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2 years ago

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Einstein and Lorentz, being avid tennis players, play a fast-paced game on a court where they stand 20.0 m from each other. Bein

g very skilled players, they play without a net. The tennis ball has mass 0.0580 kg. You can ignore gravity and assume that the ball travels parallel to the ground as it travels between the two players. Unless otherwise specified, all measurements are made by the two men.
(a) Lorentz serves the ball at 80.0 m/s. What is the ball's kinetic energy?
(b) Einstein slams a return at 1.80 x 10^8 m/s. What is the ball's kinetic energy?

1 answer:

kobusy [5.1K]2 years ago

4 0

Answer:

Explanation:

a ) Kinetic energy of ball = 1/2 m v² , m is mass of the ball and v is its velocity

= 1/2 x .058 x 80²

= 185.6 J .

b ) The velocity of ball is very high so we shall find out the relativistic kinetic energy , the formula is given below

kinetic energy = mc² ( $\frac{1}{\sqrt{1-\frac{v^2}{c^2} } } -1$ )

= mc² x $\frac{1}{\sqrt{1-\frac{v^2}{c^2} } }$ - mc²

= .058 x 9 x 10¹⁶ x $\frac{1}{\sqrt{1-\frac{1.8^2}{3^2} } }$ - .058 x 9 x 10¹⁶

= .058 x 9 x 10¹⁶ x 1 / .8 - .058 x 9 x 10¹⁶

.058 x 9 x 10¹⁶ ( 1.25 - 1 )

= .058 x 9 x 10¹⁶ x .25

= .1305 x 10¹⁶

= 13.05 x 10¹⁴ J .

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A particle has a velocity of v→(t)=5.0ti^+t2j^−2.0t3k^m/s.

Makovka662 [10]

Answer:

a) $a=5 i+2t j - 6\ t^2k$

b) $a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2$

Explanation:

Given that

v(t) = 5 t i + t² j - 2 t³ k

We know that acceleration a is given as

$a=\dfrac{dv}{dt}$

$\dfrac{dv}{dt}=5 i+2t j - 6\ t^2k$

$a=5 i+2t j - 6\ t^2k$

Therefore the acceleration function a will be

$a=5 i+2t j - 6\ t^2k$

The acceleration at t = 2 s

a= 5 i + 2 x 2 j - 6 x 2² k m/s²

a=5 i + 4 j -24 k m/s²

The magnitude of the acceleration will be

$a=\sqrt{5^2+4^2+24^2}\ m/s^2$

a= 24.83 m/s²

The direction of the acceleration a is given as

$a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2$

a) $a=5 i+2t j - 6\ t^2k$

b) $a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2$

5 0

2 years ago

Consider a double Atwood machine constructed as follows: A mass 4m is suspended from a string that passes over a massless pulley

kenny6666 [7]

Answer:

Hello your question is incomplete attached below is the complete question

Answer : x ( acceleration of mass 4m ) = $\frac{g}{7}$

The top pulley rotates because it has to keep the center of mass of the system at equilibrium

Explanation:

Given data:

mass suspended = 4 meters

mass suspended at other end = 3 meters

first we have to express the kinetic and potential energy equations

The general kinetic energy of the system can be written as

T = $\frac{4m}{2} x^2 + \frac{3m}{2} (-x+y)^2 + \frac{m}{2} (-x-y)^2$

T = $4mx^2 + 2my^2 -2mxy$

also the general potential energy can be expressed as

U = $-4mgx-3mg(-x+y)-mg(-x-y)+constant=-2mgy +constant$

The Lagrangian of the problem can now be setup as

$L =4mx^2 +2my^2 -2mxy +2mgy + constant$

next we will take the Euler-Lagrange equation for the generalized equations :

Euler-Lagrange equation = $4x-y =0\\-2y+x +g = 0$

solving the equations simultaneously

x ( acceleration of mass 4m ) = $\frac{g}{7}$

The top pulley rotates because it has to keep the center of mass of the system at equilibrium

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2 years ago

An observer O is standing on a platform of length L = 90 m on a station. A rocket train passes at a relative (constant) speed of

Natali [406]

Answer:

Explanation:

Since the front and back of the rocket simultaneously line up with forward and backward end of the platform respectively .

Then length of the platform = length of the train rocket .

A )

Time to cross a particular point on the platform

= length of rocket train / .96 x 3 x 10⁸

= 90 / .96 x 3 x 10⁸

= 31.25 x 10⁻⁸ s

B) Rest length of the rocket = length of platform = 90 m

C ) length of platform as viewed by moving observer =

$\frac{90}{\sqrt{1-\frac{v^2}{c^2 } } }$

= $\frac{90}{\sqrt{1-\frac{0.92}{1 } } }$

= 321 m

D ) For the observer on platform time taken = 31.25 x 10⁻⁸ s

for the observer in the rocket , time will be dilated so time recorded by observer in motion ,

$31.25\times10^{-8} \times \sqrt{1-\frac{.96^2}{1} }$

8.75 x 10⁻⁸ s .

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1 year ago

If two waves with identical crests and troughs meet, what is happening?

Nastasia [14]

<span>If two waves with identical crests and troughs meet, what is happening?
</span>C. Constructive interference is occurring.

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2 years ago

Read 2 more answers

: Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at

ser-zykov [4K]

Complete Question

Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A=0.0155 m. The air pressure in Denver is PD = 79000 Pa. and in New Orleans is PNo = 100250 Pa. Assume the lid is weightless.

Part (a) Write an expression for the force FNo required to remove the container lid in New Orleans.

Part (b) Calculate the force FNo required to lift off the container lid in New Orleans, in newtons.

Part (c) Calculate the force Fp required to lift off the container lid in Denver, in newtons.

Part (d) is more force required to lift the lid in Denver (higher altitude, lower pressure) or New Orleans (lower altitude, higher pressure)?

Answer:

a

The expression is $F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

b

$F_{No}= 7771.125 \ N$

c

$F_p = 2.2*10^{6} N$

d

From the value obtained we can say the that the force required to open the lid is higher at Denver

Explanation:

The altitude of container in Denver is $d_D = 6000 \ ft = 6000 * 0.3048 = 1828.8m$

The surface area of the container lid is $A = 0.0155m^2$

The altitude of container in New Orleans is sea-level

The air pressure in Denver is $P_D = 79000 \ Pa$

The air pressure in new Orleans is $P_{ro} = 100250 \ Pa$

Generally force is mathematically represented as

$F_{No} = \Delta P A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

So the $\Delta P$ which is the difference in pressure within and outside the container is

$\Delta P = P_{No} - \frac{P_{sea}}{2}$

Therefore

$F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

Now substituting values

$F_{No} = 0.0155 [100250 - \frac{101000}{2}]$

$F_{No}= 7771.125 \ N$

The force to remove the lid in Denver is

$F_p = \Delta P_d A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

At sea level the air pressure in Denver is mathematically represented as

$P_D = \rho g h$

=> $g = \frac{P_D}{\rho h}$

Let height at sea level is h = 1

The air pressure at height $d_D$

$P_d__{D}} = \rho gd_D$

=> $g = \frac{P_d_D}{\rho d_D}$

Equating the both

$\frac{P_D}{\rho h} = \frac{P_d_D}{\rho d_D}$

$P_d_D = P_D * d_D$

Substituting value

$P_d__{D}} = 1828.2 * 79000$

$P_d__{D}} = 1.445*10^{8} Pa$

So

$\Delta P_d = P_{d} _D - \frac{P_{sea}}{2}$

=> $\Delta P_d = 1.445 *10^{8} - \frac{101000}{2}$

$\Delta P_d = 1.44*10^{8}Pa$

So

$F_p = \Delta P_d A$

$= 1.44*10^8 * 0.0155$

$F_p = 2.2*10^{6} N$

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2 years ago