Ohm's Law => V = I/R where
V stands for voltage,
I stands for Current
R stands for resistance.
ATQ
we know V = 9 volts
and R = 1.5 ohm's
so we need to find I.
Lets modify our formula little bit
we know V = IR which we can write as I = V/R
so lets substitute the values in the formula.
I = 9/1.5
I = 6 A
Correct Option is A.
Answer:
stored in a state of readiness.
Explanation:
Potential energy is stored in a state of readiness.
Potential energy can be defined as an energy possessed by an object or body due to its position.
Mathematically, potential energy is given by the formula;

Where, P.E represents potential energy measured in Joules.
m represents the mass of an object.
g represents acceleration due to gravity measured in meters per seconds square.
h represents the height measured in meters.
Explanation:
Boyle's law can be stated as the "volume of a fixed mass of a gas varies inversely as the pressure changes if the temperature is constant". It is mathematically expressed as;
P1 V1 = P2 V2
P1 is the initial pressure
V1 is the initial volume
P2 is the final pressure
V2 is the final volume
Charles's law states that the volume of a fixed mass of a gas varies directly as its absolute temperature if the pressure is constant.
It is mathematically expressed as;
=
V and T are volume and temperature respectively
1 and 2 are the initial and final states
The x-component of a vector are < 106.6, 43.07 >
Depending on the angle we are provided, the x-component of a vector can either be cos or sin. Cos always corresponds to the right triangle's side that contacts the specified angle.
If a vector v with magnitude ||v|| makes an angle θ with the positive x-axis then,
v = ||v|| cos θi + ||v|| sin θj
= < ||v|| cos θ , ||v|| sin θ >
Magnitude p = 115 km
Angle = 22°
p = ||p|| < cos θ, sin θ >
p = 115 < cos 22°, sin 22° >
p = 115 < 0.927, 0.3746 >
p = < 106.6, 43.07 >
Therefore, the x-component of a vector are < 106.6, 43.07 >
Learn more about vectors here:
brainly.com/question/8043832
#SPJ1
Answer:
Work done is 12.3 J
Explanation:
We have,
Mass of puck, m = 0.35 kg
Force of friction acting on the puck when it slides is 0.15 N
Distance travelled by the puck is 82 m.
It is required to find the work done on the puck. Finally the puck comes to rest and the force of friction is acting on it. It means the applied force is 0.15 N. Work done is given by

The work done on the puck is 12.3 J.