Maximum power transfer theorem iz related to which chapter of physics Class 12

3. A microwave oven draws 12 A of current on a 110 V household circuit. What is its power

steposvetlana [31]

Answer:

W = 1320Watts

Explanation:

W = I*V

W = 12A*110V

W = 1320Watts

7 0

2 years ago

An ideal fluid flows through a pipe of variable cross section without any friction. The fluid completely fills the pipe. At any

ehidna [41]

Answer:

4. total energy

Explanation:

According to Bernoulli's principle at any two points along a streamline flow The total energy that is sum of pressure energy , Kinetic energy and potential energy of the liquid all taken in per unit volume remains constant. Therefore,

for ideal fluid flows through a pipe of variable cross section without any friction. The fluid completely fills the pipe. At any given point in the pipe, the fluid has a constant Total Energy.

3 0

3 years ago

A roller coaster car is traveling at a constant 3 m/s when it reaches a downward slope. On the slope, the car accelerates at a c

kirill [66]

Answer:

Velocity of the car at the bottom of the slope: approximately $20.3\; \rm m \cdot s^{-2}$ .

It would take approximately $3.9\; \rm s$ for the car to travel from the top of the slope to the bottom.

Explanation:

The time of the travel needs to be found. Hence, make use of the SUVAT equation that does not include time.

Let $v$ denote the final velocity of the car.
Let $u$ denote the initial velocity of the car.
Let $a$ denote the acceleration of the car.
Let $x$ denote the distance that this car travelled.

$v^2 - u^2 = 2\, a\cdot x$ .

Given:

$u = 3\; \rm m \cdot s^{-1}$ .
$a = 4.5\; \rm m \cdot s^{-2}$ .
$x = 45\; \rm m$ .

Rearrange the equation $v^2 - u^2 = 2\, a\cdot x$ and solve for $v$ :

$\begin{aligned}v &= \sqrt{2\, a \cdot x + u^2} \\ &= \sqrt{2 \times 4.5\; \rm m \cdot s^{-2} \times 45\; \rm m + \left(3\; \rm m \cdot s^{-1}\right)^{2}} \\ &\approx 20.3\; \rm m \cdot s^{-1}\end{aligned}$ .

Calculate the time required for reaching this speed from $u = 3\; \rm m \cdot s^{-1}$ at $a = 4.5\; \rm m \cdot s^{-2}$ :

$\begin{aligned}t &= \frac{v - u}{a} \\ &\approx \frac{20.3\; \rm m \cdot s^{-1} - 3\; \rm m \cdot s^{-1}}{4.5\; \rm m \cdot s^{-2}} \approx 3.9\; \rm m \cdot s^{-1}\end{aligned}$ .

3 0

2 years ago

A 0.5 kg air-track car is attached to the end of a horizontal spring of constant k = 20 N/m. The car is displaced 15 cm from its

Dimas [21]

Find the amount of work that the spring does. This can be found using the equation 1/2kx^2. Then, you must set that equal to the amount of kinetic energy the car has. This is possible thanks to the work-energy theorem.

1/2kx^2 = 1/2mv^2

Solve to find velocity. Remember, the spring is displaced .15 m, not 15!

To find the acceleration, use F = ma. The force being applied to the car is kx, and you know the mass. You do the math.

For problem C I don't know, haven't done that yet in my class. Sorry!

3 0

2 years ago

6. Two blocks are released from rest at the same height. Block A slides down a steeper ramp than Block B. Both ramps are frictio

Stolb23 [73]

Answer:

a. the work done by the gravitational force on Block A is less than the work done by the gravitational force on Block B.

b. the speed of Block A is equal to the speed of Block B.

c. the momentum of Block A is less than the momentum of Block B.

Explanation:

a. The work done by the gravitational force is equal to:

w = m*g*h

where m is mass, g is the standard gravitational acceleration and h is height. Given that both blocks are released from rest at the same height, then, the bigger the mass, the bigger the work done.

b. With ramps frictionless, the final speed of the blocs is:

v = √(2*g*h)

which is independent of the mass of the blocks.

c. The momentum is calculated as follows:

momentum = m*v

Given that both bocks has the same speed, then, the bigger the mass, the bigger the momentum.

8 0

2 years ago