Sky diving involves free fall under gravity along with the drag due to air molecules pushing against the body slowing the rate of fall of a body. This is actually a significant amount of force. The drag force depends on the contact surface area and weight of the body. More the surface area in contact, more would be the drag. The sitting position of the skydiver would experience less drag than the chest down position. This is because of the less contact surface area of the body with the air molecules while in the former case. Since no two persons have identical body shape and weight, the rate of fall can be made nearly equal but not exactly equal. This is would be possible when they are having same body position.
Answer:
The time it will take for the car to reach a velocity of 28 m/s is 7 seconds
Explanation:
The parameters of the car are;
The acceleration of the car, a = 4 m/s²
The final velocity of the car, v = 28 m/s
The initial velocity of the car, u = 0 m/s (The car starts from rest)
The kinematic equation that can be used for finding (the time) how long it will take for the car to reach a velocity of 28 m/s is given as follows;
v = u + a·t
Where;
v = The final velocity of the car, v = 28 m/s
u = The initial velocity of the car = 0 m/s
a = The acceleration of the car = 4 m/s²
t = =The time it will take for the car to reach a velocity of 28 m/s
Therefore, we get;
t = (v - u)/a
t = (28 m/s - 0 m/s)/(4 m/s²) = 7 s
The time it will take for the car to reach a velocity of 28 m/s, t = 7 seconds.
Answer:
Approximate height of the building is 23213 meters.
Explanation:
Let the height of the building be represented by h.
0.02 radians = 0.02 × 
= 0.02 x (180/
)
0.02 radians = 1.146°
10.5 km = 10500 m
Applying the trigonometric function, we have;
Tan θ = 
So that,
Tan 1.146° = 
⇒ h = Tan 1.146° x 10500
= 2.21074 x 10500
= 23212.77
h = 23213 m
The approximate height of the building is 23213 m.
Answer:
Strong nuclear force is 1-2 order of magnitude larger than the electrostatic force
Explanation:
There are mainly two forces acting between protons and neutrons in the nucleus:
- The electrostatic force, which is the force exerted between charged particles (therefore, it is exerted between protons only, since neutrons are not charged). The magnitude of the force is given by
where k is the Coulomb's constant, q1 and q2 are the charges of the two particles, r is the separation between the particles.
The force is attractive for two opposite charges and repulsive for two same charges: therefore, the electrostatic force between two protons is repulsive.
- The strong nuclear force, which is the force exerted between nucleons. At short distance (such as in the nucleus), it is attractive, therefore neutrons and protons attract each other and this contributes in keeping the whole nucleus together.
At the scale involved in the nucleus, the strong nuclear force (attractive) is 1-2 order of magnitude larger than the electrostatic force (repulsive), therefore the nucleus stays together and does not break apart.
Answer:
H = 1/2 g t^2 where t is time to fall a height H
H = 1/8 g T^2 where T is total time in air (2 t = T)
R = V T cos θ horizontal range
3/4 g T^2 = V T cos θ 6 H = R given in problem
cos θ = 3 g T / (4 V) (I)
Now t = V sin θ / g time for projectile to fall from max height
T = 2 V sin θ / g
T / V = 2 sin θ / g
cos θ = 3 g / 4 (T / V) from (I)
cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ
tan θ = 2/3
θ = 33.7 deg
As a check- let V = 100 m/s
Vx = 100 cos 33.7 = 83,2
Vy = 100 sin 33,7 = 55.5
T = 2 * 55.5 / 9.8 = 11.3 sec
H = 1/2 * 9.8 * (11.3 / 2)^2 = 156
R = 83.2 * 11.3 = 932
R / H = 932 / 156 = 5.97 6 within rounding
