Answer:
Explanation:
Calories to be burnt = 3500 - 2500 = 1000 Cals .
Efficiency of conversion to mechanical work is 25 % .
Work needed to burn this much of Cals = 1000 x 100 / 25 = 4000 Cals.
4000 Cals = 4.2 x 4000 = 16800 J .
Work done in one jump = kinetic energy while jumping
= 1/2 m v²
= .5 x 70 x 3.3²
= 381.15 J .
Number of jumps required = 16800 / 381.15
= 44 .
Answer:
aaaaa
Explanation:
M = Mass of the Earth
m = Mass of satellite
r = Radius of satellite
G = Gravitational constant




Answer:
The decelerating force is 
Solution:
As per the question:
Frontal Area, A = 
Speed of the spaceship, v = 
Mass density of dust, 
Now, to calculate the average decelerating force exerted by the particle:
(1)
Volume, 
Thus substituting the value of volume, V in eqn (1):

where
A = Area
v = velocity
t = time
(2)

From Newton's second law of motion:

Thus differentiating w.r.t time 't':

where
= average decelerating force of the particle
Now, substituting suitable values in the above eqn:

Answer:
3000 kg.m/s
Explanation:
Momentum, p is a product of mass and velocity hence
p=mv where m is mass and v is velocity.
Change in momentum is given by
where subscripts f and i represent final and initial respectively. Since the lorry finally comes to rest then the final velocity is zero. Substituting the given figures then
Change in momentum= 6000(0-0.5)=-3000 kg.m/s
Answer:
6.32 m/s 18.43° northeast
Explanation:
We express the velocity of hawk as:

We consider positive x towards east and positive y due north. So the magnitude is simply the square root of the square components:
≈
And the angle with respect to the east should be with:
