1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
WITCHER [35]
3 years ago
14

A ship is flying away from Earth at 0.9c (where c is the speed of light). A missile is fired that moves toward the Earth at a sp

eed of 0.5c relative to the ship. How fast does the missile move relative to the Earth
Physics
1 answer:
Sati [7]3 years ago
5 0

Answer:

the required speed with which the missile move relative to the Earth is -0.727c

Explanation:

Given the data in the question;

relative velocity relation;

u' = u-v / 1 - \frac{uv}{c^2}

so let V_B represent the velocity as seen by an external reference frame; u=V_B

and let V_A represent the speed of the secondary reference frame; v=V_A

hence, u' is the speed of B as seen by A

so

u' = V_B-V_A / 1 - \frac{V_BV_A}{c^2}

now, given that; V_A = 0.9c  and V_B  = 0.5c

we substitute

u' = ( 0.5c - 0.9c ) / 1 - \frac{(0.5c)(0.9c)}{c^2}

u' = ( 0.5c - 0.9c ) / 1 - \frac{c^2(0.5)(0.9)}{c^2}

u' = ( 0.5c - 0.9c ) / 1 - (0.5 × 0.9)

u' = ( -0.4c ) / 1 - 0.45

u' = -0.4c / 0.55

u' = -0.727c

Therefore, the required speed with which the missile move relative to the Earth is -0.727c

You might be interested in
A potentialiterence of 12 volts is induced across a straight wire 0.20 meters long as it is moved at a constant speed of 3.0 met
shtirl [24]

Strength of the magnetic field: 20 T

Explanation:

For a conductive wire moving perpendicular to a magnetic field, the electromotive force (voltage) induced in the wire due to electromagnetic induction is given by

\epsilon=BvL

where

B is the strength of the magnetic field

v is the speed of the wire

L is the length of the wire

For the wire in this problem, we have:

\epsilon=12 V (induced emf)

L = 0.20 m (length of the wire)

v = 3.0 m/s (speed)

Solving  for B, we find the strength of the magnetic field:

B=\frac{\epsilon}{vL}=\frac{12}{(0.20)(3.0)}=20 T

Learn more about magnetic fields:

brainly.com/question/3874443

brainly.com/question/4240735

#LearnwithBrainly

6 0
3 years ago
Classify the quadrilateral. parallel sides are indicated with arrows.
Sonbull [250]
A quadrilateral is a four-sided polygon that may be classified as: trapezoid, rectangle, square, rhombus, etc. This classification is based on the measurement of the four angles, the four sides, and their being parallel. Unfortunately though, I think you missed to include here the diagram that we are about to classify. 
4 0
3 years ago
F a forklift raises a 76-kg load a distance of 2.5 m, about how much work has it done?
sleet_krkn [62]

W= FxD. Since the weight is not put into force you’re going to convert it by multiplying it by 9.8 (gravity). 76kg x 9.8 = 7448. Then multiply that by the distance (2.5). Your answer is 1862.

6 0
4 years ago
If the atomic radius of lead is 0.175 nm, calculate the volume of its unit cell in cubic meters.
Genrish500 [490]
Before any calculations, we need to determine first the crystal structure of the lead metal. From literature, the lead metal assumes an FCC structure. So, it would have 4 atoms per units cell where the three atoms is the sum of all the portion of an atoms in each face of the cell and the 1 atom is the sum of all the portion of the corner atoms. The volume of the unit cell is equal to the edge length raise to the power three or V = a^3. The edge length can be calculated from the radius of the atoms by the pythagorean theorem. We do as follows:

V = a^3
   a^2 + a^2 = (4r)^2
  2a^2 = (4r)^2
    a = 2r
V = (2r)^3
V = 16r^3
V = 16 (0.175x10^-9)^3
V = 1.21 x 10^-28 m^3
5 0
3 years ago
After a day of testing race cars, you decide to take your own 1550 kg car onto the test track. While moving down the track at 10
quester [9]

Answer:

F = 2325 N

Explanation:

given,

mass of the car, m = 1550 Kg

initial speed, u = 10 m/s

final speed, v = 25 m/s

time, t = 10 s

Average net force = ?

acceleration of the car

a = \dfrac{v-u}{t}

a = \dfrac{25-10}{10}

a = 1.5 m/s²

we know

F = m a

F = 1550 x 1.5

F = 2325 N

Net average force applied on the car is equal to 2325 N.

6 0
3 years ago
Read 2 more answers
Other questions:
  • How are physical and chemical changes similar?
    12·1 answer
  • What is the speed of a bobsled whose distance-time graph indicates that it traveled 119m in 29s?
    8·2 answers
  • Consider a system two point charges. One has charge +q at (x, y,z) -(a,0,0) and another of charge-q at (x, y, z) = (-a, 0,0). 5.
    9·1 answer
  • Does the size of a paper airplane affect how far it flies
    8·1 answer
  • How many meters in 2.50 miles? (Use these two conversions: 1000 m = 1 km and 1.00 km = .621 mi )
    8·1 answer
  • Which Of The Following Statements Are True?
    8·1 answer
  • Which hypothesis could be used in a scientific experiment and possibly lead to more hypotheses and experimentation? If the amoun
    8·2 answers
  • Two stones, one of mass m and the other of mass 2m, are thrown directly upward with the same velocity at the same time from grou
    9·1 answer
  • It is necessary to to secure an inflated balloon tightly give reason
    6·1 answer
  • Which range is the approximate audible range for humans?
    8·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!