What does the principal quantum number determine?

What is the weight of an object when the object has a mass of 22kg

Savatey [412]

Assuming the object is on earth the objects weight would be equal to its mass multiplied by the gravitational field constant

mass=22kg
g=9.80665N/kg

weight=(22 kg) (9.80665 N/kg)=215.7463N

generally g is rounded to be 10 N/kg so for any question where it asks the weight given the mass just multiply by 10 and that should suffice. In this case the answer would be 220 N

5 0

3 years ago

How do you think car makers can design cars to limit cell phone distractions?

Dafna11 [192]

I have two (2) brilliant ideas:

1). Inside the metal that the body of the car is made of, and also between the two sheets of glass that the windows are made of, install a thin layer of material that absorbs RF (radio-wave) energy . . . like the material in the glass window of your microwave oven. Then, no radio waves from the cellular base station can get INTO the car, and no radio waves from your phone can get OUT of the car. The phone can't make a connection to the cellular network, you can't make or receive calls, and you can't connect to Instagram or Brainly, so you might as well just turn it off and save your battery until next time you're outside your car.

2). Somewhere inside the car, like under the dash or in the glove box, install a teeny tiny radio receiver that can recognize the signals coming OUT of your phone. Connect it to the car's electrical system so that when it hears signals from phones inside the car, it it shuts down the car's motor so you can't start or drive. The car only works when phones inside the car are either turned off or in Airplane Mode.

My ideas are so brilliant that I really should patent them, or copyright them, or whatever you do so that other people have to pay you to use your idea. But if you want to use them, that's OK. Just go ahead. I won't mind.

8 0

3 years ago

If earth's mass were half its actual value but its radius stayed the same, the escape velocity of earth would be:________

siniylev [52]

If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be $V_e = \sqrt{\dfrac{GM}{r}}$ .

<h3>What is an escape velocity?</h3>

The ratio of the object's travel distance over a specific period of time is known as its velocity. As a vector quantity, the velocity requires both the magnitude and the direction. the slowest possible speed at which a body can break out of the gravitational pull of a certain planet or another object.

The formula to calculate the escape velocity of earth is given below:-

$V_e=\sqrt{\dfrac{2GM}{r}}$

Given that earth's mass was half its actual value but its radius stayed the same. The escape velocity will be calculated as below:-

$V_e=\sqrt{\dfrac{2GM}{r\times 2}}$

$V_e = \sqrt{\dfrac{GM}{r}}$ .

Therefore, If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be $V_e = \sqrt{\dfrac{GM}{r}}$ .

To know more about escape velocity follow

brainly.com/question/14042253

#SPJ4

8 0

1 year ago

A mass of 1 slug, when attached to a spring, stretches it 2 feet and then comes to rest in the equilibrium position. Starting at

Vesna [10]

Answer:

$Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t$

Explanation:

Given that m= 1 slug and given that spring stretches by 2 feet so we can find the spring constant K

mg=k x

1 x 32= k x 2

K=16

And also give that damping force is 8 times the velocity so damping constant C=8.

We know that equation for spring mass system

my''+Cy'+Ky=F

Now by putting the values

1 y"+8 y'+ 16y=6 cos 4 t ----(1)

The general solution of equation Y=CF+IP

Lets assume that at steady state the equation of y will be

y(IP)=A cos 4t+ B sin 4t

To find the constant A and B we have to compare this equation with equation 1.

Now find y' and y" (by differentiate with respect to t)

y'= -4A sin 4t+4B cos 4t

y"=-16A cos 4t-16B sin 4t

Now put the values of y" , y' and y in equation 1

1 (-16A cos 4t-16B sin 4t)+8( -4A sin 4t+4B cos 4t)+16(A cos 4t+ B sin 4t)=6sin4 t

So by comparing the coefficient both sides

-16A+32B+16A=0 So B=0

-16 B-32 A+16B=6 So A=-3/16

y=-3/16 cos 4t

Now to find the CF of differential equation 1

y"+8 y'+ 16y=6 cos 4 t

Homogeneous version of above equation

$m^2+8m+16=0$

So $CF =(C_1+tC_2)e^{-2t}$

So the general equation

$Y=(C_1+tC_2)e^{-2t}-3/16 cos 4t$

Given that t=0 Y=0 So

$C_1=\dfrac{3}{16}$

t=0 Y'=0 So

$C_2 =\dfrac{3}{8}$

$Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t$

The above equation is the general equation for motion.

3 0

3 years ago

A 25-kg wagon has a momentum of 300kg m/s. What is it’s acceleration?

arsen [322]

The answer is 12 ....

8 0

3 years ago