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oksian1 [2.3K]
3 years ago
8

During nuclear fission, great amounts of energy are produced from

Physics
2 answers:
stiks02 [169]3 years ago
6 0
D.very small amounts of mass.
True [87]3 years ago
6 0

During nuclear fission, great amounts of energy are produced from very small amounts of mass. <em>(d)</em>

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Nuclear energy is released during: fission. radioactive decay. man-induced splitting of atoms. Match the basic components of a nuclear reactor with their descriptions. 1. slows down neutrons -> moderator. 2. absorb emitted neutrons -> control rods.
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Compute the final velocity of a ski jumper just as he leaves the ramp if he has a mass of 60 kg and the tower is 70 meters high?
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According to galilo in the absence of air resistance fraction then the body mass 60kg and 40kg will reach the earth surface at the same time....
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Which sphere of Earth includes the continental ice sheets and glaciers? Biosphere Cryosphere Geosphere Hydrosphere
REY [17]

Answer:

C. Cryosphere

Explanation:

The definition for Cryo is:

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If you ever need to think of "Cryo" just think of Cryo Freezing.

4 0
3 years ago
A cannon is fired from a castle wall at some unknown height above the ground. The cannonball leaves the cannon with speed 30.0m/
Sauron [17]

Answer:

Part a)

t = 3.85 s

Part b)

h = 72.67 m

Part C)

v_x = 25.98 m/s

v_y = 0

Part d)

In horizontal direction velocity will remain constant

v_x = 30 cos30 = 25.98 m/s

in vertical direction we have

v_y = -22.77 m/s

Explanation:

Part a)

Horizontal speed of the cannon

v = 30.0 m/s

angle of projection

\theta = 30^o

now we have

horizontal speed = v_x = vcos30 = 30 cos30 =25.98 m/s

vertical speed = v_y = vsin30 = 30 sin30 = 15 m/s

now the time taken by it to cover the distance 100 m from the wall

x = v_x t

100 = 25.98 t

t = 3.85 s

Part b)

Since it hits the ground in the same time

so the height of the castle is given as

h = \frac{1}{2}gt^2

h = \frac{1}{2}(9.81)(3.85^2)

h = 72.67 m

Part C)

At highest point of the projection

the vertical component of the velocity will become zero

so we will have

v_x = 25.98 m/s

v_y = 0

Part d)

In horizontal direction velocity will remain constant

so we have

v_x = 30 cos30 = 25.98 m/s

in vertical direction we have

v_y = v_i + at

v_y = 15 - 9.81(3.85)

v_y = -22.77 m/s

Part e)

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3 years ago
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What is the minimum tangential velocity a space station would need to simulate earth gravity if the radius is 50 meters ?
Aneli [31]

Explanation:

angular velocity is given by

w =  \sqrt{ \frac{g}{r} }

w =  \sqrt{ \frac{9.8}{25} }

w = 0.626

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= 15.65 m/s

5 0
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