Can someone help me figure out how to find tension in a rope? The question is:

A 50 kg person receives an absorbed dose of gamma radiation of 20 millirads. What is the total energy absorbed?.

Masteriza [31]

For a 50 kg person receives an absorbed dose of gamma radiation of 20 millirads, the total energy absorbed is mathematically given as

E=0.1457J

<h3>What is the total energy absorbed?</h3>

Generally, the equation for the total energy absorbed is mathematically given as

E=mass*gamma radiation

Therefore

E=50*20*19^{-3}

E=0.1457J

In conclusion, the total energy absorbed

E=0.1457J

Read more about Energy

brainly.com/question/13439286

4 0

2 years ago

Can someone solve this problem and explain to me how you got it

evablogger [386]

Answer:

question5: F=74312.5N

question6: charge at the end of antenna=0.37N

Explanation:

Coulomb's law: the magnitude of the force of attraction or repulsion due to two charges is proportional to the product of the magnitude of the charges and inversely proportional to the square of distance between the charges.

⇒ $F\alpha\frac{q1*q2}{r^{2}}$

∴ $F=k\frac{q1*q2}{r^{2}}$

where $F$ is the force of attraction or repulsion

$k$ is Coulumb's constant= $9*10^{9}Nm^{2}C^{-2}$

$q1$ and $q2$ are the magnitude of the charges

$r$ is the distance between two charges

The force between the two charges is attractive if they are of different polarity

The force between the two charges is repulsive if they are of same polarity

Question5:

Given: q1=0.041 C, q2=0.029 C, r=12 m

therefore by Coulumb's law,

$F=9*10^{9}*\frac{0.041*0.029}{12^{2}}$

$F=74312.5N$

Question6:

Given: q1= $3*10^{-18}C$ , r=5 m, F= $4*10^{-11}N$

therefore by Coulumb's law,

$4*10^{-11}=9*10^{9}*\frac{3*10^{-18}*q2}{5^{2}}$

⇒ $q2=\frac{4*10^{-11}*25}{9*10^{9}*3*10^{-18}} \\=0.37C$

4 0

3 years ago

A car has a force of 2000N and a mass of 1000kg. What is the acceleration

Anastaziya [24]

Answer:

2m/s/s

Explanation:

The formula goes- F=MA

F-Force M-Mass & A-Acceleration

We need to rearrange this formula to find the acceleration-

A=F/M

All we need to do now is substitute the values in

A=2000N/1000kg

A=2m/s^2

In the given option the last option (2m/s/s) would be the ans, as it's the same as 2m/s^2

So ya, I guess that's all

6 0

2 years ago

Mohs scale is used in describing which characteristic? luster toughness hardness crystal form

Lady_Fox [76]

The answer is hardness not luster

3 0

3 years ago

Read 2 more answers

A disk rotates around an axis through its center that is perpendicular to the plane of the disk. The disk has a line drawn on it

natka813 [3]

Answer:

$t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}$

Explanation:

The rotated angle is given by:

$\theta=\omega0*t+1/2*\alpha*t^2$

Since this is a quadratic equation it can be solved using:

$x=\frac{-b \± \sqrt{b^2-4*a*c} }{2*a}$

Rewriting our equation:

$1/2*\alpha*t^2+\omega0*t-\theta=0$

$t = \frac{\±\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}$

Since $\sqrt{\omega0^2+2*\theta*\alpha} >\omega0$ we discard the negative solution.

$t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}$

8 0

3 years ago