Answer:
B) R1 = 6 V and R2 = 6V
Explanation:
In series, both resistors will carry the same current.
that current will be I = V/R = 12 / (10 + 10) = 0.6 A
The voltage drop across each resistor is V = IR = 0.6(10) = 6 V
Answer:
Explanation:
The tidal current flows to the east at 2.0 m/s and the speed of the kayaker is 3.0 m/s.
Let Vector 
is the tidal current velocity as shown in the diagram.
In order to travel straight across the harbor, the vector addition of both the velocities (i.e the resultant velocity, 
must be in the north direction.
Let 
is the speed of the kayaker having angle \theta measured north of east as shown in the figure.
For the resultant velocity in the north direction, the tail of the vector 
and head of the vector must lie on the north-south line.
Now, for this condition, from the triangle OAB
Hence, the kayaker must paddle in the direction of in the north of east direction.
Answer: V = 15 m/s
Explanation:
As stationary speed gun emits a microwave beam at 2.10*10^10Hz. It reflects off a car and returns 1030 Hz higher. The observed frequency the car will be experiencing will be addition of the two frequency. That is,
F = 2.1 × 10^10 + 1030 = 2.100000103×10^10Hz
Using doppler effect formula
F = C/ ( C - V) × f
Where
F = observed frequency
f = source frequency
C = speed of light = 3×10^8
V = speed of the car
Substitute all the parameters into the formula
2.100000103×10^10 = 3×10^8/(3×10^8 -V) × 2.1×10^10
2.100000103×10^10/2.1×10^10 = 3×108/(3×10^8 - V)
1.000000049 = 3×10^8/(3×10^8 - V)
Cross multiply
300000014.7 - 1.000000049V = 3×10^8
Collect the like terms
1.000000049V = 14.71429
Make V the subject of formula
V = 14.71429/1.000000049
V = 14.7 m/s
The speed of the car is 15 m/s approximately.
1) 0N... friction opposes the motion of an object, since the block is at rest there is no motion thus no friction
2) F=ma
= (5.5kg)(30m/s)
=165 N
Answer:
I = 0.2 A
Explanation:
Lamp is rated at 300 mA
I_lamp = 0.3 A
Voltage is; V = 3V
Thus; Resistance is given by;
R = V/I
R = 3/0.3
R = 10 ohms
Now, since the ammeter of 5 ohms is connected in series with the lamp. Thus equivalent resistance;
R_eq = 10 + 5
R_eq = 15 ohms
Ammeter current will be;
I = V/R_eq
I = 3/15
I = 0.2 A
