All categories

3 years ago

Which quantity can be calculated using the equation E = mc2?

Physics

2 answers:

Mariulka [41]3 years ago

6 0

Explanation :

Einstein's equation is given by :

$E=m\times c^2$ ..............(1)

m is the mass

c is the speed of light $c=3\times 10^8\ m/s$

The energy needed to split an atom into separate protons, neutrons, and electrons can be calculated by using equation (1).

The energy needed to split the nucleus into separate protons and neutrons is called nuclear binding energy.

The energy needed to remove an electron from an atom is called ionization energy.

The energy needed to add an electron to an atom is called electron affinity.

So, the correct option is (A).

Goryan [66]3 years ago

3 0

Answer;

-The energy needed to split an atom into separate protons, neutrons, and electrons

Explanation;

E = mc² The equation is derived directly from Einstein’s Special Theory of Relativity .

Each of the letters of E = mc2 stands for a particular physical quantity. Writing them out in full we get:

Energy = mass × the speed of light squared

In other words: E = energy (measured in joules, J), m = mass (measured in kilograms, kg), c = the speed of light (measured in metres per second, ms-1), but this needs to be squared.

You might be interested in

The mass of the sun is 1.99x10^30 kg . Jupiter is 7.79x10^8 km away from the sun and the mass of 1.90x10^27 kg

maks197457 [2]

Question is missing:

"What is the gravitational force between the Sun and Jupiter?"

Answer:

$4.16\cdot 10^{23} N$

Explanation:

The gravitational force between two objects is given by

$F=G\frac{m_1m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between the objects

In this problem, we have

$m_1 = 1.99\cdot 10^{30} kg$ is the mass of the sun

$m_2 = 1.90\cdot 10^{27} kg$ is the mass of Jupiter

$r=7.79\cdot 10^8 km = 7.79\cdot 10^{11} m$ is their separation

Solving the equation, we find

$F=(6.67\cdot 10^{-11})\frac{(1.99\cdot 10^{30})(1.90\cdot 10^{27})}{(7.79\cdot 10^{11})^2}=4.16\cdot 10^{23} N$

6 0

2 years ago

A 200 g air-track glider is attached to a spring. The glider is pushed in 9.8 cm against the spring, then released. A student wi

Likurg_2 [28]

Answer:

k = 5.05 N/m

Explanation:

In order to calculate the spring mass of the system, you use the following formula:

$T=2\pi \sqrt{\frac{m}{k}}$ (1)

T: period of oscillation of the system

m: mass of the air-track glider = 200g = 0.200 kg

k: spring constant = ?

You first calculate the period of oscillation:

$T=\frac{1}{f}=\frac{1}{12/15.0s}=1.25s$

Next, you solve the equation (1) for k, and then you replace the values of the other parmateres:

$k=4\pi^2 \frac{m}{T^2}\\\\k=4\pi^2 \frac{0.200kg}{(1.25s)^2}=5.05\frac{N}{m}$

The spring constant of the spring is 5.05 N/m

6 0

2 years ago

An object is thrown with a force of 30 Newtons and ends up with an acceleration of 3 m/s ^ 2 due to that throw. What is the mass

jonny [76]

Answer:

F=30N

a= 3m/s^2

m=?

F=ma

30=m(3)

30/3=m

m=10kg

The mass of the ball is 10kg

4 0

3 years ago

Calculate the Schwarzschild radius (in kilometers) for each of the following.1.) A 1 ×108MSun black hole in the center of a quas

Westkost [7]

Answer:

(I). The Schwarzschild radius is $2.94\times10^{8}\ km$

(II). The Schwarzschild radius is 17.7 km.

(III). The Schwarzschild radius is $1.1\times10^{-7}\ km$

(IV). The Schwarzschild radius is $7.4\times10^{-29}\ km$

Explanation:

Given that,

Mass of black hole $m= 1\times10^{8} M_{sun}$

(I). We need to calculate the Schwarzschild radius

Using formula of radius

$R_{g}=\dfrac{2MG}{c^2}$

Where, G = gravitational constant

M = mass

c = speed of light

Put the value into the formula

$R_{g}=\dfrac{2\times6.67\times10^{-11}\times1\times10^{8}\times1.989\times10^{30}}{(3\times10^{8})^2}$

$R_{g}=2.94\times10^{8}\ km$

(II). Mass of block hole $m= 6 M_{sun}$

We need to calculate the Schwarzschild radius

Using formula of radius

$R_{g}=\dfrac{2MG}{c^2}$

Put the value into the formula

$R_{g}=\dfrac{2\times6.67\times10^{-11}\times6\times1.989\times10^{30}}{(3\times10^{8})^2}$

$R_{g}=17.7\ km$

(III). Mass of block hole m= mass of moon

We need to calculate the Schwarzschild radius

Using formula of radius

$R_{g}=\dfrac{2MG}{c^2}$

Put the value into the formula

$R_{g}=\dfrac{2\times6.67\times10^{-11}\times7.35\times10^{22}}{(3\times10^{8})^2}$

$R_{g}=1.1\times10^{-7}\ km$

(IV). Mass = 50 kg

We need to calculate the Schwarzschild radius

Using formula of radius

$R_{g}=\dfrac{2MG}{c^2}$

Put the value into the formula

$R_{g}=\dfrac{2\times6.67\times10^{-11}\times50}{(3\times10^{8})^2}$

$R_{g}=7.4\times10^{-29}\ km$

Hence, (I). The Schwarzschild radius is $2.94\times10^{8}\ km$

(II). The Schwarzschild radius is 17.7 km.

(III). The Schwarzschild radius is $1.1\times10^{-7}\ km$

(IV). The Schwarzschild radius is $7.4\times10^{-29}\ km$

8 0

3 years ago

Two vehicles, a car and a truck, leave an intersection at the same time. the car heads east at an average speed of 50 miles pe

olya-2409 [2.1K]

The car heads east at an average speed of 50 miles per hour from the intersection point towards East. The truck heads east at an average speed of 60 miles per hour from the intersection point towards South.

The distance of car from the intersection point after t hours is $50t$ .

The distance of truck from the intersection point after t hours is $60t$ .

Since these distances are perpendicular to each other, distance apart d (in miles) at the end of t hours is

$d=\sqrt{(50t)^2+(60t)^2} \\ d=10\sqrt{61} t\\ d=78.1t$

Thus the distance apart is $d=78.1t \;miles$

5 0

2 years ago