Answer:
Explanation:
Normal length of spring = 28.3 cm
stretched length of spring = 38.2 cm
length of extension = 38.2 - 28.3 = 9.9 cm
= 9.9 x 10⁻² m
force applied to stretch = .55 x 9.8 ( mg )
= 5.39 N
Force constant = force applied / extension
= 5.39 / 9.9 x 10⁻²
= .5444 x 10² N /m
= 54.44 N/m
Answer:
C. Add all the force vectors
Explanation:
The net force acting on an object is the vector sum of all the forces on the object.
Remember, Newton's first law tells us a body at rest will remain at rest or that in uniform motion will continue in motion unless acted by unbalanced forces.These unbalanced forces act in all direction towards the body thus to get the net force you require a summation of all these force with respect to their magnitudes and directions.
For example a force of 3N towards the East direction acting on a body and another force of 2N towards the West direction on the same body will generate a net force of 1N towards the East direction.
Answer:
Explanation:
If we assume there is a sharp boundary between the two masses of air, there will be a refraction. The refractive index of each medium will depend on the relative speeds of light.
n = c / v
If light travels faster in warmer air, it will have a lower refractive index
nh < nc
Snell's law of refraction relates angles of incidence and refracted with the indexes of refraction:
n1 * sin(θ1) = n2 * sin(θ2)
sin(θ2) = sin(θ1) * n1/n2
If blue light from the sky passing through the hot air will cross to the cold air, then
n1 = nh
n2 = nc
Then:
n1 < n2
So:
n1/n2 < 1
The refracted light will come into the cold air at angle θ2 wich will be smaller than θ1, so the light is bent upwards, creating the appearance of water in the distance, which is actually a mirror image of the sky.
Answer:

Explanation:
When an amount of energy Q is supplied to a substance of mass m, the temperature of the substance increases by
, according to the equation

where
is the specific heat capacity of the substance.
In this problem, we have:
is the amount of heat supplied to the sample of gold
m = 0.1 kg = 100 g is the mass of the sample
is the specific heat capacity of gold
Solving for
, we find the change in temperature

And since the final temperature was

The initial temperature was
