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Jet001 [13]
3 years ago
7

A spaceship moves at a speed of 0.95 c away from the Earth. It shoots a star wars torpedo toward the Earthat a speed of 0.90 c r

elative to the ship. What is the velocity of the torpedo relative to the Earth? (The direction in which the spaceship move
Physics
1 answer:
wariber [46]3 years ago
5 0

Answer:

0.345 c

Explanation:

v_{se} = velocity of spaceship relative to earth = 0.95 c

v_{ts} = velocity of torpedo relative to spaceship = - 0.90 c

v_{te} = velocity of torpedo relative to earth

Velocity of torpedo relative to earth is given as

v_{te}=\frac{v_{se} + v_{ts}}{1 + \frac{v_{se}v_{ts}}{c^{2}}}

v_{te}=\frac{0.95c + (- 0.90 c)}{1 + \frac{(0.95 c)(- 0.90 c))}{c^{2}}}

v_{te} = 0.345 c

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If a melon has a a mass of 1 kg, how much does the melon weigh?
My name is Ann [436]

Explanation:

1kg = approximately 10N

6 0
2 years ago
A comet of mass 1.20 × 10¹⁰kg moves in an elliptical orbit around the Sun. Its distance from the Sun ranges between 0.500 AU and
irina [24]

The eccentricity of its orbit is $$U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

<h3>What is mass?</h3>
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The length of the semi-major axis is calculated as follows:

where, $G=6.67 \times 10^{-1} \mathrm{~m}^3 / \mathrm{kgs}$

$M=1.99 \times 10^{30} \mathrm{~kg}=$mass of sur

$m=1.20 \times 10^{10} \mathrm{~kg}$ - a mass of the comet

$$\begin{aligned}\therefore \quad \text { At aphelion, } r &=50 \times U \\&=50 \times 1.496 \times 10^{11} \mathrm{~m} . \\U=-\frac{6.67 \times 10^{-11} \times \mathrm{m} 1.99 \times 10^{30} \times 1.20 \times 10^{10}}{50 \times 1.496 \times 10^{11}} \\U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

$$U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

To learn more about mass, refer to:

brainly.com/question/3187640

#SPJ4

4 0
1 year ago
A 2.13-kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 5.00 N
ANTONII [103]

Answer:

17.54N in -x direction.

Explanation:

Amplitude (A) = 3.54m

Force constant (k) = 5N/m

Mass (m) = 2.13kg

Angular frequency ω = √(k/m)

ω = √(5/2.13)

ω = 1.53 rad/s

The force acting on the object F(t) = ?

F(t) = -mAω²cos(ωt)

F(t) = -2.13 * 3.54 * (1.53)² * cos (1.53 * 3.50)

F(t) = -17.65 * cos (5.355)

F(t) = -17.57N

The force is 17.57 in -x direction

5 0
3 years ago
A hiker walked a hiking trail according to the chart below. What is a possible explanation of the hiker's movement?
joja [24]
<span>a.The hiker had an easy, level trail from 11:00-12:00 and was able to travel the fastest during that time period.---> may be because this was indeed fastest  stage


b.The hiker got tired and walked the slowest from 1:00-2:00.---> no, because this was not the slowest stage


c.The hiker stopped for lunch from 11:00-12:00 and that slowed him down.---> no because this was the fastest stage


d.The hiker ended up in the same place that he started.---> no, because the hiker walked more toward east than toward west and more toward south than toward north.

Answer: option a)
</span>
6 0
3 years ago
A train pulls away from a station with a constant acceleration of 0.42 m/s2. A passenger arrives at a point next to the track 6.
Rina8888 [55]

Answer:

2.69 m/s

Explanation:

Hi!

First lets find the position of the train as a function of time as seen by the passenger when he arrives to the train station. For this state, the train is at a position x0 given by:

x0 = (1/2)(0.42m/s^2)*(6.4s)^2 = 8.6016 m

So, the position as a function of time is:

xT(t)=(1/2)(0.42m/s^2)t^2 + x0 = (1/2)(0.42m/s^2)t^2 + 8.6016 m

Now, if the passanger is moving at a constant velocity of V, his position as a fucntion of time is given by:

xP(t)=V*t

In order for the passenger to catch the train

xP(t)=xT(t)

(1/2)(0.42m/s^2)t^2 + 8.6016 m = V*t

To solve this equation for t we make use of the quadratic formula, which has real solutions whenever its determinat is grater than zero:

0≤ b^2-4*a*c = V^2 - 4 * ((1/2)(0.42m/s^2)) * 8.6016 m =V^2 - 7.22534(m/s)^2

This equation give us the minimum velocity the passenger must have in order to catch the train:

V^2 - 7.22534(m/s)^2 = 0

V^2 = 7.22534(m/s)^2

V = 2.6879 m/s

4 0
3 years ago
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