Why do some clusters of galaxies emit x-rays?

A proton (mass=1.67x10^-27 kg, charge= 1.60x10^-19 C) moves from point A to point under the influence of an electrostatic force

Tom [10]

Answer:

$VB - VA = - 33.4$

Explanation:

Generally the workdone in moving the proton is mathematically represented as

$W = KE_f - KE_i$

Where $KE_i \ and \ KE_f \ are\ the\ initial \ and \ final \ kinetic \ energy$

So

$KE_i = \frac{1}{2} m v_a^2$

Here $v_a$ is the velocity at A with value 50 m/s

So

$KE_i = \frac{1}{2} (1.67*10^{-27}) * 50^2$

$KE_i = 2.09 *10^{-24} \ J$

Also

$KE_f = \frac{1}{2} m v_b^2$

Here $v_a$ is the velocity at A with value $80 km/s = 80000 m/s$

=> $KE_f = \frac{1}{2} (1.67*10^{-27}) * 80000^2$

=> $KE_f = 5.34 *10^{-18} \ J$

So

$W = 5.34 *10^{-18} - 2.09 *10^{-24}$

$W = 5.34 *10^{-18} m/s$

Now this workdone is also mathematically represented as

$W = q * V$

So

$q * V = 5.34 *10^{-18}$

Here $q = 1.60*10^{-19} C$

So

$V = \frac{5.34 *10^{-18} }{1.60*10^{-19}}$

$V = 33.4 \ V$

Generally proton movement is in the direction of the electric field it means that $VA>VB$

So

$VB - VA = - 33.4$

8 0

3 years ago

A charged object is suspended motionless in the air by the gravitational force pulling it down and an electric force pushing it

Savatey [412]

The charge of the object must be $1.11 \times e^{-5} \text { coulomb }$

Answer: Option C

<u>Explanation:</u>

Suppose an electric charge can be represented by the symbol Q. This electric charge generates an electric field; Because Q is the source of the electric field, we call this as source charge. The electric field strength of the source charge can be measured with any other charge anywhere in the area. The test charges used to test the field strength.

Its quantity indicated by the symbol q. In the electric field, q exerts an electric, either attractive or repulsive force. As usual, this force is indicated by the symbol F. The electric field’s magnitude is simply defined as the force per charge (q) on Q.

$Electric field, E=\frac{\text { Force }(F)}{q}$

Here, given E = 4500 N/C and F = 0.05 N.

We need to find charge of the object (q)

By substituting the given values, we get

$q=\frac{F}{E}=\frac{0.05 N}{4500 \mathrm{N} / \mathrm{c}}=1.11 \times e^{-5} \text { coulomb }$

6 0

3 years ago

Help pls i need this right now

pantera1 [17]

Answer:

The x-component of $F_{3}$ is 56.148 newtons.

Explanation:

From 1st and 2nd Newton's Law we know that a system is at rest when net acceleration is zero. Then, the vectorial sum of the three forces must be equal to zero. That is:

$\vec F_{1} + \vec F_{2} + \vec F_{3} = \vec O$ (1)

Where:

$\vec F_{1}$ , $\vec F_{2}$ , $\vec F_{3}$ - External forces exerted on the ring, measured in newtons.

$\vec O$ - Vector zero, measured in newtons.

If we know that $\vec F_{1} = (70.711,70.711)\,[N]$ , $\vec F_{2} = (-126.859, 46.173)\,[N]$ , $F_{3} = (F_{3,x},F_{3,y})$ and $\vec O = (0,0)\,[N]$ , then we construct the following system of linear equations: