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Andru [333]
2 years ago
9

Meghan and Kailey are attempting to push a piano down the hallway, but as they turn up the 8thgrade hallway they are faced with

a hill. Meghan and Kailey are pushing with a force of 10N each, but the heavy piano is exerting a force of 50N in the opposite direction. Will the girls be able to push the piano? What happens if two other students begin to help by pulling on the piano. Danny pulls with a force of 17N and Liam pulls with a force of 16N. How much force is being exerted on the piano, and will the piano move?
Physics
2 answers:
Alex Ar [27]2 years ago
6 0

Answer: GIVE ME MIN

Explanation:

stich3 [128]2 years ago
5 0
Meghan and Kailey make a pushing force of 20 newtons together, so no they will not push the piano. However with Danny and Liam they make a total force of 53 Newtons so they should be able to push it up the hill.
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When the resistance of a circuit is doubled, and no other changes occur, what effect does this have on this current in the circu
liraira [26]

As a result of doubled resistance, current will be halved.

As we know, Ohm's law inversely relates current and resistance to each other. The equation is as follows -

V = I*R, where V is voltage, I is current and R is resistance.

As the question mentions no change in any other parameter, let's assume I₁ and I₂ be initial and final current and R₁ and R₂ initial and final resistance.

So, R₂ will be 2R₁ as it is doubled.

Now, equating the values and finding out the value of final current.

I₁R₁ =  I₂R₂

Keep the value of R₂

I₁R₁ =  I₂2R₁

Cancelling R₁ as it is common on both sides of equation.

So, we get - I₁ =  2I₂

I₂ = I₁/2

Thus, as can be seen above, final current will be half of initial current.

Learn more about Ohm's law and relation between current, voltage and resistance -

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2 years ago
Consider the following statements. A. Heat flows from an object at higher temperature to an object at lower temperature; B. Heat
Monica [59]

Answer:

A. Heat flows from an object at higher temperature to an object at lower temperature

Explanation:

The option A obeys the 2nd law of thermodynamics. The heat will flow from the object at higher temperature to the object at Lower temperature till they reach an equilibrial state.

Heat doesn’t necessarily flow from an object with higher thermal energy to an object with lower thermal energy because an object has a higher thermal energy when it’s mass is more than the other. This makes B wrong.

C is wrong because heat moves from an object with higher temperature to objects with Lower temperature regardless of the state of matter.

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3 years ago
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dem82 [27]
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Middle adulthood marks the time period in which some adults face the fabled midlife crisis. What exactly is that? Do you think y
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6 0
3 years ago
Find the quantity of heat needed
krok68 [10]

Answer:

Approximately 3.99\times 10^{4}\; \rm J (assuming that the melting point of ice is 0\; \rm ^\circ C.)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}

The energy required comes in three parts:

  • Energy required to raise the temperature of that 0.100\; \rm kg of ice from (-10\; \rm ^\circ C) to 0\; \rm ^\circ C (the melting point of ice.)
  • Energy required to turn 0.100\; \rm kg of ice into water while temperature stayed constant.
  • Energy required to raise the temperature of that newly-formed 0.100\; \rm kg of water from 0\; \rm ^\circ C to 10\;\ rm ^\circ C.

The following equation gives the amount of energy Q required to raise the temperature of a sample of mass m and specific heat capacity c by \Delta T:

Q = c \cdot m \cdot \Delta T,

where

  • c is the specific heat capacity of the material,
  • m is the mass of the sample, and
  • \Delta T is the change in the temperature of this sample.

For the first part of energy input, c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1} whereas m = 0.100\; \rm kg. Calculate the change in the temperature:

\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}.

Calculate the energy required to achieve that temperature change:

\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}.

Similarly, for the third part of energy input, c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1} whereas m = 0.100\; \rm kg. Calculate the change in the temperature:

\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}.

Calculate the energy required to achieve that temperature change:

\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}.

The second part of energy input requires a different equation. The energy Q required to melt a sample of mass m and latent heat of fusion L_\text{f} is:

Q = m \cdot L_\text{f}.

Apply this equation to find the size of the second part of energy input:

\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}.

Find the sum of these three parts of energy:

\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}.

3 0
3 years ago
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