Answer:
D- Repulsion
Explanation:
A positively charged object will exert a repulsive force upon a second positively charged object.
Answer:
563.86 N
Explanation:
We know the buoyant force F = weight of air displaced by the balloon.
F = ρgV where ρ = density of air = 1.29 kg/m³, g = acceleration due to gravity = 9.8 m/s² and V = volume of balloon = 4πr/3 (since it is a sphere) where r = radius of balloon = 2.20 m
So, F = ρgV = ρg4πr³/3
substituting the values of the variables into the equation, we have
F = 1.29 kg/m³ × 9.8 m/s² × 4π × (2.20 m)³/3
= 1691.58 N/3
= 563.86 N
Given that,
Mass of a tribble, m = 2.5 kg
Radius, r = 1.4 m
The force on the tribble from the bucket does not exceed 10 times its weight.
To find,
The maximum tangential speed.
Solution,
The force acting on the tribble is equal to the centripetal force.
F = 10mg
The formula for the centripetal force is given by :
v is maximum tangential speed
So, the maximum tangential speed is 11.7 m/s.
Answer:
garden
Explanation: All the other habitats are aquatic
Answer:
the velocity of the bullet-wood system after the collision is 2.48 m/s
Explanation:
Given;
mass of the bullet, m₀ = 20 g = 0.02 kg
velocity of the bullet, v₀ = 250 m/s
mass of the wood, m₁ = 2 kg
velocity of the wood, v₁ = 0
Let the velocity of the bullet-wood system after collision = v
Apply the principle of conservation of linear momentum to calculate the final velocity of the system;
Initial momentum = final momentum
m₀v₀ + m₁v₁ = v(m₀ + m₁)
0.02 x 250 + 2 x 0 = v(2 + 0.02)
5 + 0 = v(2.02)
5 = 2.02v
v = 5/2.02
v = 2.48 m/s
Therefore, the velocity of the bullet-wood system after the collision is 2.48 m/s
