Answer:
A) 31 kJ
B) 1.92 KJ
C) 40 , 2.48
Explanation:
weight of person ( m ) = 79 kg
height of jump ( h ) = 0.510 m
Compression of joint material ( d ) = 1.30 cm ≈ 0.013 m
A) calculate the force
Fd = mgh
F = mgh / d
W = mg
F(net) = W + F = mg ( 1 + 
= 79 * 9.81 ( 1 + (0.51 / 0.013) )
= 774.99 ( 40.231 ) ≈ 31 KJ
B) calculate the force when the stopping distance = 0.345 m
d = 0.345 m
Fd = mgh hence F = mgh / d
F(net) = W + F = mg ( 1 + 
= 79 * 9.81 ( 1 + (0.51 / 0.345) )
= 774.99 ( 2.478 ) = 1.92 KJ
C) Ratio of force in part a with weight of person
= 31000 / ( 79 * 9.81 ) = 31000 / 774.99 = 40
Ratio of force in part b with weight of person
= 1920 / 774.99 = 2.48
Answer:
D. Upward force on the shuttle
Explanation:
The hot gas from space shuttles released downward causes an upward force on the shuttle and propels it up the more.
- This hot gas is produced from super cooled oxygen and hydrogen tanks within the shuttle.
- The upward force on the shuttle allows the craft to escape the gravitational pull of the earth on the shuttle
- Special level of rapid acceleration must be attained for the shuttle to escape the earth pull.
Answer:
-32.5 * 10^-5 J
Explanation:
The potential energy of this system of charges is;
Ue = kq1q2/r
Where;
k is the Coulumb's constant
q1 and q2 are the magnitudes of the charges
r is the distance of separation between the charges
Substituting values;
Ue = 9.0×10^9 N⋅m2/C2 * 5.5 x 10^-8 C *( -2.3 x10^-8) C/(3.5 * 10^-2)
Ue= -32.5 * 10^-5 J
You are Thomas Alva Edison. You also invented the phonograph
and the first practical movie camera. Sadly, you died almost exactly
9 years before I was born.
Answer:
28.73 m from the base of the cliff collide happen
Explanation:
Equation for ball dropped from 50 cliff is reaches distance x in t seconds isi given by

stone is thrown up from the bottom with speed u=24 m/s . it reaches distance y when stone collide with ball.(g is negative here)

we know that total distance traveled by ball and stone is 50 m

adding equation 1 and 2, we get time t

substitute this time in equation 2, we can get the required distance where they collide

28.73 m from the base of the cliff collide happen