Finding acceleration= final speed-initial speed/time taken (or A=V-U\T)
Finial speed= 27.8s
Initial speed= 0s
Time taken= 5.15
So..
27.8-0/5.15= 5.40m/s (rounded to two decimal places)
Answer:
F_Balance = 46.6 N ,m' = 4,755 kg
Explanation:
In this exercise, when the sphere is placed on the balance, it indicates the weight of the sphere, when another sphere of opposite charge is placed, they are attracted so that the balance reading decreases, resulting in
∑ F = 0
Fe –W + F_Balance = 0
F_Balance = - Fe + W
The electric force is given by Coulomb's law
Fe = k q₁ q₂ / r₂
The weight is
W = mg
Let's replace
F_Balance = mg - k q₁q₂ / r₂
Let's reduce the magnitudes to the SI system
q₁ = + 8 μC = +8 10⁻⁶ C
q₂ = - 3 μC = - 3 10⁻⁶ C
r = 0.3 m = 0.3 m
Let's calculate
F_Balance = 5 9.8 - 8.99 10⁹ 8 10⁻⁶ 3 10⁻⁶ / (0.3)²
F_Balance = 49 - 2,397
F_Balance = 46.6 N
This is the balance reading, if it is calibrated in kg, it must be divided by the value of the gravity acceleration.
Mass reading is
m' = F_Balance / g
m' = 46.6 /9.8
m' = 4,755 kg
Answer:
B. the stars to come back to the same positions in the sky.
Explanation:
In fact, the solar day is equivalent to more than a rotation, because when the point has turned completely, it is not, as it should, in the same position with respect to the Sun.
The reason for this is that while performing the rotation, the Earth simultaneously moved following its orbit around the Sun.
When the reference point completed its rotation, the Earth already moved almost 2,500,000 km., So that to see the Sun again it will be necessary to turn a little more.
Solar day is more than a rotation. The sidereal or sidereal day, commonly used by astronomers, is also based on the rotation of the Earth; but in this case a distant star is taken as a reference (sidereal comes from the Latin sidus which means "star").
Heat required to change the phase of ice is given by
Q = m* L
here
m = mass of ice
L = latent heat of fusion
now we have
m = 45 kg
L = 334 KJ/kg
now by using above formula


In KJ we can convert this as

so the correct answer is D option
If the spaceship's Physicist happens to be hanging out of one side
of the ship, and he measures the speed of the photons as they pass
him and leave the ship, he'll see them passing him at 'c' ... the speed
of light.
When those photons pass somebody who happens to be in their
path, and he decides to measure their speed, he'll see them move
past him at 'c' ... the speed of light.
It doesn't matter whether the observer who measures them is
moving, or at what speed.
And it doesn't matter what source the photons come from, or
whether the source is moving, or at what speed.
And it doesn't matter what the photons' wavelength/frequency is ...
anything from radio to gamma rays.
The photons pass everybody at 'c' ... the speed of light.
Yes, I hear you. That can't be true. It's crazy.
Maybe it's crazy, but it's true.