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rusak2 [61]
3 years ago
13

How does the mass of an electron compare to the mass of a proton?

Physics
1 answer:
harina [27]3 years ago
6 0

Answer:

The mass of an electron is only about 1/2000 the mass of a proton.

Explanation:

You might be interested in
How is air resistance similar to gravity? give me two ways.
OLga [1]

Answer:

1. they both act on an object in free fall

Explanation:

2. both help determine how fast the object will accelerate

4 0
2 years ago
An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary
Talja [164]

Answer:

A) 5.2 x 10³ N

B) 8.8 x 10³ N

Explanation:

Part A)

F_{g} = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

T = Tension force in upward direction

F_{d} = Drag force in upward direction = 1800 N

Force equation for the motion of craft is given as

F_{g} - F_{d} - T = 0

7000 - 1800 - T = 0

T = 5200 N

T = 5.2 x 10³ N

Part B)

F_{g} = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

T = Tension force in upward direction

F_{d} = Drag force in downward direction = 1800 N

Force equation for the motion of craft is given as

T  - F_{g} - F_{d} = 0

T - 7000 - 1800  = 0

T = 8800 N

T = 8.8 x 10³ N

4 0
3 years ago
How does tightening a string on a guitar affect its natural frequency?
iragen [17]
By tightening a string you are actually putting more stress on the string you are giving it a new frequency that isn't natural.

Hope this helps
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3 0
3 years ago
Read 2 more answers
video for A door 1 m wide, of mass 15 kg, is hinged at one side so that it can rotate without friction about a vertical axis. It
natka813 [3]

Answer:

\omega_f = 0.4\ rad/s

Explanation:

given,

width of door dimension  = 1 m

mass of door = 15 Kg

mass of bullet = 10 g = 0.001 Kg

speed of bullet = 400 m/s

I_{total} =I_{door} + I_{bullet}

I_{total} =\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2

a) from conservation of angular momentum  

L_i = L_f  

mv\dfrac{W}{2} = I_{total}\omega_f  

mv\dfrac{W}{2}= (\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2)\omega_f

\omega_f= \dfrac{mv\dfrac{W}{2}}{\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2}

\omega_f= \dfrac{\dfrac{mv}{2}}{\dfrac{MW }{3}+(\dfrac{mW}{4})}

\omega_f= \dfrac{\dfrac{0.01\times 400}{2}}{\dfrac{15\times 1 }{3}+(\dfrac{0.01\times 1}{4})}

\omega_f = 0.4\ rad/s

8 0
3 years ago
Compared to a plane gravitational pull on the ground to 7 miles in sky is what
Citrus2011 [14]
When you're in an airplane that's 7 miles up off the ground, the strength of gravity plunges to only 99.6 percent of its strength all the way down on the ground. A big heavy person, who weighs 200 pounds down at the airport, weighs only 199 pounds 4.7 ounces in a plane at the altitude of 7 miles.
4 0
2 years ago
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