Ask question

Ask question

All categories

2 years ago

12

Problem 6.13 The point of this question is to compare rest energy and kinetic energy at high speeds. An alpha particle (a helium

nucleus) is moving at a speed of 0.9998 times the speed of light. Its mass is 6.40 × 10-27 kg. (a) What is its rest energy?

1 answer:

Kryger [21]2 years ago

3 0

Answer: E = 5.76×10^-10 J

Explanation: rest energy formula is given below as

E = mc²

m = mass of object = 6.40×10^-27 kg

c = speed of light = 3×10^8 m/s

E = 6.40×10^-27 × (3×10^8)²

E = 6.40×10^-27 × 9×10^16

E = 57.6 ×10^-11

E = 5.76×10^-10 J

You might be interested in

A motorist inflates the tires of her car to a pressure of 180 kPa on a day when the temperature is -8.0° C. When she arrives at

GuDViN [60]

Apply Gay-Lussac's law:

P/T = const.

P = pressure, T = temperature, the quotient of P/T must stay constant.

Initial P and T values:

P = 180kPa, T = -8.0°C = 265.15K

Final P and T values:

P = 245kPa, T = ?

Set the initial and final P/T values equal to each other and solve for the final T:

180/265.15 = 245/T

T = 361K

4 0

2 years ago

The fastest pitched baseball was clocked at 47 m/s. Assume that the pitcher exerted his force (assumed to be horizontal and cons

arsen [322]

Answer:

Explanation:

F ×1 = 0.5×0.145×47×47

F = 160.15 N

4 0

2 years ago

What mass needs to be attached to a spring with a force constant of 7N/m in order to make a simple harmonic oscillator oscillate

Alex_Xolod [135]

Answer:

Mass will be 4.437 kg

Explanation:

We have given force constant k = 7 N/m

Time period of oscillation T = 5 sec

So angular frequency $\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{5}=1.256rad/sec$

We know that angular frequency is given by

$\omega =\sqrt{\frac{k}{m}}$

$1.256 =\sqrt{\frac{7}{m}}$

Squaring both side

$1.577 =\frac{7}{m}$

m = 4.437 kg

6 0

2 years ago

A solid sphere is released from the top of a ramp that is at a height h1 = 2.30 m. It rolls down the ramp without slipping. The

Oksi-84 [34.3K]

Answer:

The horizontal distance d does the ball travel before landing is 1.72 m.

Explanation:

Given that,

Height of ramp $h_{1}=2.30\ m$

Height of bottom of ramp $h_{2}=1.69\ m$

Diameter = 0.17 m

Suppose we need to calculate the horizontal distance d does the ball travel before landing?

We need to calculate the time

Using equation of motion

$h_{2}=ut+\dfrac{1}{2}gt^2$

$t=\sqrt{\dfrac{2h_{2}}{g}}$

$t=\sqrt{\dfrac{2\times1.69}{9.8}}$

$t=0.587\ sec$

We need to calculate the velocity of the ball

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$

$K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times(\dfrac{2}{5}mr^2)\times(\dfrac{v}{r})^2$

$K.E=\dfrac{7}{10}mv^2$

Using conservation of energy

$K.E=mg(h_{1}-h_{2})$

$\dfrac{7}{10}mv^2=mg(h_{1}-h_{2})$

$v^2=\dfrac{10}{7}\times g(h_{1}-h_{2})$

Put the value into the formula

$v=\sqrt{\dfrac{10\times9.8\times(2.30-1.69)}{7}}$

$v=2.922\ m/s$

We need to calculate the horizontal distance d does the ball travel before landing

Using formula of distance

$d =vt$

Where. d = distance

t = time

v = velocity

Put the value into the formula

$d=2.922\times 0.587$

$d=1.72\ m$

Hence, The horizontal distance d does the ball travel before landing is 1.72 m.

8 0

2 years ago

Suppose Earth's mass increased but Earth's diame-

navik [9.2K]

Answer: It would increase.

Explanation:

The equation for determining the force of the gravitational pull between any two objects is:

$F = G \frac{m1m2}{r^2}$

Where G is the universal gravitational constant, m1 is the mass of one body, m2 is the mass of the other body, and r^2 is the distance between the two objects' centers squared.

Assuming the Earth's mass but not its diameter increased, in the equation above m1 (the term usually indicative of the object of larger mass) would increase, while the r^2 would not.

Thus, it goes without saying that, with some simple reasoning about fractions, an increasing numerator over a constant denominator would result in a larger number to multiply by G, thus also meaning a larger gravitational strength between Earth and whatever other object is of interest.

7 0

2 years ago