What is the magnitude of the resultant vector? Round

A lamp can work on 50V mains taking 2 amps. What value of resistance must be connected in series with it, so that it can be oper

wolverine [178]

The resistance of the lamp is apparently 50V/2A = 25 ohms.

When the circuit is fed with more than 50V, we want to add
another resistor in series with the 25-ohm lamp so that the
current through the combination will be 2A.

In order for 200V to cause 2A of current, the total resistance
must be 200V/2A = 100 ohms.

The lamp provides 25 ohms, so we want to add another 75 ohms
in series with the lamp. Then the total resistance of the circuit is
(75 + 25) = 100 ohms, and the current is 200V/100 ohms = 2 Amps.

The power delivered by the 200V mains is (200V) x (2A) = 400 watts.

The lamp dissipates ( I² · R ) = (2² · 25 ohms) = 100 watts.

The extra resistor dissipates ( I² · R) = (2² · 75 ohms) = 300 watts.

Together, they add up to the 400 watts delivered by the mains.

CAUTION:
300 watts is an awful lot of power for a resistor to dissipate !
Those little striped jobbies can't do it.
It has to be a special 'power resistor'.
300 watts is even an unusually big power resistor.
If this story actually happened, it would be cheaper, easier,
and safer to get three more of the same kind of lamp, and
connect THOSE in series for 100 ohms. Then at least the
power would all be going to provide some light, and not just
wasted to heat the room with a big moose resistor that's too
hot to touch.

3 0

3 years ago

A train travels 55 kilometers in 2 hours, and then 52 kilometers in 5 hours. What is its average speed?

bulgar [2K]

The formula for working out speed is distance ÷ time.

55 km ÷ 2 hours = 27.5 km/h (average speed for first part of journey)

52km ÷ 5 hours = 10.4 km/h (average speed for second part of journey)

(27.5 + 10.4) ÷ 2 = 18.95 km/h (average speed throughout the journey)

4 0

3 years ago

Read 2 more answers

according to newton's third law, when a horse pulls on a cart, the cart pulls back on the horse with an equal force on the horse

Debora [2.8K]

Based on Newton's principle, whenever objects A and B interact with each other, they exert forces upon each other.

When a horse pulls on a cart, the horse exerts a force only to the cart. But that force applies only to the cart, not to the horse.

The cart in turn exerts a force on the horse. But that force applies only to the horse, not the cart also.

There are two forces resulting from this interaction - a force on the horse and a force on the cart. The net force on the cart remains as it was --- a positive force in the direction of the horse's movement. Therefore, the cart begins to accelerate and move.

8 0

3 years ago

A metal rod has a moves with a constant velocity of 40 cm/s along two parallel metal rails through a magnetic field of 0.575 T.

love history [14]

Answer:

2.12/R mW

Explanation:

The electrical power, P generated by the rod is

P = B²L²v²/R where B = magnetic field = 0.575 T, L = length of metal rod = separation of metal rails = 20 cm = 0.2 m, v = velocity of metal rod = 40 cm/s = 0.4 m/s and R = resistance of rod = ?

So, the induced emf on the conductor is

E = BLv

= 0.575 T × 0.2 m × 0.4 m/s

= 0.046 V

= 46 mV

The electrical power, P generated by the rod is

P = B²L²v²/R

= B²L²v²/R

So, P = (0.575 T)² × (0.2 m)² × (0.4 m/s)²

= 0.002116/R W

= 2.12/R mW

3 0

3 years ago

a ball kicked with a velocity of 8m/s at an angle of 30 degree to horizontal. calculate the time of flight of the ball. (g=10ms^

posledela

Answer:

Approximately $0.8\; \rm s$ (assuming that air resistance is negligible.)

Explanation:

Let $v_0$ denote the initial velocity of this ball. Let $\theta$ denote the angle of elevation of that velocity.

The initial velocity of this ball could be decomposed into two parts:

Initial vertical velocity: $v_0(\text{vertical}) = v_0 \cdot \sin(\theta)$ .
Initial horizontal velocity: $v_0(\text{vertical}) = v_0 \cdot \cos(\theta)$ .

If air resistance on this ball is negligible, $v_0(\text{vertical})$ alone would be sufficient for finding the time of flight of this ball.

Calculate $v_0(\text{vertical})$ given that $v_0 = 8 \; \rm m \cdot s^{-1}$ and $\theta = 30^\circ$ :

$\begin{aligned}& v_0(\text{vertical}) \\ &= v_0 \cdot \sin(\theta) \\ &= \left(8 \; \rm m \cdot s^{-1} \right) \cdot \sin\left(30^{\circ}\right) \\ &= 4\;\rm m \cdot s^{-1} \end{aligned}$ .

Assume that air resistance on this ball is zero. Right before the ball hits the ground, the vertical velocity of this ball would be exactly the opposite of the value when the ball was launched.

Since $v_0(\text{vertical}) = 4\; \rm m \cdot s^{-1}$ , the vertical velocity of this ball right before landing would be $v_1(\text{vertical}) = -4\; \rm m \cdot s^{-1}$ .

Calculate the change to the vertical velocity of this ball:

$\begin{aligned}& \Delta v(\text{vertical}) \\ & = v_1(\text{vertical}) - v_0(\text{vertical}) \\ &= -8\; \rm m \cdot s^{-1}\end{aligned}$ .

In other words, the vertical velocity of this ball should have change by $8\; \rm m \cdot s^{-1}$ during the entire flight (from the launch to the landing.)

The question states that the gravitational field strength on this ball is $g = 10\; \rm m \cdot s^{-2}$ . In other words, the (vertical) downward gravitational pull on this ball could change the vertical velocity of the ball by $10\; \rm m\cdot s^{-1}$ each second. What fraction of a second would it take to change the vertical velocity of this ball by $8\; \rm m \cdot s^{-1}$ ?

$\begin{aligned}t &= \frac{\Delta v(\text{initial})}{g} \\ &= \frac{8\; \rm m \cdot s^{-1}}{10\; \rm m \cdot s^{-2}} = 0.8\; \rm s\end{aligned}$ .

In other words, it would take $0.8\; \rm s$ to change the velocity of this ball from the initial velocity at launch to the final velocity at landing. Therefore, the time of the flight of this ball would be $0.8\; \rm s\!$ .

5 0

3 years ago