A person on a rocket traveling at 0.47 c (with respect to the Earth) observes a meteor come from behind and pass her at a speed she measures as0.47 c.How fast is the meteor moving with respect to the Earth?

Answers

Answer 1

Answer:

The concept to solve this problem is related to the relativistic physics for which the speed of the object in different frames of reference is related. This concept is called Velocity-addition formula

and can be written as,

$u = (v+u')/(1+(vu')/(c^2))$

Where,

u = Velocity of a body within a Lorentz Frame

v = Velocity of a second frame

u'= The transformed velocity of the body within the second frame

c = speed of light

Replacing we have to

$u = (v+u')/(1+(vu')/(c^2))$

$u = (0.47c+0.47c)/(1+(0.47c*0.47c)/(c^2))$

$u = 0.769c$

$u \approx 230'700.000m/s$

Therefore the meteor moving with respect to the Earth to 230'700.000m/s

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A tennis player tosses a tennis ball straight up and then catches it after 1.77 s at the same height as the point of release. (a) What is the acceleration of the ball while it is in flight? magnitude m/s2 direction (b) What is the velocity of the ball when it reaches its maximum height? magnitude m/s direction (c) Find the initial velocity of the ball. m/s upward (d) Find the maximum height it reaches. m'

Answers

(a) 9.8 m/s^2, downward

There is only one force acting on the ball while it is in flight: the force of gravity, which is

F = mg

where

m is the mass of the ball

g is the gravitational acceleration

According to Newton's second law, the force acting on the ball is equal to the product between the mass of the ball and its acceleration, so

F = mg = ma

which means

a = g

So, the acceleration of the ball during the whole flight is equal to the acceleration of gravity:

g = -9.8 m/s^2

where the negative sign means the direction is downward.

(b) v = 0

Any object thrown upward reaches its maximum height when its velocity is zero:

v = 0

In fact, at that moment, the object's velocity is turning from upward to downward: that means that at that instant, the velocity must be zero.

The initial velocity of the ball can be found by using the equation:

v = u + at

Where

v = 0 is the velocity at the maximum height

u is the initial velocity

a = g = -9.8 m/s^2 is the acceleration

t is the time at which the ball reaches the maximum height: this is half of the time it takes for the ball to reach again the starting point of the motion, so

$t=(1.77 s)/(2)=0.89 s$

So we can now solve the equation for u, and we find:

$u=v-at=0-(-9.8 m/s^2)(0.89 s)=8.72 m/s$

(d) 3.88 m

The maximum height reached by the ball can be found by using the equation:

$v^2 - u^2 = 2ad$

where

v = 0 is the velocity at the maximum height

u = 8.72 m/s is the initial velocity

a = g = -9.8 m/s^2 is the gravitational acceleration

d is the maximum height reached

Solving the equation for d, we find

$d=(v^2-u^2)/(2a)=(0^2-(8.72 m/s)^2)/(2(-9.8 m/s^2))=3.88 m$

A novice scuba diver practicing in a swimming pool takes enough air from his tank to fully expand his lungs before abandoning the tank at depth L and swimming to the surface. He ignores instructions and fails to exhale during his ascent. When he reaches the surface, the difference between the external pressure on him and the air pressure in his lungs is 9.3 kPa. From what depth does he start?

Answers

Answer:

11.625 m.

Explanation:

Difference of pressure will be due to hydro-static pressure due to column of water of height L.

Pressure of water column = L d g , where L is depth ,

d is density of water = 1000kg /m³

g = 9.8 ms²

Pressure difference = 9.3 kPa = 9300 Pa

So Ldg = 9300

L X 1000 X 0.8 =9300

800 L = 9300

L = 11.625 m.

A capacitor consists of two closely spaced metal conductors of large area, separated by a thin insulating foil. It has an electrical capacity of 3000.0 μF and is charged to a potential difference of 60.0 V. Calculate the amount of energy stored in the capacitor. Tries 0/20 Calculate the charge on this capacitor when the electrical energy stored in the capacitor is 6.53 J. Tries 0/20 If the two plates of the capacitor have their separation increased by a factor of 5 while the charge on the plates remains constant, by what factor is the energy stored in the capacitor increased?

Answers

Answer:

1 = 5.4 J

2 = 0.1979 C

3 = 5

Explanation:

Energy in a capacitor, E is

E = 1/2 * C * V²

E = 1/2 * 3000*10^-6 * 60²

E = 1/2 * 3000*10^-6 * 3600

E = 1/2 * 10.8

E = 5.4 J

E = Q²/2C = 6.53 J

E * 2C = Q²

Q² = 6.53 * 2 * 3000*10^-6

Q² = 13.06 * 3000*10^-6

Q² = 0.03918

Q = √0.03918

Q = 0.1979 C

The Capacitor, C is inversely proportional to the distance of separation, D. Thus, if D is increased by 5 to be 5D, then C would be C/5. And therefore, our energy stored in the capacitor is increased by a factor of 5.

Suppose that a barometer was made using oil with rho=900 kg/m3. What is the height of the barometer at atmospheric pressure?

Answers

Hey there!

The pressure under a liquid column can be , calculated using the following formula :

P = p x g x h

P atm = 1.013 x 10⁵ Pa

g = 9.8 m/s²

h = ?

h = P / ( p x g ) =

h= ( 1.013 x 10⁵ Pa ) / ( 900 x 9.8 ) =

h = ( 1.013 x 10⁵ ) / ( 8820 ) =

h = 11.48 m ≈ 11.50 m

Hope this helps!

wo parallel plates of area 100cm2are given charges of equal magnitudes 8.9 ×10−7C but opposite signs. The electric field within the dielectric material filling the space between the plates is 1.4 ×106V/m. (a) Calculate the dielectric constant of the material. (b) Determine the magnitude of the charge induced on each dielectric surface.

Answers

(a) 7.18

The electric field within a parallel plate capacitor with dielectric is given by:

$E=(\sigma)/(k \epsilon_0)$ (1)

where

$\sigma$ is the surface charge density

k is the dielectric constant

$\epsilon_0$ is the vacuum permittivity

The area of the plates in this capacitor is

$A=100 cm^2 = 100\cdot 10^(-4) m^2$

while the charge is

$Q=8.9\cdot 10^(-7)C$

So the surface charge density is

$\sigma = (Q)/(A)=(8.9\cdot 10^(-7) C)/(100\cdot 10^(-4) m^2)=8.9\cdot 10^(-5) C/m^2$

The electric field is

$E=1.4\cdot 10^6 V/m$

So we can re-arrange eq.(1) to find k:

$k=(\sigma)/(E \epsilon_0)=(8.9\cdot 10^(-5) C/m^2)/((1.4\cdot 10^6 V/m)(8.85\cdot 10^(-12) F/m))=7.18$

(b) $7.66\cdot 10^(-7)C$

The surface charge density induced on each dielectric surface is given by

$\sigma' = \sigma (1-(1)/(k))$

where

$\sigma=8.9\cdot 10^(-5) C/m^2$ is the initial charge density

k = 7.18 is the dielectric constant

Substituting,

$\sigma' = (8.9\cdot 10^(-5) C/m^2) (1-(1)/(7.18))=7.66\cdot 10^(5) C/m^2$

And by multiplying by the area, we find the charge induced on each surface:

$Q' = \sigma' A = (7.66\cdot 10^(-5) C/m^2)(100 \cdot 10^(-4)m^2)=7.66\cdot 10^(-7)C$

When two point charges are a distance d part, the electric force that each one feels from the other has magnitude F. In order to make this force twice as strong, the distance would have to be changed toA) √2d
B) d/√2
C) d/4
D) 2d
E) d/2