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2 years ago

An 8.00- W resistor is dissipating 100 watts. What are the current through it and the difference of potential across it?

Physics

1 answer:

hjlf2 years ago

8 0

Answer:

I= 3.5 amps

Explanation:

Step one:

given data

rating of resistor R= 8 ohms

power P= 100W

Required

The current I

Step two

Yet this power is also given by

$P = I^2R$

make I subject of the formula we have

$I= \sqrt{\frac{P}{R} }$

substitute

$I= \sqrt{\frac{100}{8} }\\\\I=\sqrt{12.5}\\\\I= 3.5 amps$

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A pebble is dropped from rest from the top of a tall cliff and falls 53.4 m after 3.3 s has elapsed. How much farther does it dr

jeka94

Answer:

426.84 m

Explanation:

initial velocity u = 0

time t = 3.3 s

distance travelled s = 53.4 m

acceleration due to gravity = g

s = ut + 1/2 g t²

53.4 = 0 + 1/2 g x 3.3²

g = 9.8 m /s²

For the whole length of fall

distance travelled = h

total time = 6.6 + 3.3 = 9.9 s

h = ut + 1/2 g t²

u again = 0

h = .5 x 9.8 x 9.9²

= 480.24 m

distance travelled in last 6.6 s

= 480.24 - 53.4

= 426.84 m

6 0

2 years ago

A wheel has a radius of 5.9 m. How far (path length) does a point on the circumference travel if the wheel is rotated through an

posledela

Answer:

(a). The path length is 3.09 m at 30°.

(b). The path length is 188.4 m at 30 rad.

(c). The path length is 1111.5 m at 30 rev.

Explanation:

Given that,

Radius = 5.9 m

(a). Angle $\theta=30°$

We need to calculate the angle in radian

$\theta=30\times\dfrac{\pi}{180}$

$\theta=0.523\ rad$

We need to calculate the path length

Using formula of path length

$Path\ length =angle\times radius$

$Path\ length=0.523\times5.9$

$Path\ length =3.09\ m$

(b). Angle = 30 rad

We need to calculate the path length

$Path\ length=30\times5.9$

$Path\ length=177\ m$

(c). Angle = 30 rev

We need to calculate the angle in rad

$\theta=30\times2\pi$

$\theta=188.4\ rad$

We need to calculate the path length

$Path\ length=188.4\times5.9$

$Path\ length =1111.56\ m$

Hence, (a). The path length is 3.09 m at 30°.

(b). The path length is 188.4 m at 30 rad.

(c). The path length is 1111.5 m at 30 rev.

8 0

2 years ago

The robot HooRU is lost in space, floating around aimlessly, and radiates heat into the depths of the cosmos at the rate of 13.1

ahrayia [7]

Answer:

The temperature is $T = 168.44 \ K$

Explanation:

From the question ewe are told that

The rate of heat transferred is $P = 13.1 \ W$

The surface area is $A = 1.55 \ m^2$

The emissivity of its surface is $e = 0.287$

Generally, the rate of heat transfer is mathematically represented as

$H = A e \sigma T^{4}$

=> $T = \sqrt[4]{\frac{P}{e* \sigma } }$

where $\sigma$ is the Boltzmann constant with value $\sigma = 5.67*10^{-8} \ W\cdot m^{-2} \cdot K^{-4}.$

substituting value

$T = \sqrt[4]{\frac{13.1}{ 0.287* 5.67 *10^{-8} } }$

$T = 168.44 \ K$

7 0

3 years ago

Which situation is an example of increasing potential energy? Question 4 options: A. a cat jumping from a tree B. pulling a wago

jeka94

Pulling an wagon uphill I believe.

4 0

2 years ago

Read 2 more answers

A ball is thrown vertically upwards with an initial velocity of 20.00 m/s. Neglecting air resistance, how long is the ball in th

algol [13]

Answer:

1) The greatest height attained by the ball equals 20.387 meters.

2) The time it takes for the ball to reach 15 meters approximately equals 1 second.

Explanation:

The greatest height will be attained when the ball stop's in the air and starts falling back to the earth.

thus using third equation of kinematics we obtain the height attained as

$v^2=u^2+2as$

where

'v' is the final speed of the ball

'u' is the initial speed of the ball

'a' is the acceleration that the ball is under which in this case equals 9.81 $m/s^{2}$

's' is the distance it covers

Thus for maximum height applying the values in the equation we get

$0=20^{2}-2\times 9.81\times h\\\\\therefore h=\frac{20^{2}}{2\times 9.81}=20.387meters$

Using the same equation we can find the speed of the ball when it reaches 15 meters of height as

$v^2=20^{2}-2\times 9.81\times 15\\\\v^{2}=105.7\\\\\therefore v=10.28m/s$

the time it takes to reduce the velocity to this value can be found by first equation of kinematics as

$v=u+at\\\\t=\frac{v-u}{a}\\\\t=\frac{10.28-20}{-9.81}=0.991seconds\approx 1second$

4 0

2 years ago