Uranium-235 undergoes fission, forming krypton-92, barium-141, and 3neutrons. The mass of the uranium-235 is greater than the total mass of the
products. Which statement explains this difference in mass?
A. Some of the mass was transformed into neutrons during the
process.
O B. Mass was destroyed and disappeared during the process.
C. Some of the mass was transformed into gases during the
process.
D. Mass was transformed into energy during the process.

Final answer:

The force of gravity that the space shuttle experiences is 9.8 x 10^5 Newtons.

Explanation:

To calculate the force of gravity that the space shuttle experiences, we can use the equation F = mg, where F represents the force of gravity, m is the mass of the object, and g is the acceleration due to gravity (approximately 9.8 m/s² on Earth). In this case, the mass of the space shuttle is given as 1.0 x 10^5 kg. However, we need to convert the altitude of the shuttle into meters, so 200.0 km becomes 200,000 meters.

Now we can calculate the force of gravity:

F = (1.0 x 10^5 kg)(9.8 m/s²)

F = 9.8 x 10^5 N

Therefore, the space shuttle experiences a force of gravity of 9.8 x 10^5 Newtons.

Learn more about force of gravity here:

brainly.com/question/12753714

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(664.2 km) · (1,000 m/km) · (100 cm/m) =

(664.2 · 1,000 · 100) (km·m·cm/km·m) =

66,420,000 cm

For metric conversion, you can remember this acronym for help:

King Henry died unusually drinking chocolate milk. Which stand for:

Kilo - unit * 1000

Hecto - unit * 100

Deca - unit * 10

Unit - unit * 1

Deci - unit *

Centi - unit *

Milli - unit *

Kilometers and centimeters are five places apart apart, so you move the decimal point in 664.2 to the right five times, which means 664.2 km = 66420000 cm.

To avoid confusion on which direction to move the decimal point, imagine two shapes on each end of a scale. On each end, there is one large shape and one small shape. There has to be one of each on either side for it to balance. For this problem, a kilometer is a larger unit than a centimeter, so this means that the blank space needs to have a number greater than 664.2, or else the scale won't balance. Hope this helped.

Answer:

61.3 cm

Explanation:

Radial acceleration of the object in circular motion is given by formula

Given:

Plugging in the values in the formula

so length of his arm is 61.3 cm

The height of the building is 60 m.

calculation of building height:

The velocity of the ball should be provided by

v = u + gt

here,

u is the initial velocity of the ball = 0

v = 0 + 9.8 x 3.5

v = 34.3 m/s

Now

When the ball hits the ground, energy is conserved;

mgh = ¹/₂mv²

gh = ¹/₂v²

h = (0.5 v²) / g

h = (0.5 x 34.3²) / (9.8)

h = 60.025 m

h = 60 m

Learn more about friction here: brainly.com/question/14455351

Answer:

The height of the building is 60 m.

Explanation:

Given;

mass of the mass of the ball, m = 3 kg

time of motion, t = 3.5 s

The velocity of the ball is given by;

v = u + gt

where;

u is the initial velocity of the ball = 0

v = 0 + 9.8 x 3.5

v = 34.3 m/s

When the ball hits the ground, energy is conserved;

mgh = ¹/₂mv²

gh = ¹/₂v²

h = (0.5 v²) / g

h = (0.5 x 34.3²) / (9.8)

h = 60.025 m

h = 60 m

Therefore, the height of the building is 60 m.

The angular speed should be 17.18  rad / s

Calculation of the angular speed:

Since

moment of inertia of the disc I = 1/2 m r²

= .5 x 8 x .25²

= .25 kg m²

Now the work done by force should be converted into the rotational kinetic energy

F x d = 1/2 I ω²

here,

F is the force applied,

d is displacement,

I is moment of inertia of disc

and ω is angular velocity of disc

So,

41 x .9 = 1/2 x .25 ω²

ω² = .25

ω = 17.18  rad / s

Learn more about speed here: brainly.com/question/18742396

Answer:

Explanation:

moment of inertia of the disc I = 1/2 m r²

= .5 x 8 x .25²

= .25 kg m²

The wok done by force will be converted into rotational kinetic energy

F x d = 1/2 I ω²

F is force applied , d is displacement , I is moment of inertia of disc and ω

is angular velocity of disc

41 x .9 = 1/2 x .25 ω²

ω² = .25

ω = 17.18  rad / s

Answer:

1. 1 s
2. 19.6 m
3. 2 s
4. 0.8 m/s^2
5. 28 m/s
6. 79 m/s
7. 0.37 s
8. 26 m/s
9. 242 m/s
10. 19,930 m

Explanation:

In physics, many of the relationships between speed, distance, and acceleration are tied up in the equations for potential and kinetic energy. For an object of mass M* at height h in a gravity field with acceleration g, the potential energy is

  PE = Mgh

At velocity v, the kinetic energy of the object is ...

  KE = 1/2Mv^2

When an object is dropped or launched from rest, the height and velocity are related by the fact that kinetic energy gets translated to potential energy, or vice versa. This gives rise to ...

  PE = KE

  Mgh = (1/2)Mv^2

The mass (M) can be factored out of this, so we have ...

  2gh = v^2

This can be solved for height:

  h = v^2/(2g) . . . . [eq1]

or for velocity:

  v = √(2gh) . . . . [eq2]

__

When acceleration is constant, as assumed here, the velocity changes linearly (to/from 0). So, over the time of travel, the average velocity is half the final velocity. That is,

  t = 2h/v

Depending on whether you start with h or with v, this resolves to two more equations:

  t = 2(v^2/(2g))/v = v/g . . . . [eq3]

  t = 2h/(√(2gh)) = √(4h^2/(2gh)) = √(2h/g) . . . . [eq4]

The last of these can be rearranged to give distance as a function of time:

  h = gt^2/2 . . . . [eq5]

or acceleration as a function of time and distance:

  g = 2h/t^2 . . . . [eq6]

__

These 6 equations can be used to solve the problems posed. Just "plug and chug." For problems in Earth's gravity, we use g=9.8 m/s^2. (You may want to keep these equations handy. Be aware of the assumptions they make.)

_____

* M is used for mass in these equations so as not to get confused with m, which is used for meters.

_____

1) Use [eq4]: t = √(2·6 m/(9.8 m/s^2)) ≈ 1.107 s ≈ 1 s

__

2) Use [eq5]: h = (9.8 m/s^2)(2 s)^2/2 = 19.6 m

__

3) Use [eq4]: t = √(25 m/(4.9 m/s^2)) ≈ 2.259 s ≈ 2 s

__

4) Use [eq6]: g = 2(10 m)/(5 s)^2 = 0.8 m/s^2

__

5) Use [eq2]: v = √(2·9.8 m/s^2·40 m) = 28 m/s

__

6) Use [eq2]: v = √(2·9.8 m/s^2·321 m) ≈ 79.32 m/s ≈ 79 m/s

__

7) Using equation [eq3], we will find the time until Tina reaches her maximum height. Her actual off-the-ground total time is double this value. Using [eq3]: t = v/g = (1.8 m/s)/(9.8 m/s^2) = 9/49 s. Tina is in the air for double this time:

  2(9/49 s) ≈ 0.37 s

__

8) Use [eq2]: v = √(2·9.8 m/s^2·33.5 m) ≈ 25.624 m/s ≈ 26 m/s

__

9) Use [eq2]: v = √(2·9.8·3000) m/s ≈ 242.49 m/s ≈ 242 m/s

(Note: the terminal velocity in air is a lot lower than this for an object like a house.)

__

10) Use [eq1]: h = (625 m/s)^2/(2·9.8 m/s^2) ≈ 19,930 m

_____

Additional comment

Since all these questions make use of the same equation development, I have elected to answer them. Your questions are more likely to be answered if you restrict your posts to 3 or fewer questions each.