Answer:
The astronaut can throw the hammer in a direction away from the space station. While he is holding the hammer, the total momentum of the astronaut and hammer is 0 kg • m/s. According to the law of conservation of momentum, the total momentum after he throws the hammer must still be 0 kg • m/s. In order for momentum to be conserved, the astronaut will have to move in the opposite direction of the hammer, which will be toward the space station.
Explanation:
Answer:
(a) 
(b) 
Explanation:
Given:
*p = charge on proton = 
*e = magnitude of charge on an electron =
*r = distance between the proton and the electron in the Rutherford's atom = 
Part (a):
Since two unlike charges attract each other.
According to Coulomb's law:
Hence, the attractive electrostatic force of attraction acting between an electron and a proton of Rutherford's atom is 
.
Part (b):
Potential energy between two charges separated by a distance r is given by:
So, the potential energy between the electron and the proton of the Rutherford's atom is given by:
Hence, the electrostatic potential energy of the atom is .
Answer:
(a) 8Ω (b) Ratio = Parra/P8 ohm = 1
Explanation:
Solution
Recall that,
An high-fidelity amplifier has one output for a speaker of resistance of = 8 Ω
Now,
(a) How can two 8-Ω speakers be arranged, when one = 4-Ω speaker, and one =12-Ω speaker
The Upper arm is : 8 ohm, 8 ohm
The Lower arm is : 12 ohm, 4 ohm
The Requirement is = (16 x 16)/(16 + 16) = 8 ohm
(b) compare your arrangement power output of with the power output of a single 8-Ω speaker
The Ratio = Parra/P8 ohm = 1
Answer:
Explanation:
How long does the football need to rise?
4.70/3 = 2.35 s
What height will the football reach?
h = ½(9.81)2.35² = 27.1 m
With what speed does the punter need to kick the football?
vy = g•t = 9.81(2.35) = 23.1 m/s
vx = d/t = 56.0/4.70 = 11.9 m/s
v = √(vx²+vy²) = 26.0 m/s
At what angle (θ), with the horizontal, does the punter need to kick the football?
θ = arctan(vy/vx) = 62.7°
Answer:
Explanation:
Let the equilibrium position of third charge be x distance from q₁.
Force on third charge due to q₁
= 9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x²
Force on third charge due to q₂
= 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²
Both the force will act in opposite direction and for balancing , they should be equal.
9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x² = 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²
5 / x² = 2 / ( .4 - x )²
Taking square root on both sides
2.236 / x = 1.414 / .4 - x
2.236 ( .4 - x ) = 1.414 x
.8944 - 2.236 x = 1.414 x
.8944 = 3.65 x
x = .245 m
24.5 cm
So the third charge should be at a distance of 24.5 cm from q₁ .
