All categories

3 years ago

Does anyone know? Please help! Thank you!

Physics

2 answers:

Vedmedyk [2.9K]3 years ago

7 0

B density will increase

daser333 [38]3 years ago

6 0

B it increased because it put more pressure and volume that makes it more

You might be interested in

Convert the following to gram. a,250 kg b373 mg c,10 quanital,15 ton

rusak2 [61]

Explanation:

250 KG = 250000 Grams

373 MG = 0.373 Grams

10 Quintals = 1000000 Grams

15 Ton = 15000000 Grams

Please Follow and inbox me

4 0

2 years ago

A cyclotron with a magnetic field of 0.47 t is designed to accelerate protons in a circle with a radius 0.68 m. What is the angu

Ghella [55]

The angular frequency of the cyclotron is 0.07 x $10^{8}$ Hz.

<h3>What is angular frequency?</h3>

Angular frequency, abbreviated "ω" is a scalar measure of rotation rate in physics.
It describes the rate of change of the argument of the sine function, the rate of change of the phase of a sinusoidal waveform, or the angular displacement per unit of time.

<h3>What is cyclotron?</h3>

The cyclotron device is made to accelerate charge particles to extremely high speeds by applying crossed electric and magnetic fields.

<h3>Calculation of angular frequency:</h3>

Given,

B = 0.47 T

r = 0.68

mass of proton = 1.6x $10^{-27}$

q = 1.6 x $10^{-19}$

so, the frequency is:

f = qB/2 $\pi$ m

f = 1.6 x $10^{-19}$ x 0.47/2x3.14x1.6x $10^{-27}$

f = 0.07 x $10^{8}$

Hence, the angular frequency of the cyclotron is 0.07 x $10^{8}$ Hz.

Learn more about angular frequency here:

brainly.com/question/14244057

#SPJ4

7 0

2 years ago

A 1500-kg car locks its brakes and skids to a stop on a slippery horizontal road, leaving skid marks that are 15 m long. How muc

Harman [31]

Answer:

$E=88200\ J$

Explanation:

Given:

mass of car, $m=1500\ kg$

distance of skidding after the application of brakes, $d=15\ m$

coefficient of kinetic friction, $\mu_k=0.4$

<u>So, the energy dissipated during the skidding of car:</u>

<em>Frictional force:</em>

$f=\mu_k.N$

where N = normal reaction by ground on the car

$f=0.4\ties 1500\times 9.8$

$f=5880\ N$

<em>Now from the work-energy equivalence:</em>

$E=f.d$

$E=5880\times 15$

$E=88200\ J$ is the dissipated energy.

3 0

3 years ago

How wood helps in home insulation plz answer fast<br>plz answwweeeeeeerrrrrrr

Usimov [2.4K]

Wood is a poor conductor and therefore a good insulator keeping heat inside

3 0

3 years ago

Read 2 more answers

A 0.00275 kg air‑inflated balloon is given an excess negative charge q1 =−3.50×10−8 C by rubbing it with a blanket. It is found

kati45 [8]

Answer:

1) $\rm q_2$ is<u> positive.</u>

<u></u>

2) $\rm q_2=4.56\times 10^{-10}\ C.$

Explanation:

<u></u>

The charged rod is held above the balloon and the weight of the balloon acts in downwards direction. To balance the weight of the balloon, the force on the balloon due to the rod must be directed along the upwards direction, which is only possible when the rod exerts an attractive force on the balloon and the electrostatic force on the balloon due to the rod is attractive when the polarities of the charge on the two are different.

Thus, In order for this to occur, the polarity of charge on the rod must be positive, i.e., $\rm q_2$ is <u>positive.</u>

<u></u>

<u></u>

<u>Given:</u>

Mass of the balloon, m = 0.00275 kg.
Charge on the balloon, $\rm q_1 = -3.50\times 10^{-8}\ C.$
Distance between the rod and the balloon, d = 0.0640 m.
Acceleration due to gravity, $\rm g = 9.81\ m/s^2.$

In order to balloon to be float in air, the weight of the balloom must be balanced with the electrostatic force on the balloon due to rod.

Weight of the balloon, $\rm W = mg = 0.00275\times 9.81=2.70\times 10^{-2}\ N.$

The magnitude of the electrostatic force on the balloon due to the rod is given by

$\rm F_e = \dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}.$

$\rm \dfrac{1}{4\pi \epsilon_o}$ is the Coulomb's constant.

For the elecric force and the weight to be balanced,

$\rm F_e = W\\\dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}=W\\8.99\times 10^9\times \dfrac{3.50\times10^{-8}\times |q_2| }{0.0640^2}=2.70\times 10^{-2}\\|q_2| = \dfrac{2.70\times 10^{-2}\times 0.00640^2}{8.99\times 10^9\times 2.70\times 10^{-7}}=4.56\times 10^{-10}\ C.$

3 0

3 years ago