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Ber [7]
3 years ago
15

Does water expand more as the temperature drops even more below 0 deg? I am worried about the coolant in my rad. as the temp.is

40deg below.
Physics
1 answer:
alex41 [277]3 years ago
4 0
Water is, indeed, an exception. Normally when temperature drops, material shrinks. Water doesn't, water expands instead.
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A student is preparing to bake a cake at home and places several ingredients on the countertop to use. First he preheats the ove
Evgen [1.6K]

Answer:

baking the cake batter

Explanation:

Baking the cake batter will indicate that chemical change has occurred here. What is a chemical change?

  • A chemical change is one in which a new kind of matter is formed.
  • It is usually accompanied by energy either evolution or absorption of energy in form of heat or light or both.
  • The process is irreversible.
  • When the batter bakes, a new substance different from the cake mix is obtainable.
  • We cannot get back the ingredient from this baked cake. It is impossible.
  • This is good indicator of chemical change.
7 0
3 years ago
A 20.0-N weight slides down a rough inclined plane which makes an angle of 30 degree with the horizontal. The weight starts from
Ulleksa [173]

Answer:

1270.64\ \text{J}

Explanation:

m = Mass of object = \dfrac{mg}{g}

mg = Weight of object = 20 N

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

v = Final velocity = 15 m/s

u = Initial velocity = 0

d = Distance moved by the object = 150 m

\theta = Angle of slope = 30^{\circ}

f = Force of friction

fd = Work done against friction

The force balance of the system is

\dfrac{1}{2}m(v^2-u^2)=(mg\sin\theta-f)d\\\Rightarrow \dfrac{1}{2}mv^2=mg\sin\theta d-fd\\\Rightarrow fd=mg\sin\theta d-\dfrac{1}{2}mv^2\\\Rightarrow fd=20\times \sin 30^{\circ}\times 150-\dfrac{1}{2}\times \dfrac{20}{9.81}\times 15^2\\\Rightarrow fd=1270.64\ \text{J}

The work done against friction is 1270.64\ \text{J}.

8 0
3 years ago
A proton is moving in a circular orbit of radius 12 cm in a uniform 0.31-T magnetic field perpendicular to the velocity of the p
PIT_PIT [208]

Answer:

Explanation:

A proton of charge

q=+1.609×10^-19C

Orbit a radius of 12cm

r=0.12m

Magnetic Field of 0.31T

Angle between velocity and field is 90°

a. Because the magnetic force F supplies the centripetal force Fc.

The magnitude of the magnetic force F on a charge q moving at a speed v in a magnetic field of strength B is given by

F = qvB sin θ

And the centripetal force is given as

Fc=mv²/r

Where m is mass of proton

m=1.673×10^-27kg

Then, F=Fc

qvB sin θ=mv²/r

qBSin90=mv/r

rqB=mv

Then, v=rqB/m

v=0.12×1.609×10^-19×0.31/1.673×10^-23

v=3577692.78m/s

v=3.58×10^6m/s

b. Since,

F=qVBSin90

F=1.609×10^-19×3.58×10^6×0.31

F=1.785×10^-13 N.

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3 years ago
What force is needed to propel a 421-kg car with an acceleration of 3.77 m/s2?
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What is a successful result of science?
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A discovery not the others
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2 years ago
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