = 2E\\\\v^2-u^2 = \frac{2E}{m} \\\\v^2 =  \frac{2E}{m}  + u^2\\\\v = \sqrt{\frac{2E}{m}  + u^2} \\\\v = \sqrt{\frac{2(350)}{35}  + (4)^2}\\\\v = 6 \ m/s

Therefore, the final velocity of the wagon is 6 m/s.

8 0
3 years ago
Objects with masses of 255 kg and a 555 kg are separated by 0.390 m.
eimsori [14]

Answer:

(a). The net gravitational force is 4.20\times10^{-6}\ N

(b). The position is at 0.232 m.

Explanation:

Given that,

Mass of one object M = 255 kg

Mass of another object M'= 555 kg

Separation = 0.390 m

(a). We need to calculate the net gravitational force

Using formula of force

F_{net}=\dfrac{GM'm}{r^2}-\dfrac{GMm}{r^2}

F_{net}=\dfrac{Gm(M'-M)}{r^2}

Put the value into the formula

F_{net}=\dfrac{6.67\times10^{-11}\times32.0(555-255)}{(0.390)^2}

F_{net}=4.20\times10^{-6}\ N

The net gravitational force is 4.20\times10^{-6}\ N

(b). We need to calculate the position

Force from 555 mass = Force from 255 mass

\dfrac{Gm\times555}{x^2}=\dfrac{Gm\times255}{(0.390-x)^2}

555(0.390-x)^2=255x^2

300x^2-0.78x+84.4155=0

x=0.232\ m

The position is at 0.232 m.

Hence, (a). The net gravitational force is 4.20\times10^{-6}\ N

(b). The position is at 0.232 m.

5 0
3 years ago
Read 2 more answers
PLEASE HELP I NEED THIS DONE BY 11:59!<br> After how long do the cars pass each other???
ivann1987 [24]

Answer:

They pass after roughly 5 hours and 35 minutes.

3 0
3 years ago
Read 2 more answers
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