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3 years ago

14

When testing a plugged-in but partly disassembled toaster, you should push the rack down with a

1 answer:

Sphinxa [80]3 years ago

4 0

A wooden stick would be best.
The wire and the wrench are both metal and therefore are able to conduct electricity.
Using your finger is obviously dangerous.
Wood is not a conductor so it is the best answer.

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Earth's tides are primarily caused by _____.

Fiesta28 [93]

B.) The moon

There you go.

8 0

4 years ago

Read 2 more answers

24 POINTS!!!!!!!!!!!!!!!

TEA [102]

Potential Energy (Initial one) = m * g * h
P.E. = 60 * 9.8 * 10
P.E. = 5880

Kinetic Energy (Final One) = 1/2 mv²
K.E. = 1/2 * 60 * (10)²
K.E. = 6000/2
K.E. = 3000

Lost Energy = 5880 - 3000 = 2880 J

In short, Your Answer would be 2880 Joules

Hope this helps!

7 0

3 years ago

Read 2 more answers

Two kilograms of air is contained in a rigid wellinsulated tank with a volume of 0.6 m3 . The tank is fitted with a paddle wheel

wolverine [178]

Answer:

$0.3\ \text{m}^3/\text{kg}$

$36\ \text{kJ}$

$18\ \text{kJ/kg}$

Explanation:

V = Volume of air = $0.6\ \text{m}^3$

P = Power = 10 W

t = Time = 1 hour

m = Mass of air = 2 kg

Specific volume is given by

$v=\dfrac{V}{m}\\\Rightarrow v=\dfrac{0.6}{2}\\\Rightarrow v=0.3\ \text{m}^3/\text{kg}$

The specific volume at the final state is $0.3\ \text{m}^3/\text{kg}$

Work done is given by

$W=Pt\\\Rightarrow W=10\times 60\times 60\\\Rightarrow W=36000\ \text{J}=36\ \text{kJ}$

The energy transfer by work, is $36\ \text{kJ}$

Change in specific internal energy is given by

$\Delta u=\dfrac{Q}{m}+\dfrac{W}{m}\\\Rightarrow \Delta u=0+\dfrac{36}{2}\\\Rightarrow \Delta u=18\ \text{kJ/kg}$

The change in specific internal energy of the air is $18\ \text{kJ/kg}$

8 0

3 years ago

a boy steps off the end of a high diving board and drops into the water below. it takes the boy 0.78 seconds to reach the water.

Mice21 [21]

Cuz as effvsdcbh ftjhj

6 0

3 years ago

In Example 2.12, two circus performers rehearse a trick in which a ball and a dart collide. Horatio stands on a platform 6.4 m a

pickupchik [31]

Answer:

time of collision is

t = 0.395 s

$h = 5.63 m$

so they will collide at height of 5.63 m from ground

Explanation:

initial speed of the ball when it is dropped down is

$v_1 = 0$

similarly initial speed of the object which is projected by spring is given as

$v_2 = 16.2 m/s$

now relative velocity of object with respect to ball

$v_r = 16.2 m/s$

now since we know that both are moving under gravity so their relative acceleration is ZERO and the relative distance between them is 6.4 m

$d = v_r t$

$6.4 = 16.2 t$

$t = 0.395 m$

Now the height attained by the object in the same time is given as

$h = v_2 t - \frac{1}{2}gt^2$

$h = 16.2(0.395) - \frac{1}{2}(9.81).395^2$

$h = 5.63 m$

so they will collide at height of 5.63 m from ground

4 0

4 years ago