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Goryan [66]
3 years ago
7

A 200 g piece of iron is heated to 100°C. It is then dropped into water to bring its temperature down to 22°C. What is the amoun

t of heat transferred to water? (ciron = 0.444 J/g°C) A) 1.9 kJ B) 6.9 kJ C) 8.9 kJ Eliminate D) 10.9 kJ
Physics
2 answers:
Amiraneli [1.4K]3 years ago
8 0
The piece of iron is brought from a temperature of 100C to a temperature of 22C. In this process, the heat released by this piece of iron is
Q=m C_s \Delta T
where m=200 g is the mass of the piece of iron, C_s = 0.444 J/(g C) is the iron's specific heat, and the temperature variation is \Delta T = 22^{\circ}-100^{\circ}=-78^{\circ}C. Using these values, the amount of heat released by the iron is
Q=(200 g)(0.441 J/(gC))(-78C)=-6926J=-6.9 kJ
With negative sign because it is amount of heat released. This heat is then trasferred to the water, so the amount of heat absorbed by the water is Q=6.9 kJ.
ASHA 777 [7]3 years ago
5 0

The answer is B) 6.9 kJ.

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Galactic Alliance Junior Mission Officer (GAJMO) Bundit Nermalloy is predicting the kinetic energy of a supply spacecraft, which
antiseptic1488 [7]

Answer:

the ship's energy is greater than this and the crew member does not meet the requirement

Explanation:

In this exercise to calculate kinetic energy or final ship speed in the supply hangar let's use the relationship

                W =∫ F dx = ΔK

                 

Let's replace

             

          ∫ (α x³ + β) dx = ΔK

         α x⁴ / 4 + β x = ΔK

           

Let's look for the maximum distance for which the variation of the energy percent is 10¹⁰ J

         x (α x³ + β) = K_{f} - K₀

          K_{f}  = K₀ + x (α x³ + β)

Assuming that the low limit is x = 0, measured from the cargo hangar

     

Let's calculate

        K_{f}  = 2.7 10¹¹ + 7.5 10⁴ (6.1 10⁻⁹ (7.5 10⁴) 3 -4.1 10⁶)

        Kf = 2.7 10¹¹ + 7.5 10⁴ (2.57 10⁶ - 4.1 10⁶)

        Kf = 2.7 10¹¹ - 1.1475 10¹¹

        Kf = 1.55 10¹¹ J

In the problem it indicates that the maximum energy must be 10¹⁰ J, so the ship's energy is greater than this and the crew member does not meet the requirement

We evaluate the kinetic energy if the System is well calibrated

                W = x F₀ = K_{f} –K₀

                K_{f} = K₀ + x F₀

We calculate

              K_{f} = 2.7 10¹¹ -7.5 10⁴ 3.5 10⁶

               K_{f} = (2.7 -2.625) 10¹¹

              K_{f} = 7.5 10⁹ J

5 0
3 years ago
Select all of the answers that apply.
Svetllana [295]
Researchers found the "cosmic microwave background radiation", which is a heat imprint left over from the big bang.

The redshift of light emitted by most galaxies indicates the universe is expanding.

5 0
3 years ago
Read 2 more answers
A student hangs a weight on a newtonmeter. The energy currently stored in the spring in the newton meter is 0.045N. The student
castortr0y [4]

Answer:

5x10^-3

Explanation:

Hooke's Law states that the force needed to compress or extend a spring is directly proportional to the distance you stretch it.

Hooke's Law can be represented as

<h3> F = kx, </h3>

<em>where F is the force </em>

<em>            k is the spring constant</em>

<em>            x is the extension of the material </em>

<em />

Plug values in the equation

Step 1 find the original extension

0.045 = (400)x

x = 1.125x 10^-4 m d

Step 2 find the new extension

0.045+2 = 400(x)

2.045 = 400x

x = 5.1125x10^-3

Step 3 subtract the new extension with original

Total extension of the spring =  5.1125x10^-3 - 1.125x 10^-4 m = 5x10^-3

8 0
3 years ago
HELP URGENT!!!!!!!!!!!!!!!!!!!!!!!!
nirvana33 [79]
D
Because the rest of the answers are illogical
5 0
3 years ago
In an old-fashioned amusement park ride, passengers stand inside a 4.9-m-diameter hollow steel cylinder with their backs against
Viefleur [7K]

Answer:

24.07415 rpm

Explanation:

\mu = Coefficient of friction = 0.63

v = Velocity

d = Diameter = 4.9 m

r = Radius = \frac{d}{2}=\frac{4.9}{2}=2.45\ m

m = Mass

g = Acceleration due to gravity = 9.81 m/s²

Here the frictional force balances the rider's weight

f=\mu F_n

The centripetal force balances the weight of the person

\mu m\frac{v^2}{r}=mg\\\Rightarrow \mu \frac{v^2}{r}=g\\\Rightarrow v=\sqrt{\frac{gr}{\mu}}\\\Rightarrow v=\sqrt{\frac{9.81\times 2.45}{0.63}}\\\Rightarrow v=6.17656\ m/s

Velocity is given by

v=\omega r\\\Rightarrow \omega=\frac{v}{r}\\\Rightarrow \omega=\frac{6.17656}{2.45}\\\Rightarrow \omega=2.52104\ rad/s

Converting to rpm

2.52104\times \frac{60}{2\pi}=24.07415\ rpm

The minimum angular speed for which the ride is safe is 24.07415 rpm

4 0
3 years ago
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