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2 years ago

9

If the chain reaction is not controlled, what can happen?

1 answer:

Pie2 years ago

6 0

Answer: An uncontrolled chain reaction can become highly explosive. That is because these reactions can release a very high amount of energy at a time. For example, the radioactive isotope Uranium-235 can undergo nuclear fission be releasing neutron slowly.

Explanation: Yw and pls mark me as brainiest

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What is the minimum speed necessary to make it safely??

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Answer:

10-15mph

Explanation:

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2 years ago

Isabela is an astronomer studying stars A and B. Star A is farther away from Isabela and yet when Isabela observes the stars, th

kakasveta [241]

Answer: star A is more luminous

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3 years ago

A house is wired so that one electrical source comes to a room but many outlets and lights work from that source . when one ligh

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3 years ago

The figure shows a 100-kg block being released from rest from a height of 1.0 m. It then takes it 0.90 s to reach the floor. Wha

Anna007 [38]

Answer:

The mass of the another block is 60 kg.

Explanation:

Given that,

Mass of block M= 100 kg

Height = 1.0 m

Time = 0.90 s

Let the mass of the other block is m.

We need to calculate the acceleration of each block

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Put the value into the formula

$1.0=0+\dfrac{1}{2}\times a\times(0.90)^2$

$a=\dfrac{2\times1.0}{(0.90)^2}$

$a=2.46\ m/s^2$

We need to calculate the mass of the other block

Using newton's second law

The net force of the block M

$Ma=Mg-T$

$T=Mg-Ma$ ....(I)

The net force of the block m

$ma=T-mg$

Put the value of T from equation (I)

$ma=Mg-Ma-mg$

$m(a+g)=M(g-a)$

$m=\dfrac{M(g-a)}{(a+g)}$

Put the value into the formula

$m=\dfrac{100(9.8-2.46)}{2.46+9.8}$

$m=59.8\ \approx60\ kg$

Hence, The mass of the another block is 60 kg.

8 0

3 years ago

A string is wrapped around a pulley with a radius of 2.0 cm. The pulley is initially at rest. A constant force of 50 N is applie

Ray Of Light [21]

Answer:

0.20kg-m^2

Explanation:

Let the linear velocity of the rope(=of pulley) is v m/s

Using kinematic equation

=> v = u + at

=>v = 0 + 4.9a

=>v = 4.9a ------------ eq1

By v^2 = u^2 + 2as

=>v^2 = 0 + 2 x v/4.9 x 1.2

=>4.9v^2 - 2.4v = 0

=>v(4.9v - 2.4) = 0

=>v = 2.4/4.9 = 0.49 m/s

Thus by v = r x omega

=>omega = v/r = 0.49/0.02 = 24.49 rad/sec

BY W = F x s = 50 x 1.2 = 60 J

=>KE(rotational) = W = 1/2 x I x omega^2

=>60 = 1/2 x I x (24.49)^2

=>I = 0.20 kg-m^2

5 0

3 years ago