Newton's 2nd law of motion:
Force = (mass) x (acceleration)
= (0.314 kg) x (164 m/s²)
= 51.5 newtons
(about 11.6 pounds) .
Notice that the ball is only accelerating while it's in contact with the racket.
The instant the ball loses contact with the racket, it stops accelerating, and
sails off in a straight line at whatever speed it had when it left the strings.
Answer:
11.) g = 3.695 m/s^2
12.) g = 8.879 m/s^2
13.) E = 8127 N/C
Explanation:
11.) Given that the
Mercury mass M = 3.3 × 10^23kg
Radius r = 2.44 ×10^6 m
Gravitational constant G = 6.67408 × 10^-11 m3kg-1 s^-2
Gravitational field strength g can be calculated by using the formula below
g = GM/r^2
Substitutes all the parameters into the formula
g = (6.67408 × 10^-11 × 3.3 × 10^23)/(2.44×10^6)^2
g = 2.2×10^13/5.954×10^12
g = 3.695 m/s^2
12.) Given that the
Venus mass M = 4.87×10^24kg
Radius r = 6.05 × 10^6 m
Using the same formula for gravitational field strength g
g = GM/R2
Substitute all the parameters into the formula
g = (6.67408 × 10^-11 × 4.87×10^24)/(6.05×10^6)^2
g = 3.25×10^14/3.66×10^13
g = 8.879 m/s^2
13.) Given that the
Charge = 2.26 nC = 2.26×10^-9
Distance d = 0.05m
Electric field strength E can be calculated by using the formula below
E = Kq/d^2
Where
K = electrostatic constant 8.99 × 10^9 Nm2/C2
Substitutes all the parameters into the formula
E = (8.99 × 10^9 × 2.26×10^-9)/0.05^2
E = 20.3174/2.5×10^-3
E = 8126.96 N/C
Im positive that the answer is b
Answer:
the one above the surface of earth
Explanation:
earth has gravity the ball of the moon would float away
Answer:The speed is 54 mm/s, the units are mm/s.
Explanation:
1) Data:
a) λ = 6mm
b) f = 9 Hz = 9 s⁻¹
c) v = ?
2) Formula:
wave equation: v = f × λ
3) Solution:
v = 9 s⁻¹ × 6mm
v = 54 mm/s
The speed is 54 mm/s, the units are mm/s.
