Let us assume the upstream rowing rate of Alicia = x
Let us assume the downstream rowing rate of Alicia = y
We already know that
Travelling time = Distance traveled/rowing rate
Then
6/(x + 3) = 4/x
6x = 4x + 12
6x - 4x = 12
2x = 12
x = 6
Then
Rowing rate of Alicia going upstream = 6 miles per hour
Rowing rate of Alicia going downstream = 9 miles per hour.
Answer:
4.02 km/hr
Explanation:
5 km/hr = 1.39 m/s
The swimmer's speed relative to the ground must have the same direction as line AC.
The vertical component of the velocity is:
uᵧ = us cos 45
uᵧ = √2/2 us
The horizontal component of the velocity is:
uₓ = 1.39 − us sin 45
uₓ = 1.39 − √2/2 us
Writing a proportion:
uₓ / uᵧ = 121 / 159
(1.39 − √2/2 us) / (√2/2 us) = 121 / 159
Cross multiply and solve:
159 (1.39 − √2/2 us) = 121 (√2/2 us)
220.8 − 79.5√2 us = 60.5√2 us
220.8 = 140√2 us
us = 1.115
The swimmer's speed is 1.115 m/s, or 4.02 km/hr.
Explanation:
If a positive test charge is placed in an electric field, it will exert the force in the test charge in the direction of electric field vector. We know that the direction of electric field is given by electric field lines. The field lines for a positive charge is outwards. The electric force acting on the charge is given by :
F = q E
Hence, this is the required solution.
Given Information:
KEa = 9520 eV
KEb = 7060 eV
Electric potential = Va = -55 V
Electric potential = Vb = +27 V
Required Information:
Charge of the particle = q = ?
Answer:
Charge of the particle = +4.8x10⁻¹⁸ C
Explanation:
From the law of conservation of energy, we have
ΔKE = -qΔV
KEb - KEa = -q(Vb - Va)
-q = KEb - KEa/Vb - Va
-q = 7060 - 9520/27 - (-55)
-q = 7060 - 9520/27 + 55
-q = -2460/82
minus sign cancels out
q = 2460/82
Convert eV into Joules by multiplying it with 1.60x10⁻¹⁹
q = 2460(1.60x10⁻¹⁹)/82
q = +4.8x10⁻¹⁸ C
