Answer:
it's right you did a great job
Answer:
Explanation:We should know that weight = mass * gravity.
That is weight equals mass times gravity.
Gravity is a force of attraction between any two bodies in the universe. It is directly proportional to product of their masses and inversely proportional to the square of the distance between them.
Gravity is generally measured in terms of acceleration due to gravity, denoted as g. For Earth it is, 9.8 m/s². And for moon, it is about 1.62 m/s².
On Earth, your weight is 70 kg = W
W = mass x 9.8
70 = mass x 9.8
Your mass is 70/ 9.8
i.e approximately 7.14
Weight at the Moon, W' = 7.14 x 1.62
Hence, your weight on the surface of the moon is just 11.56 kg.
Congratulations, you've lost about 58.14 kilograms without any hard exercise. And you're as light as a Sweedish Vallhund! Cheers!
Answer:
Δμ = hΔf/B
Explanation:
If the photon energy , ΔE = hΔf where Δf = small frequency shift and since the potential energy change of the magnetic dipole moment μ in magnetic field B from parallel to anti-parallel state is ΔU = ΔμB. where Δμ = small shift in magnetic moment.
Since the magnetic energy change equals the photon energy,
ΔE = ΔU
hΔf = ΔμB
Δμ = hΔf/B
The correct answer is C) becuse without certain medicine they will dieeeee
Answer:
The tank is losing

Explanation:
According to the Bernoulli’s equation:
We are being informed that both the tank and the hole is being exposed to air :
∴ P₁ = P₂
Also as the tank is voluminous ; we take the initial volume
≅ 0 ;
then
can be determined as:![\sqrt{[2g (h_1- h_2)]](https://tex.z-dn.net/?f=%5Csqrt%7B%5B2g%20%28h_1-%20h_2%29%5D)
h₁ = 5 + 15 = 20 m;
h₂ = 15 m
![v_2 = \sqrt{[2*9.81*(20 - 15)]](https://tex.z-dn.net/?f=v_2%20%3D%20%5Csqrt%7B%5B2%2A9.81%2A%2820%20-%2015%29%5D)
![v_2 = \sqrt{[2*9.81*(5)]](https://tex.z-dn.net/?f=v_2%20%3D%20%5Csqrt%7B%5B2%2A9.81%2A%285%29%5D)
as it leaves the hole at the base.
radius r = d/2 = 4/2 = 2.0 mm
(a) From the law of continuity; its equation can be expressed as:
J = 
J = πr²
J =
J =
b)
How fast is the water from the hole moving just as it reaches the ground?
In order to determine that; we use the relation of the velocity from the equation of motion which says:
v² = u² + 2gh
₂
v² = 9.9² + 2×9.81×15
v² = 392.31
The velocity of how fast the water from the hole is moving just as it reaches the ground is : 
