All categories

3 years ago

A woman holds a frozen smoothie in her hand on a warm

Physics

2 answers:

GaryK [48]3 years ago

6 0

Answer:

Explanation:

The thermal energy from her hand will go up into the smoothie.

nata0808 [166]3 years ago

3 0

Answer:

C bois

Explanation:

You might be interested in

PLEASE HELP MEE <br> Read the question below and pick the correct answer.

wel

I would, but what is the question?

5 0

2 years ago

Read 2 more answers

What are the stages of stellar evolution

Rudik [331]

Giant Molecular Cloud

Protostar

T-Tauri

Main Sequence

Subgiant, Red Giant, Supergiant

Core fusion

Red Giant, Supergiant

Planetary Nebula or Supernova or if they are big enough a black hole

Remnant

3 0

4 years ago

Read 2 more answers

a) If a proton moved from a location with a 5.0 V potential to a location with 7.5 V potential, would its potential energy incre

tekilochka [14]

Answer:

A. The potential energy of the positively charge proton from 5v to 7.5v will increase because in an electric field work is done on the proton to move it to a point of higher potential

B. when an electron is moved from 7.5v to 5v it loses potential energy due to the charges lost so as to bring it to a point of lower potential

Explanation:

3 0

3 years ago

Please someone derive 3rd equation of motion u- v= 2sa

zlopas [31]

We know

$\\ \sf\longmapsto a=\dfrac{dv}{dt}$

$\\ \sf\longmapsto a=\dfrac{dv}{dx}.\dfrac{dx}{dt}$

$\\ \sf\longmapsto a=v\dfrac{dv}{dx}$

$\\ \sf\longmapsto adx=vdv$

Integrate

$\\ \sf\longmapsto a{\displaystyle{\int}^x_{x_0}}dx=\displaystyle{\int}_u^v vdv$

$\\ \sf\longmapsto a(x-x_0)=\dfrac{v^2-u^2}{2}$

Here

x-x_0=s

$\\ \sf\longmapsto v^2-u^2=2as$

6 0

3 years ago

Read 2 more answers

A charged particle enters into a uniform magnetic field such that its velocity vector is perpendicular to the magnetic field vec

Anit [1.1K]

Answer:

The charged particle will follow a circular path.

Explanation:

The magnetic force exerted on the charged particle due to the magnetic field is given by:

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the particle

B is the magnetic field

$\theta$ is the angle between v and B

In this problem, the velocity is perpendicular to the magnetic field, so $\theta = 90^{\circ}, sin \theta=1$ and the force is simply

$F=qvB$

Moreover, the force is perpendicular to both B and v, according to the right-hand rule. Therefore, we have:

- a force that is always perpendicular to the velocity, v

- a force which is constant in magnitude (because the magnitude of v or B does not change)

--> this means that the force acts as a centripetal force, so it will keep the charged particle in a uniform circular motion. So, the correct answer is

The charged particle will follow a circular path.

8 0

4 years ago