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Triss [41]
2 years ago
12

With respect to the earth, object 1 is moving at speed 0.80 c to the right. Object 2 is moving in the same direction at speed 0.

80 c with respect to object 1. How fast is object 2 moving with respect to the earth?
Physics
1 answer:
viktelen [127]2 years ago
6 0

Answer:

0.976 c

Explanation:

v_{1e} = velocity of object 1 relative to earth = 0.80 c

v_{21} = velocity of object 2 relative to object 1 = 0.80 c

v_{2e} = velocity of object 2 relative to earth

Velocity of object 2 relative to earth is given as

v_{2e}= \frac{v_{1e} + v_{21}}{1 + \frac{v_{1e}v_{21}}{c^{2}}}

v_{2e}= \frac{0.80 c + 0.80 c}{1 + \frac{(0.80c)(0.80c)}{c^{2}}}

v_{2e} = 0.976 c

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The famous scientist Galileo Galilei did several experiments with sloping planes, which he rolled metal balls down so that he co
KatRina [158]
Hi there!

The answer would be B. the slope of the plane.

Changing the slope of the plane would show how fast the ball went when Galileo changed the steepness of the slope. If he didn’t change the slopes steepness he would have the same results each time.

Hope this helps !
5 0
3 years ago
What do you think we call this graphical representation based on your prior experience with electric fields and electric field l
slavikrds [6]

Answer:

Explanation:

The strengthcompassion field is proportional to the closeness of the field lines—more precisely, it is proportional to the number of lines per unit area perpendicular to the lines. The direction of the electric field is tangent to the field line at any point in space. Field lines can never cross. These pattern of lines, sometimes referred to as electric field lines, point in the direction that a positive test charge would accelerate if placed upon the line. As such, the lines are directed away from positively charged source charges and toward negatively charged source charges.

Rules for drawing electric field lines

1. Electric field lines are always drawn from High potential to

low potential.

2. Two electric field lines can never intersect each other.

3. The net electric field inside a Conductor is Zero.

4. Electric field line from a positive charge is drawn radially outwards and from a negative charge radially inwards.

5. The density of electric field lines tells the strength of the electric field at that region.

6. Electric field lines terminate Perpendicularly to the surface of a conductor.

A vector quantity has a direction and a magnitude, while a scalar has only a magnitude. You can tell if a quantity is a vector by whether or not it has a direction associated with it.

So, electric fields are vector quantity due to the fact any student can tell you that a compass is used to determine which direction is north.

Since the compass always point northward, then it has a direction and magnitude and so it is a vector quantity

6 0
3 years ago
A speedboat is approaching a dock at 25 m/s (56 mph). When the dock is 150 m away, the driver begins to slow down. a) What accel
trasher [3.6K]

Answer:

a) -2.038 m/s²

b) 40.33 mph

c) 312.5 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-25^2}{2\times 150}\\\Rightarrow a=-2.083\ m/s^2

Acceleration of the boat is -2.083 m/s² if the boat will stop at 150 m.

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -1\times 150+25^2}\\\Rightarrow v=18.03\ m/s

Speed of the boat by when it will hit the dock is 18.03 m/s

Converting to mph

1\ mile=1609.34\ m

1\ h=3600\ seconds

18.03\times \frac{3600}{1609.34}=40.33\ mph

Speed of the boat by when it will hit the dock is 40.33 mph

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-25^2}{2\times -1}\\\Rightarrow s=312.5\ m

The distance at which the boat will have to start decelerating is 312.5 m

5 0
3 years ago
An empty truck traveling at 10 km/h has kinetic energy. How much kinetic energy does it have when loaded so that its mass and it
Katena32 [7]

Answer:

8 time increase in K.E.

Explanation:

Consider Mass of truck = m kg and speed = v m/s then

K.E. = 1/2 ×mv²

If mass and speed both are doubled i.e let m₀ = 2m and v₀ = 2v then

(K.E.)₀ =  1/2 ×2m(2v)²

(K.E.)₀ = 8 (1/2 × mv²) = 8 × K.E.

7 0
3 years ago
Problem I Marcella (see warmup problem, above) gets her car moving steadily at 4m/s but suddenly someone stops ahead to assist h
Nady [450]

Answer:

Explanation:

a) KE = (1/2) * m * (v^{2}) = F * d = 14m * 200N = 2800 m/N or 2.8 * 10^{3} m/N

b) 0J and 0m/s (if Marcella stopped after going 14 meters)

c)  Known from part (a) that KE = 2800 J = F1 * d1,

    2800J = F1 * (14m - 1m)  => F1 = 2800J/13m = 215.384 N

   

3 0
3 years ago
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