V = 8 * 10^2 km/h = 800km/h
S= 1,8* 10^3 km = 1800km
t = ?
v = S/t
t = S/v
t = 1800km/ 800km/h
t ≈ 2,25h (135min)
Answer:
Explanation:
Gravitational law states that, the force of attraction or repulsion between two masses is directly proportional to the product of the two masses and inversely proportional to the square of their distance apart.
So,
Let the masses be M1 and M2,
F ∝ M1 × M2
Let the distance apart be R
F ∝ 1 / R²
Combining the two equation
F ∝ M1•M2 / R²
G is the constant of proportional and it is called gravitational constant
F = G•M1•M2 / R²
So, to increase the gravitational force, the masses to the object must be increased and the distance apart must be reduced.
So, option c is correct
C. Both objects have large masses and are close together.
-- If the field were inclined to the surface, then it would have
some component parallel to the surface.
-- Then, since we're talking about a conductor, the charges
on the object would move in response to that component
of the field, until there was no longer any component of the
field trying to move them.
Answer:
a) Diffusion coefficient, D = 1.5 in/hr
b) Mean jump frequency, f = 0.0833 Hz
Explanation:
a) The relationship between the diffusion coefficient, time and mean displacement and can be given by the expression:
..........(1)
Where <r> = mean displacement
D = Diffusion coefficient
t = time = 12 hrs
sum of the squares of the distance divided by 100 is 36 in2.
<r>²= 36 in²
Substituting these values into equation (1) above
b) Mean jumping distance, <r> = 0.1 inches
Applying equation (1) again
Where D = 1.5 in/hr
The mean jump frequency, f = 1/t
f = 1/12
f = 0.0833 Hz
Answer:
Pressure of woman will be
Pressure of the elephant will be
Explanation:
We have given that mass of the woman m = 80 kg
Acceleration due to gravity
Diameter of shoes = 1 cm =0.01 m
So radius
So area
We know that force is given F = mg
So
Now we know that pressure is given by
Now mass of elephant m = 5500 kg
So force of elephant = 5500×9.8 = 53900 N
Diameter = 20 cm
So radius r = 10 cm
So area will be
So pressure will be