Tension in the rod is responsible for balancing force due to gravity.
You need to know the mass number to work it out. I can see you have it but it's not visible on your photo.
Every proton weighs 1 and every neutron weighs 1 so if you know the total mass of the nucleus and the number of protons, then you can do:
Mass Number - Proton Number = Neutron Number
because the rest of the mass (that isn't from the protons) must come from the neutrons. Make sense?
Answer:
823.46 kgm/s
Explanation:
At 9 m above the water before he jumps, Henri LaMothe has a potential energy change, mgh which equals his kinetic energy 1/2mv² just as he reaches the surface of the water.
So, mgh = 1/2mv²
From here, his velocity just as he reaches the surface of the water is
v = √2gh
h = 9 m and g = 9.8 m/s²
v = √(2 × 9 × 9.8) m/s
v = √176.4 m/s
v₁ = 13.28 m/s
So his velocity just as he reaches the surface of the water is 13.28 m/s.
Now he dives into 32 cm = 0.32 m of water and stops so his final velocity v₂ = 0.
So, if we take the upward direction as positive, his initial momentum at the surface of the water is p₁ = -mv₁. His final momentum is p₂ = mv₂.
His momentum change or impulse, J = p₂ - p₁ = mv₂ - (-mv₁) = mv₂ + mv₁. Since m = Henri LaMothe's mass = 62 kg,
J = (62 × 0 + 62 × 13.28) kgm/s = 0 + 823.46 kgm/s = 823.46 kgm/s
So the magnitude of the impulse J of the water on him is 823.46 kgm/s
Answer: 40m/s
Explanation:
Given that,
speed of car (v) = ?
Mass of car (m) = 500.0 kilogram
Kinetic energy (KE) = 400,000 joules
Recall that kinetic energy is possessed by a moving object, and it depends on the mass and speed by which the object moves
i.e Kinetic energy = 1/2mv²
400,000J = 1/2 x 500.0kg x v²
400,000J = 250.0kg x v²
v² = 400,000J / 250.0kg
v² = 1600
Find the square root of v²
v = √1600
v = 40m/s
Thus, the speed of the car is 40m/s
Answer:
a body weigh more at the equator since earth is not completely round it is an oblate spheroid – a sphere with a bulge around the equator. that is why the places at the equator are more in contact with the core rather than the places at the poles.
