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AVprozaik [17]
3 years ago
15

While riding in a moving car, you toss an egg directly upward. Does the egg tend to land behind you, in front of you, or back in

your hands if the car is (a) traveling at a constant speed, (b) increasing in speed, and (c) decreasing in speed?
Physics
1 answer:
bazaltina [42]3 years ago
4 0

Answer:

Explanation:

The movement of a body can be analyzed using New's first law. In an inertial frame (without acceleration) every body is kept at rest or moving at constant speed until there is an external force that changes this state

Let's analyze these cases in the framework of this first law

a) If the vehicle is going at constant speed the two bodies (the egg and the hands) do not change movement so he had returned to the hands

b) If the vehicle accelerates the passenger goes faster, but the egg that is not subject to anything does not change the movement, so it falls behind the passenger

c) If the vehicle slows down, the passenger reduces its speed and the distance traveled in time, but the egg that is not attached follows its movement and falls in front of the passenger.

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A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision
marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

x \approx 103,46x10^{9}m.

B) The space time coordinate t of the collision in Earth's reference frame is

t=377,29s

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

<em>Lorentz transformation</em>

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

                          x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                                y'=y

                                z'=z

                          t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

                       x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                           y=y'

                           z=z'

                        t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

First we calculate the expression in the denominator

                            \frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}

                                \sqrt{1-\frac{v^{2}}{c^{2}}} =0,714

then we calculate t

                      t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                      t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}

                      t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}

                      t=\frac{190s+79,388s}{0,714}

finally we get that

                                     t=377,29s

then we calculate x

                         x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                         x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}

                         x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}

                         x=\frac{3,4x10^{10}m+39872396914m}{0,714}}

                         x=\frac{73872396914m}{0,714}}

                         x=103462740775,91m

finally we get that

                                     x \approx 103,46x10^{9} m

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3 years ago
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5 0
3 years ago
Help me to solve it . It’s urgent
Artist 52 [7]

Answer: 0°

Explanation:

Step 1: Squaring the given equation and simplifying it

Let θ be the angle between a and b.

Given: a+b=c

Squaring on both sides:

... (a+b) . (a+b) = c.c

> |a|² + |b|² + 2(a.b) = |c|²

> |a|² + |b|² + 2|a| |b| cos 0 = |c|²

a.b = |a| |b| cos 0]

We are also given;

|a+|b| = |c|

Squaring above equation

> |a|² + |b|² + 2|a| |b| = |c|²

Step 2: Comparing the equations:

Comparing eq( insert: small n)(1) and (2)

We get, cos 0 = 1

> 0 = 0°

Final answer: 0°

[Reminders: every letters in here has an arrow above on it]

7 0
2 years ago
Describe what happens to chemical<br> bonds during a chemical reaction
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Explanation:

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