All categories

3 years ago

Uranium-235 undergoes fission, forming krypton-92, barium-141, and 3

Physics

2 answers:

Ahat [919]3 years ago

8 0

Answer:

Explanation:

Some of the mass

GarryVolchara [31]3 years ago

6 0

Answer:

D. Mass was transformed into energy during the process.

You might be interested in

Two ships of equal mass are 110 m apart. What is the acceleration of either ship due to the gravitational attraction of the othe

dusya [7]

Answer:

Acceleration of the ship, $a=2.14\times 10^{-7}\ m/s^2$

Explanation:

It is given that,

Mass of both ships, $m=39000\ metric\ tons=39\times 10^6\ kg$

Distance between two ships, d = 110 m

The gravitational force between two ships is given by :

$F=G\dfrac{m^2}{d^2}$

$F=6.67\times 10^{-11}\ Nm^2/kg^2\times \dfrac{(39\times 10^6\ kg)^2}{(110\ m)^2}$

F = 8.38 N

Let a is the acceleration. Now, using second law of motion as :

$a=\dfrac{F}{m}$

$a=\dfrac{8.38\ N}{39\times 10^6\ kg}$

$a=2.14\times 10^{-7}\ m/s^2$

So, the acceleration of either ship due to the gravitational attraction of the other is $2.14\times 10^{-7}\ m/s^2$ . Hence, this is the required solution.

7 0

3 years ago

A 5.75 mm high firefly sits on the axis of, and 11.3 cm in front of, the thin lens A, whose focal length is 5.77 cm . Behind len

weeeeeb [17]

Answer

given,

focal length of lens A = 5.77 cm

focal length of lens B= 27.9 cm

flies distance from mirror = 11.3 m

now,

Using lens formula

$\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q}$

$\dfrac{1}{5.77} = \dfrac{1}{11.3} + \dfrac{1}{q}$

q =11.79 cm

image of lens A is object of lens B

distance of lens = 59.9 - 11.79 = 48.11

now, Again applying lens formula

$\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q'}$

$\dfrac{1}{27.9} = \dfrac{1}{48.11} + \dfrac{1}{q'}$

q' =66.41 cm

hence, the image distance from the second lens is equal to q' =66.41 cm

6 0

3 years ago

Find the radius of the circle formed by the intersection of a sphere of diameter 26 units and a plane that is 5 units away from

horrorfan [7]

Answer:

12 units

Explanation:

This problem can be solved if we take into account the equation for a sphere

$x^{2}+y^{2}+z^{2}=r^{2}\\x^{2}+y^{2}+z^{2}=(\frac{26}{2})^{2}=13^{2}$

where we took that the radius is 13 units. If we take z=5 and we replace this value in the equation of the sphere we have

$x^{2}+y^{2}+(5)^{5}=13^{2}\\x^{2}+y^{2}=144=(12)^{2}$

where we have taken x2 +y2 because if the equation of a circunference.

In this case the intersection is made when we take z=5, for this value the sphere and the plane coincides in values.

Hence, the radius is 12 units

I hope this is useful for you

regards

8 0

3 years ago

Hai tiếp điểm hình trụ bằng đồng có đầu hình cầu bán kính R = 70mm, lực ép tiếp điểm F = 100N, Mô đun đàn hồi của đồng là 11.8 *

lisabon 2012 [21]

Answer:

5dhydydyeze6ese6z7zuezuze6ezyezyezzeze

6 0

3 years ago

Read 2 more answers

3.00 m^3 of water is at 20.0°C.

romanna [79]

Answer:

$\triangle V = 0.02484m^3$

Explanation:

Given

$V_1 = 3.00m^3$ --- initial volume

$T_1 = 20.0^oC$ --- initial temperature

$T_2 = 60.0^oC$ --- final temperature

$\gamma = 207*10^{-6$ --- coefficient of thermal expansion:

Required

The change in volume

To do this, we make use of cubic expansivity formula

$\triangle V = \gamma * V_2 * (T_2 - T_1)$

So, we have:

$\triangle V = 207 * 10^{-6} * 3.00 * (60.0 - 20.0)$

$\triangle V = 207 * 10^{-6} * 3.00 * 40.0$

$\triangle V = 0.02484m^3$

The volume will expand by $0.02484m^3$

7 0

2 years ago