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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
A graduated cylinder.measures 15.3 mL. Convert this measurement to DaL
ololo11 [35]

Answer:

0.000153DaL  

Explanation:

We have been given:

         15.3mL to convert to DaL

DaL is a unit of volume which indicates a decaliter.

 This implies that;

             1 Da L  = 1 x 10²L

So:

               1 mL  = 1 x 10⁻³L

       So 15.3mL will give 15.3 x 10⁻³L

So;

           1 x 10²L   =  1 DaL  

      15.3 x 10⁻³L  will give \frac{15.3 x 10^{-3} }{1 x 10^{2} }   = 15.3 x 10⁻⁵DaL

Therefore, this is 0.000153DaL  

           

5 0
3 years ago
Kate must do 39 days of training each year as a naval reservist. How will she serve that time?
elixir [45]

Answer:

c

Explanation:

i did the test

3 0
3 years ago
A racing car is travelling at 70 m/s and accelerates at -14 m/s2. What would the car’s speed be after 3 s?
Leno4ka [110]
<h3><u>Question</u><u>:</u></h3>

A racing car is travelling at 70 m/s and accelerates at -14 m/s^2. What would the car’s speed be after 3 s?

<h3><u>Statement:</u></h3>

A racing car is travelling at 70 m/s and accelerates at -14 m/s^2.

<h3><u>Solution</u><u>:</u></h3>
  • Initial velocity (u) = 70 m/s
  • Acceleration (a) = -14 m/s^2
  • Time (t) = 3 s
  • Let the velocity of the car after 3 s be v m/s
  • By using the formula,

v = u + at, we have

v = 70 + ( - 14)(3) \\  =  > v = 70 - 42 \\  =  > v = 28

  • So, the velocity of the car after 3 s is 28 m/s.
<h3><u>Answer:</u></h3>

The car's speed after 3 s is 28 m/s.

Hope it helps

3 0
3 years ago
Balance the equation​
mote1985 [20]
Is this a chemistry question ?
7 0
3 years ago
I have a question regarding friction in rolling without slipping.
Solnce55 [7]

Explanation:

They probably put "rolls without slipping" in there to indicate that there is no loss in friction; or that the friction is constant throughout the movement of the disk. So it's more of a contingency part of the explanation of the problem.

(Remember how earlier on in Physics lessons, we see "ignore friction" written into problems; it just removes the "What about [ ]?" question for anyone who might ask.)

In this case, you can't ignore friction because the disk wouldn't roll without it.

As far as friction producing a torque... I would say that friction is a result of the torque in this case. And because the point of contact is, presumably, the ground, the friction is tangential to the disk. Meaning the friction is linear and has no angular component.

(You could probably argue that by Newton's 3rd Law there should be some opposing torque, but I think that's outside of the scope of this problem.)

Hopefully this helps clear up the misunderstanding for you.

4 0
3 years ago
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