Does solid give out or take in energy when it melts?

A food processor draws 8.47 A of current when connected to a potential difference of 110 V.

Andreyy89

Answer:

27.95[kW*min]

Explanation:

We must remember that the power can be determined by the product of the current by the voltage.

$P=V*I$

where:

P = power [W]

V = voltage [volt]

I = amperage [Amp]

Now replacing:

$P=110*8.47\\P=931.7[W]$

Now the energy consumed can be obtained mediate the multiplication of the power by the amount of time in operation, we must obtain an amount in Kw per hour [kW-min]

$Energy = 931.7[kW]*30[days]*10[\frac{min}{1day} ]=279510[W*min]or 27.95[kW*min]$

3 0

3 years ago

Which of the following traits might be considered an adaptation fora rabbit living in the Arctic?

puteri [66]

Where are the following traits ?

7 0

3 years ago

Read 2 more answers

I need to speak with zuka

Alex787 [66]

Answer:

DMs are not accessible anymore. I assume Zuka is a staff member? the only way to talk to a staff member anymore is to report something, but even then, the probably won't even look at what they're deleting :/

May I have brainliest please? :)

3 0

2 years ago

Problem #2: An apple is thrown upward with an initial velocity of +24.0 m/s. a. Sketch the apple's trip and label what you know.

bogdanovich [222]

Answer:

The answer is below

Explanation:

a) The initial velocity (u) = 24 m/s

We can solve this problem using the formula:

v² = u² - 2gh

where v = final velocity, g= acceleration due to gravity = 9.8 m/s², h = height.

At maximum height, the final velocity = 0 m/s

v² = u² - 2gh

0² = 24² - 2(9.8)h

2(9.8)h = 24²

2(9.8)h = 576

19.6h = 576

h = 29.4 m

b) The time taken to reach the maximum height is given as:

v = u - gt

0 = 24 - 9.8t

9.8t = 24

t = 2.45 s

The total time needed for the apple to return to its original position = 2t = 2 * 2.45 = 4.9 s

4 0

2 years ago

A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The

Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}$

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

$\sum {F = ma}$

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

$F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg$

Here, ${\mu _{\rm{s}}}$ is the coefficient of static friction, mm is mass of the object, and $g$ is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass ${m_1}$

Substitute for in the expression of maximum static friction ${F_{\rm{f}}} = {\mu _{\rm{s}}}mg$

${F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g$

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute for , [/tex]{m_1}[/tex] for in the equation .

${F_{\rm{f}}} = {m_1}a$

Substitute ${\mu _{\rm{s}}}{m_1}g$ for ${F_{\rm{f}}}$ in the equation ${F_{\rm{f}}} = {m_1}a$

${\mu _{\rm{s}}}{m_1}g = {m_1}a$

Rearrange for a.

$a = {\mu _{\rm{s}}}g$

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right)a$

Substitute for in the above equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

5 0

3 years ago