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Yuliya22 [10]
2 years ago
7

Why is the level of glycerin initially decreases when heated ?​

Physics
1 answer:
Mila [183]2 years ago
4 0
Because glucerin is hotter so it decreases faster
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A car goes from 5 m/s to 25 m/s in 6 s. What is the acceleration of the car?
damaskus [11]

Answer:

5m/s

Explanation:

3 0
3 years ago
a bus starting from rest moves with a uniform acceleration of 0.1 metre per second square for 2 minutes find the speed acquired
blondinia [14]
S ?
U 0m/s
V ?
A 0.1m/s^2
T 2min (120 sec)

S=ut+0.5at^2
S=0(120 sec)+0.5(0.1m/s^2)(120 sec)^2
S=720m

Distance double 720m*2=1440m

V^2=u^2+2as
V^2=(0)^2+2(0.1 m/s^2)(1440m)
V^2=288
V= square root of 288=12 root 2=16.97 to 2 decimal places
6 0
3 years ago
a ball of mass 16 kg on the end of a string is spun at a constant speed of 2 m/s in a horizontal circle with a radius of 1m. Wha
miss Akunina [59]

The work done by the centripetal force during om complete revolution is 401.92 J.

<h3>What is centripetal force?</h3>

Centripetal force is a force that acts on a body undergoing a circular motion and is directed towards the center of the circle in which the body is moving.

To Calculate the work done by the centripetal force during one complete revolution, we use the formula below.

Formula:

  • W = (mv²/r)2πr
  • W = 2πmv²................... Equation 1

Where:

  • W = Work done by the centripetal force
  • m = mass of the ball
  • v = velocity of the ball
  • π = pie

From the question,

Given:

  • m = 16 kg
  • v = 2 m/s
  • π = 3.14

Substitute these values into equation 1

  • W = 2×3.14×16×2²
  • W = 401.92 J

Hence, The work done by the centripetal force during om complete revolution is 401.92 J.
Learn more about centripetal force here: brainly.com/question/20905151

5 0
2 years ago
Which phrase best defines topography?
serious [3.7K]

Answer:

mapping the land

Explanation:

4 0
3 years ago
If an electron moves in a circle of radius 21 cm perpendicular to a B field of 0.4 T, what are the speed of the electron and the
kodGreya [7K]

Answer:

a)

v = 4.048 *10^6 m/s

b)  

Angular frequency =  1.92 * 10^7

Explanation:

As we know

v =  \frac{qBr}{m}

q is the charge on the electron = 3.2 * 10^{-19} C

B is the magnetic field in Tesla = 0.4 T

r is the radius of the circle = 0.21 m

mass of the electrons = 6.64 * 10^{-27} Kg

a)

Substituting the given values in above equation, we get -

v = \frac{3.2 * 10^{-19}*0.4*0.21}{6.64 * 10^{-27}} \\v = 4.048 *10^6m/s

b)  

Angular frequency =

\frac{4.048 * 10^6 }{0.21} \\1.92 * 10^7

8 0
2 years ago
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