If a ball is 10m high with what velocity will it fall?
1 answer:
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(a)
The velocity of the meteorite just before hitting the ground is:
The loss of energy of the meteorite corresponds to the kinetic energy the meteorite had just before hitting the ground, so:
(b) 1 megaton of tnt is equal to
To find to how many megatons the meteorite energy loss
corresponds, we can set the following proportion
And so we find
So, 0.162 megatons.
(c) 1 Hiroshima bomb is equivalent to 13 kilotons (13 kT). The impact of the meteorite had an energy of
. So, to find to how many hiroshima bombs it corresponds, we can set the following proportion:
And so we find
So, the energy released by the impact of the meteorite corresponds to the energy of 12.46 hiroshima bombs.
The velocity of the meteorite just before hitting the ground is:
The loss of energy of the meteorite corresponds to the kinetic energy the meteorite had just before hitting the ground, so:
(b) 1 megaton of tnt is equal to
To find to how many megatons the meteorite energy loss
corresponds, we can set the following proportion
And so we find
So, 0.162 megatons.
(c) 1 Hiroshima bomb is equivalent to 13 kilotons (13 kT). The impact of the meteorite had an energy of
And so we find
So, the energy released by the impact of the meteorite corresponds to the energy of 12.46 hiroshima bombs.
Answer:
P = 9622.9 Pa = 9.62 KPa
Explanation:
First, we will calculate the mass of all three liquids:
m = ρV
where,
m = mass of liquid
ρ = density of liquid
V = Volume of liquid
FOR LIQUID 1:
m₁ = (2.8 x 10³ kg/m³)(2 x 10⁻³ m³) = 5.6 kg
m₂ = (1 x 10³ kg/m³)(1.5 x 10⁻³ m³) = 1.5 kg
m₃ = (0.6 x 10³ kg/m³)(1 x 10⁻³ m³) = 0.6 kg
The total mass will be:
m = m₁ + m₂+ m₃ = 5.6 kg + 1.5 kg + 0.6 kg
m = 7.7 kg
Hence, the weight of the liquids will be:
W = mg = (7.7 kg)(9.81 m/s²) = 75.54 N
Now, we calculate the base area:
A = πr² = π(0.05 m)²
A = 7.85 x 10⁻³ m²
Now the pressure will be given as:
<u>P = 9622.9 Pa = 9.62 KPa</u>