All categories

2 years ago

A spaceship of 3.20 x 10*7kg travels around another planet, of

Physics

1 answer:

vfiekz [6]2 years ago

4 0

Answer:

answer d

Explanation:

It would be asnwer d because you need. to multiople both numbers 3.20x10*7 and 6.34x10^25 then multiple.

You might be interested in

NEED HELP ASAP

Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{1} V_{o} + m_{2} U_{o}$ (2)

$p_{f}=(m_{1} + m_{2}) V_{f}$ (3)

$m_{1}=110 kg$ is the mass of the first football player

$V{o}=-7 m/s$ is the velocity of the first football player (to the south)

$m_{2}=75 kg$ is the mass of the second football player

$U_{o}=4 m/s$ is the velocity of the second football player (to the north)

$V_{f}$ is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

$m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}}$ (5)

$V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg}$ (6)

$V_{f}=-2.54 m/s$ (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy $\Delta K$ is defined as:

$\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2}$ (8)

Simplifying:

$\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2})$ (9)

$\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2})$ (10)

Finally:

$\Delta K=-2351.25 J$ (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

6 0

3 years ago

The force of gravity on a 1 kg object on the Earth's surface is approximately 9.8 N. For the same object in low-earth orbit arou

mojhsa [17]

Answer:

g = 8.61 m/s²

Explanation:

distance of the International Space Station form earth is 200 Km

mass of the object = 1 Kg

acceleration due to gravity on earth = 9.8 m/s²

mass of earth = 5.972 x 10²⁴ Kg

acceleration due to gravity = ?

r = 6400 + 200 = 6800 Km = 6.8 x 10⁶ n

using formula

$g = \dfrac{GM}{r^2}$

$g = \dfrac{6.67\times 10^{-11}\times 5.972\times 10^24}{(6.8\times 10^6)^2}$

g = 8.61 m/s²

3 0

3 years ago

A skydiver jumps out of a hovering helicopter. A few seconds later, another diver jumps out, so they both fall along the same ve

Sergio039 [100]

Answer:

distance difference would a) increase

speed difference would f) stay the same

Explanation:

Let t be the time the 2nd skydiver takes to travel, since the first skydiver jumped first, his time would be t + Δt where Δt represent the duration between the the first skydiver and the 2nd one. Remember that as t progress (increases), Δt remain constant.

Their equations of motion for distance and velocities are

$s_2 = gt^2/2$

$s_1 = g(t + \Delta t)^2/2$

$v_2 = gt$

$v_1 = g(t + \Delta t)$

Their difference in distance are therefore:

$\Delta s = s_1 - s_2 = g(t + \Delta t)^2/2 - gt^2/2$

$\Delta s = g/2((t + \Delta t)^2 - t^2)$

$\Delta s = g/2(t + \Delta t - t)(t + \Delta t + t)$ (As $A^2 - B^2 = (A-B)(A+B)$

$\Delta s = g\Delta t/2(2t + \Delta t)$

So as time progress t increases, Δs would also increases, their distance becomes wider with time.

Similarly for their velocity difference

$\Delta v = v_1 - v_2 = g(t + \Delta t) - gt$

$\Delta v = gt + g\Delta t - gt = g\Delta t$

Since g and Δt both are constant, Δv would also remain constant, their difference in velocity remain the same.

This of this in this way: only the DIFFERENCE in speed stay the same, their own individual speed increases at same rate (due to same acceleration g). But the first skydiver is already at a faster speed (because he jumped first) when the 2nd one jumps. The 1st one would travel more distance compare to the 2nd one in a unit of time.

8 0

3 years ago

Read 2 more answers

A galvanometer is used as a

agasfer [191]

Answer:

detecting and indicating an electric current

4 0

3 years ago

After which action would the concentration of a solution remain constant?(1 point)

Alika [10]

Answer:

C. Evaporating water from the container.

Explanation:

The concentration of solution changes when solvent or solute are added/removed from a solution.

5 0

3 years ago