All categories

2 years ago

Which of the following statements about the force on a charged particle due to a magnetic field are not valid?

1 answer:

5 0

The correct answer is "None of the above; all of these statements are valid." All the statements namely, it depends on the particle's charge, it depends on the strength of the external magnetic field, it depends on the particle's velocity, and it acts at right angles to the direction of the particle's motion are all valid. Thank you for posting your question. I hope this answer helped you. Let me know if you need more help.

You might be interested in

How does the movement of particles of matter change when temperature decreases

Zanzabum

INCREASE in temperature of the material practically increase the energy of the particles. which increases their motion due to increase in energy . thus when the temperature is decreased the energy level decreases which causes the particle's motion to slow down.. the motion of the particle is highly reduced when the temperature is lowered

4 0

3 years ago

An atom with more or less neutrons than expected is a what ?

salantis [7]

Answer:

Isotope it will have a different number of neutrons than normal

4 0

1 year ago

Radiant heat makes it impossible to stand close to a hot lava flow. Calculate the rate of heat loss by radiation from 1.00 m^2 o

VARVARA [1.3K]

The rate of heat loss by radiation is equal to <u>-207.5kW</u>

Why?

To calculate the heat loss rate (or heat transfer rate) by radiation, from the given situation, we can use the following formula:

$HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )$

Where,

E, is the emissivity of the body.

A, is the area of the body.

T, are the temperatures.

S, is the Stefan-Boltzmann constant, which is equal to:

$5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }$

Now, before substitute the given information, we must remember that the given formula works with absolute temperatures (Kelvin), so, we need to convert the given values of temperature from Celsius degrees to Kelvin.

We know that:

$K=Celsius+273.15$

So, converting we have:

$T_{1}=1110\°C+273.15=1383.15K\\\\T_{2}=36.2\°C+273.15=309.35K$

Therefore, substituting the given information and calculating, we have:

$HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )$

$HeatLossRate=1*5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }*1m^{2} *((309.35K)^{4} -(1383.15})^{4} )\\\\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(95697.42K^{4} -3.66x10^{12}K^{4})\\ \\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(-3.66x10^{12} K^{4})=-207522W=-207.5kW$

Hence, we have that the rate of heat loss is equal to -207.5kW.

8 0

2 years ago

An unknown galaxy has a large flattened core. Which of the following classifications would best fit this galaxy's description? I

Alja [10]

Answer:

Spiral

i know it is so dont say nun people

Explanation:

7 0

2 years ago

Read 2 more answers

If at 10m above the ground an object has 50J of Kinetic Energy, with 50J of Potential Energy. How high can the object travel?

steposvetlana [31]

Hi there!

$\large\boxed{\text{B) 20 meters}}$

We know that:

$E_T = U + K$

U = Potential Energy (J)

K = Kinetic Energy (J)

E = Total Energy (J)

At 10m, the total amount of energy is equivalent to:

U + K = 50 + 50 = 100 J

To find the highest point the object can travel, K = 0 J and U is at a maximum of 100 J, so:

100J = mgh

We know at 10m U = 50J, so we can solve for mass. Let g = 10 m/s².

50J = 10(10)m

m = 1/2 kg

Now, solve for height given that E = 100 J:

100J = 1/2(10)h

100J = 5h

<u>h = 20 meters</u>

3 0

1 year ago