All categories

3 years ago

Add me and ill add you back!! you have to bc i just gave you free coins :)

Physics

2 answers:

Rainbow [258]3 years ago

6 0

Answer:

yoooooooooooooooooooo

Explanation:

matrenka [14]3 years ago

3 0

Answer:

Okay

Explanation:

Ok now Ima sleep bye bye.

You might be interested in

Describe the sequence of events in the lithification of a sandstone

guajiro [1.7K]

Lithification is a diagenetic process in which loose sediment is converted to hard rocky compaction and cementation. So sediments (sand) are buried the increase in pressure from the weight of the overlaying material pushes the grains closer together. The volume of sediment is reduced and the fluids between the grains are also squeezed out. This leaves the sandstone tightly compressed <span>together this is lithification.</span>

6 0

3 years ago

Read 2 more answers

The magnetic fields in all planets are generated by the dynamo effect, caused by rapidly rotating and conducting material flowin

exis [7]

Answer:

Venus's very slow rotations rate does not allow a magnetic field to form

8 0

3 years ago

A spherical, non-conducting shell of inner radius = 10 cm and outer radius = 15 cm carries a total charge Q = 13 μC distributed

Vaselesa [24]

Answer:

E = 1580594.95 N/C

Explanation:

To find the electric field inside the the non-conducting shell for r=11.2cm you use the Gauss' law:

$\int EdS=\frac{Q_{in}}{\epsilon_o}$ (1)

dS: differential of the Gaussian surface

Qin: charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85 × 10-12 C2/N ∙ m2

The electric field is parallel to the dS vector. In this case you have the surface of a sphere, thus you have:

$\int EdS=ES=E(4\pi r^2)$ (2)

Qin is calculate by using the charge density:

$Q_{in}=V_{in}\rho=\frac{4}{3}(r^3-a^3)\rho$ (3)

Vin is the volume of the spherical shell enclosed by the surface. a is the inner radius.

The charge density is given by:

$\rho=\frac{Q}{V}=\frac{13*10^{-6}C}{\frac{4}{3}\pi((0.15m)^3-(0.10m)^3)}\\\\\rho=1.30*10^{-3}\frac{C}{m^3}$

Next, you use the results of (3), (2) and (1):

$E(4\pi r^2)=\frac{4}{3\epsilon_o}(r^3-a^3)\rho\\\\E=\frac{\rho}{3\epsilo_o}(r-\frac{a^3}{r^2})$

Finally, you replace the values of all parameters, and for r = 11.2cm = 0.112m you obtain:

$E=\frac{1.30*10^{-3}C/m^3}{3(8.85*10^{-12}C^2/Nm^2)}((0.112m)-\frac{(0.10)^3}{(0.112m)^2})\\\\E=1,580,594.95\frac{N}{C}$

hence, the electric field is 1580594.95 N/C

7 0

3 years ago

Adding heat to a liquid causes which of the following physical changes?

kow [346]

The density would decrease because the mass of an object deals with the amount of atoms in the object and since none of the object was reduced "a" wouldn't be the answer. Depending on the amount and period of time that the heat is applied the liquid could change into a gas so "d" wouldn't be correct. Density is the mass÷ volume, and when you add heat to an object it could take up different amounts of space because of its particles gaining energy and spreading apart. So the density would decrease because of the volume increasing. So I believe that "c" is the answer.

3 0

3 years ago

E) Thermal energy is released during

vichka [17]

Answer:

e) True, f) False

Explanation:

e) Let consider a close system, that is, a system with no mass interactions with surroundings. Then, we get the following expression by the First Law of Thermodynamics:

$Q_{net,in} - W_{net, out} = \Delta U$ (1)

Where:

$Q_{net, in}$ - Net input heat, measured in joules.

$W_{net, out}$ - Net output work, measured in joules.

$\Delta U$ - Change in thermal energy, measured in joules.

Please notice that work comprises all kind of work (i.e. mechanical, electric, magnetic), whereas heat comprises all heat interactions including chemical and radioactive phenomena.

If thermal energy is released, then $\Delta U < 0$ , which is caused by three scenarios:

(i) $Q_{net,in} < 0$ , $W_{net, out} < 0$ , $|Q_{net,in}|>|W_{net,out}|$

(ii) $Q_{net, in} > 0$ , $W_{net,out} > 0$ , $|Q_{net,in}|$

(iii) $Q_{net,in}< 0$ , $W_{net, out}>0$

In the case $Q_{net,in} > 0$ , $W_{net, out}$ , the thermal energy of the system is increased. Therefore, thermal energy is released during some energy conversions. Answer: True

f) A liquid solidifies when temperature goes below point of fusion, meaning a realease of heat with no work interactions. That is:

$Q_{net, in} = \Delta U$ , $Q_{net, in} < 0$ (2)

If $Q_{net, in} < 0$ , then $\Delta U < 0$ . Then, if a liquid absorbs heat energy, then thermal energy is increase and the liquid does not solidifies. Answer: False.

7 0

2 years ago

Add me and ill add you back!! you have to bc i just gave you free coins :)​

Add me and ill add you back!! you have to bc i just gave you free coins :)