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sattari [20]
3 years ago
9

You have to heat up a 1.000 gram ingot of aluminum from initial temperature Ti = 847 K to its melting point, 933 K. Calculate th

e number of calories required for this heat up process. Type in the numeric part of your answer to the nearest 0.1 calorie. E.g., if your answer is 31.38 calories, than type 31.4 in the answer box.
Physics
1 answer:
alexdok [17]3 years ago
3 0

Answer: 18.9 calories

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

Q=m\times c\times \Delta T

Q = Heat absorbed

m= mass of substance = 1.0 g  = 0.001 kg

c = specific heat capacity  of aluminium = 921.096J/kgK

Initial temperature = T_i = 847 K

Final temperature = T_f  = 933 K

Change in temperature ,\Delta T=T_f-T_i=(933-847)K=86K

Putting in the values, we get:

Q=0.001kg\times 921.096J/kgK\times 86K

Q=79.2J=18.9cal      (1J=0.24cal)

The number of calories required for this heat up process is 18.9 calories.

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A 5.0-kilogram sphere, starting from rest, falls freely 22 meters in 3.0 seconds near the surface of a planet. Compared to the a
Whitepunk [10]

Answer:

C) one-half as great

Explanation:

We can calculate the acceleration of gravity in that planet, using the following kinematic equation:

\Delta x=v_0t+\frac{gt^2}{2}

In this case, the sphere starts from rest, so v_0=0. Replacing the given values and solving for g':

g'=\frac{2\Delta x}{t^2}\\g'=\frac{2(22m)}{(3s)^2}\\g'=4.89\frac{m}{s^2}

The acceleration due to gravity near Earth's surface is g=9.8\frac{m}{s^2}. So, the acceleration due to gravity near the surface of the planet is approximately one-half of the acceleration due to gravity near Earth's surface.

5 0
2 years ago
What is the acceleration of a 7 kg mass if the force of 70 N is used to move it toward the Earth?
Assoli18 [71]

Answer:

<h2>10 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

m is the mass

f is the force

From the question we have

a =  \frac{70}{7}  = 10 \\

We have the final answer as

<h3>10 m/s²</h3>

Hope this helps you

4 0
3 years ago
Can someone please help me out with this quiz will give brainiest and thanks to people
Virty [35]

Answer:

Energy transferred = 28.8 Joules.

1. Energy transferred = 144 Joules.

2. The unit of potential difference, volts can also be described as Joules per Coulombs.

3. Current, I = 6.945 Amperes.  

Explanation:

<u>Part A.</u>

Given the following data;

Current, I = 1.2A

Time, t = 2 minutes

Potential difference, V = 12 volts.

To find the energy transfered;

Energy transferred = charge moved * potential difference

E = Q * V

Substituting into the equation, we have;

Energy transferred = (1.2 * 2) * 12

Energy transferred = 2.4 * 12

Energy transferred = 28.8 Joules.

<u>Part B.</u>

1. <em><u>Given the following data;</u></em>

Charge, Q = 24C

Potential difference = 6V

To find the energy transferred;

E = Q * V

Substituting into the equation, we have;

E = 24 * 6

E = 144 Joules.

2. Since we know that, Energy transferred = charge moved * potential difference

Potential \; difference = \frac {Energy \; transferred}{Charged \; moved}

The units of energy is Joules while the unit of the quantity of charge moved is Coulomb.

Therefore, the unit of potential difference becomes Joules per Coulomb.

3. <em><u>Given the following data;</u></em>

Potential difference = 18V

Energy transferred = 500J

Time, t = 4 minutes.

To find the current;

E = Q * V

Substituting into the equation, we have;

500 = Q*18

Q = 500/18

Q = 27.78C

But, Charge moved (Q) = current (I) * time (t)

Current, I = Q/t

Substituting into the equation, we have;

Current, I = 27.78/4

Current, I = 6.945 Amperes..

3 0
3 years ago
Why are all machines not 100% efficient<br> A. Mass<br> B. Gravity <br> C. Friction <br> D. Distance
Anon25 [30]
All machines are not 100% efficient because of <span>C. Friction</span>
7 0
3 years ago
Read 2 more answers
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
2 years ago
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