Answer:
the warm air would be up and cool would be down
Explanation:
warm air is less dense and cool air is more dense
Answer:
A) 350 N
B) 58.33 N
C) 35 kg
D) 35 kg
Explanation:
If we use that g = 10 m/s^2, then the acceleration of gravity on the Moon will be 10/6 m/s^2 = 5/3 m/s*2
The weight of the object on Earth is given by:
Weight = mass * g = 35 * 10 = 350 N
The weight of the object on the Moon:
Weight = mass * gmoon = 35 * 5/3 = 58.33 N
The mass of the object on Earth is 35 kg
The mass of the object on the Moon is exactly the same as on the Earth (35 kg) since the mass is a quantity inherent to the object and not to its location.
Answer:
576 joules
Explanation:
From the question we are given the following:
weight = 810 N
radius (r) = 1.6 m
horizontal force (F) = 55 N
time (t) = 4 s
acceleration due to gravity (g) = 9.8 m/s^{2}
K.E = 0.5 x MI x ω^{2}
where MI is the moment of inertia and ω is the angular velocity
MI = 0.5 x m x r^2
mass = weight ÷ g = 810 ÷ 9.8 = 82.65 kg
MI = 0.5 x 82.65 x 1.6^{2}
MI = 105.8 kg.m^{2}
angular velocity (ω) = a x t
angular acceleration (a) = torque ÷ MI
where torque = F x r = 55 x 1.6 = 88 N.m
a= 88 ÷ 105.8 = 0.83 rad /s^{2}
therefore
angular velocity (ω) = a x t = 0.83 x 4 = 3.33 rad/s
K.E = 0.5 x MI x ω^{2}
K.E = 0.5 x 105.8 x 3.33^{2} = 576 joules
Answer:
A. Zero
Explanation:
Given data,
The charge of the test charge, q = 1 C
The distance the charge moved against the filed of intensity, x = 30 cm
= 0.3 m
The electric field intensity, E = 50 N/C
The energy stored in the charge at 0.3 m is given by the formula,
V = k q/r
Where,
= 9 x 10⁹ Nm²C⁻²
The charge is moved from the potential V₁ to V₂ at 30 cm
Substituting the given values in the above equation
V₁ = 9 x 10⁹ x 30 / 0.3
= 1.5 x 10¹² J
And,
V₂ = 1.5 x 10¹² J
The energy stored in it is,
W = V₂ - V₁
= 0
Hence, the energy stored in the charge is, W = 0
