Water in the air combines with

An electric device, which heats water by immersing a resistance wire in the water, generates 50 cal of heat per second when an e

Oksi-84 [34.3K]

Answer:

0.69 ohm

Explanation:

Heat generated per second, H = 50 cal/s

Potential difference, V = 12 V

Let R is the resistance of coil.

The formula for the heat is given by

$H = \frac{V^{2}}{R}t$

$50\times 4.186 = \frac{12^{2}}{R}\times 1$

R = 0.69 ohm

3 0

3 years ago

A 220 kg crate hangs from the end of a rope of length L = 14.0 m. You push horizontally on the crate with a varying force F to m

kifflom [539]

Answer:

(a) magnitude of F = 797 N

(b)the total work done W = 0

(c)work done by the gravitational force = -1.55 kJ

(d)the work done by the pull = 0

(e) work your force F does on the crate = 1.55 kJ

Explanation:

Given:

Mass of the crate, m = 220 kg

Length of the rope, L = 14.0m

Distance, d = 4.00m

(a) What is the magnitude of F when the crate is in this final position

Let us first determine vertical angle as follows

=> $Sin \theta = \frac{d }{L}$

=> $\theta = Sin^{-1} \frac{d}{L}$ =

Now substituting thje values

=> $\theta = Sin^{-1} \frac{4}{12}$ =

=> $\theta = Sin^{-1} \frac{1}{3}$

=> $\theta = Sin^{-1}(0.333)$

=> $\theta = 19.5^{\circ}$

Now the tension in the string resolve into components

The vertical component supports the weight

=> $Tcos\theta = mg$

=> $T = \frac{mg}{cos\theta}$

=> $T = \frac{230 \times 9.8 }{cos(19.5)}$

=> $T = \frac{2254 }{cos(19.5)}$

=> $T = \frac{2254 }{0.9426}$

=>T =2391N

Therefore the horizontal force

$F = TSin(19.5)$

F = 797 N

b) The total work done on it

As there is no change in Kinetic energy

The total work done W = 0

c) The work done by the gravitational force on the crate

The work done by gravity

Wg = Fs.d = - mgh

Wg = - mgL ( 1 - Cosθ )

Substituting the values

= $-230 \times 9.8\times 12 ( 1 - cos(19.5) )$

= $-230 \times 9.8\times 12 ( 1 - 0.9426) )$

= $-230 \times 9.8\times 12 (0.0574)$

= $-230 \times 9.8\times 0.6888$

= $-230 \times 6.750$

= -1552.55 J

The work done by gravity = -1.55 kJ

d) the work done by the pull on the crate from the rope

Since the pull is perpendicular to the direction of motion,

The work done = 0

e)Find the work your force F does on the crate.

Work done by the Force on the crate

WF = - Wg

WF = -(-1.55)

WF = 1.55 kJ

(f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)

Here the work done by force is not equal to F*d

and it is equal to product of the cos angle and F*d

So, it is not equal to the product of the horizontal displacement and the answer to (a)

7 0

3 years ago

What is the substance through which sound travels?

Ann [662]

Answer:

Sound waves need to travel through a medium such as solids, liquids and gases. The sound waves move through each of these mediums by vibrating the molecules in the matter. The molecules in solids are packed very tightly. Liquids are not packed as tightly.

Explanation:

Hope this helped, Have a Great Day!!

4 0

3 years ago

2. The length of a day is based on the amount of time that

Serjik [45]

Answer:

the answer is B: earth takes to rotate once on its axis

6 0

3 years ago

Why is it important to include variety in your workouts

grigory [225]

Explanation:

Variety in to an exercise program allows the body to adapt to many demands, from high intensity exercise to slow, steady state exercise. Variety is important in a fitness program because it allows for your body to be challenged on a consistent basis and to overcome a plateau.

4 0

3 years ago