how your weight would change with time if you were on a space ship traveling away from,earth toward the moon

A proton (mass m = 1.67 × 10-27 kg) is being accelerated along a straight line at 2.50 × 1012 m/s2 in a machine. If the proton h

Veronika [31]

Answer:

(A) Speed will be $44.18\times 10^4m/sec$

(b) Change in kinetic energy = $1560\times 10^{-19}$

Explanation:

We have given mass of proton $m=1.67\times 10^{-27}kg$

Acceleration of the proton $a=2.50\times 10^{12}m/sec^2$

Initial velocity u = $1.60\times 10^4$ m/sec

Distance traveled by proton s = 3.90 cm = 0.039 m

(a) From third equation of motion we know that

$v^2=u^2+2as$

$v^2=(1.60\times 10^4)^2+2\times 2.5\times 10^{12}\times 0.039$

$v=44.18\times 10^4m/sec$

(b) Initial kinetic energy $KE_I=\frac{1}{2}mv^2=\frac{1}{2}\times 1.67\times 10^{-27}\times (1.6\times 10^4)^2$

Final kinetic energy $KE_F=\frac{1}{2}mv^2=\frac{1}{2}\times 1.67\times 10^{-27}\times (44.18\times 10^4)^2$

So change in kinetic energy $\Delta KE=KE_F-KE_I=\frac{1}{2}\times 1.6\times 10^{-27}\times 10^8\times (44.18^2-1.6^2)=1560\times 10^{-19}J$

6 0

3 years ago

A gas that is exist in outer space is made of ?

DochEvi [55]

I think the awnser is Dark Matter

3 0

3 years ago

Read 2 more answers

Which of the following statements about large scale mineral extraction is true?

Sonbull [250]

Answer: D Minarel extraction cannot be done in ways that does not completely destroy the environment

Explanation: Hope this helps !!

8 0

3 years ago

It is estimated that uranium is relatively common in the earth's crust, occurring in amounts of 4 g / metric ton. A metric ton i

timama [110]

Answer:

The mass of Uranium present in a 1.2mg sample is $4.8 \cdot 10^{-6}\,mg$

Explanation:

The ration between Uranium mass and total sample mass is: $\frac{4g}{1000kg} =\frac{4g}{1000000g}=\frac{1}{250000}$

For a sample of mass 1.2 mg, the amount of uranium is:

$1.2\, mg \cdot \frac{1}{250000}=4.8 \cdot 10^{-6}\,mg$

7 0

3 years ago

The wing of an airplane experiences the forces as depicted in the vector diagram to the right. Using both one and two dimensiona

Vedmedyk [2.9K]

Answer:

A.) 3605.6 N

B.) 33.7 degree

Explanation:

To find the result force acting on the wing of the airplane, we need to resolve the forces into x and y components

Resolving into x component :

Sum of forces = 3500 - 500 = 3000N

Resolving into y component:

Sum of forces = 2000N

Resultant force Fr = sqrt ( Fx^2 + Fy^2)

Fr = sqrt ( 3000^2 + 2000^2 )

Fr = sqrt ( 9000000 + 4000000 )

Fr = sqrt ( 13000000)

Fr = 3605.6 N

Therefore, resultant force acting on the wing is 3605.6 N

The direction of the vector will be:

Tan Ø = Fy / Fx

Substitute Fx and Fy into the formula

Tan Ø = 2000 / 3000

Tan Ø = 0.66666

Ø = tan^-1(0. 66666)

Ø = 33.7 degree.

6 0

3 years ago