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KATRIN_1 [288]
3 years ago
7

The change of motion of a body is proportional to the applied force, and it takes place in the direction of a straight line in w

hich that force is applied. This describes what law of motion?
Physics
2 answers:
7nadin3 [17]3 years ago
4 0

Answer:

Newton's Second Law of Motion  

Explanation:

According to Newton's second law of motion, the change in velocity of a body is directly proportional to the force applied on it. Velocity is a vector quantity. It measures the magnitude of the speed as well as its direction.

F = m a

where, F is the applied force, m is the mass and a is the acceleration.

It can also be expressed as:

F = \frac{dp}{dt}

where, p = mv ( momentum)

FrozenT [24]3 years ago
3 0

Answer:

Newton's second law

Explanation:

Momentum is defined as the measurement of motion in the object.

So, the change in motion means the change in momentum of the body.

According to newton's second law, the force acting on an object is directly proportional to the rate of change of momentum of the object.

So, it describes the Newton's second law.

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12.51 A parallel RLC circuit, which is driven by a variable frequency 2-A current source, has the following values: R = 1 kΩ, L
Anastaziya [24]

Answer:

BW = 100 rad/s

wlow = 452.49 rad/s

whigh = 552.49 rad/s

V(jwlow) =1414.21 < 45°V

V(jwhigh) =1414.21 <-45°V

Explanation:

To calculate bandwidth we have formula

BW = 1/RC

BW = 1/ 1000x10x10^¯6

BW = 100 rad/s

We will first calculate resonant frequency and quality factor for half power frequencies.

For resonant frequency

wo = 1/(SQRT LC)

wo = 1/SQRT 400×10¯³ × 10×10^¯6

wo = 500 rad/s

For Quality

Q = wo / BW

Q = 500/100

Q = 5

wlow = wo [-1/2Q+ SQRT (1/2Q)² + 1]

wlow = 500 [-1/2×5 + SQRT (1/2×5)² + 1]

wlow = 452.49 rad/s

whigh = wo [1/2Q+ SQRT (1/2Q)² + 1]

whigh = 500 [1/2×5 + SQRT (1/2×5)² + 1]

whigh = 552.49 rad/s

We will start with admittance at lower half power frequency

Y(jwlow) = (1/R) + (1/jwlow L) + (jwlow C)

Y(jwlow) = (1/1000) + (1/j×452.49×400×10¯³) + (j×452.49×10×10^¯6)

Y(jwlow) = 0.001 - j5.525×10¯³ + j4.525×10¯³

Y(jwlow) = (1-j).10¯³ S

Voltage across the network is calculated by ohm's law

V(jwlow) = I/Y(jwlow)

V(jwlow) = 2/(1-j).10¯³

V(jwlow) = 1414.2 < 45°V

Now we will calculate the admittance at higher half power frequency

Y(jwhigh) = (1/R) + (1/jwhigh L) + (jwhigh C)

Y(jwhigh) = (1/1000) + (1/j×552.49×400×10¯³) + (j×552.49×10×10^¯6)

Y(jwhigh) = 0.001 - j4.525×10¯³ + j5.525×10¯³

Y(jwhigh) = (1+j).10¯³ S

Voltage across network will be calculated by ohm's law

V(jwhigh) = I/Y(jwhigh)

V(jwhigh) = 2/(1+j).10¯³

V(jwhigh) = 1414.2 < - 45°V

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Answer:

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Explanation:

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Your answer is A.

I hope this helps
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