Question

modified true/false indicate whether the statement is true or false. if false, change the underlined part of sentence to make the statement true. 2 marks - k/u 1. __ the latent heat of fusion is used to calculate the energy required to turn solids into liquids. 2. substance a heats up quicker than substance b, substance a has a lower specific heat capacity __ multiple choice circle the choice that best completes the statement or answers the question. 3. the unit for power is equivalent to: a. w·s b. w/s c. n/s d. j·s e. \\( \frac{n·m}{s} \\) 4. what will happen to an objects gravitational potential energy if its height is doubled? a. halved b. divided by three c. doubled d. quadrupled e. divided by four 5. one joule is equivalent to: a. the amount of energy required to raise the temperature by 1°c. b. the amount of energy required to complete certain work in one second. c. the amount of energy required to apply one newton of force for one meter. d. the amount of force required to lift a mass by one meter. e. none of the above. 6. a box slides down and stops on a ramp due to friction. which statement is true? a. the kinetic energy throughout the motion is the same. b. the mechanical energy is conserved. c. the work done by frictional force is always greater than the work done by gravity. d. some gravitational potential energy is transformed into heat energy. e. none of the above. 7. an unknown amount of gold is melted into a puddle of liquid with 8.0 × 10⁷ j of energy. what is the mass of the gold given that \\( l_f = 1.1 × 10⁶ j/kg \\) and \\( l_v = 6.4 × 10⁴ j/kg \\) for gold, assume that the golds temperature is already at the melting point. a. 73 kg b. 1250 kg c. 8.8 × 10¹³ kg d. 5.1 × 10¹² kg e. not enough information 8. a light bulb is rated at 32.0% efficiency, how much energy is wasted if the output energy is 12.0 j? a. 37.5 j b. 3.84 j c. 25.5 j d. 33.7 j e. 0 j 9. the potential energy stored in a diver above the waters surface is 5730 j. if the diver is 9.00 meters above water. what is the mass of the diver? a. 51600 kg b. 505000 kg c. 637 kg d. 65 kg e. 36 kg

Explanation:

Step1: Analyze question 1

The latent heat of fusion is indeed used to calculate the energy required to turn solids into liquids. The formula is \(Q = mL_f\) where \(Q\) is the energy, \(m\) is the mass and \(L_f\) is the latent heat of fusion.

Step2: Analyze question 2

The specific - heat formula is \(Q=mc\Delta T\). If \(Q\) (heat energy) and \(m\) (mass) are the same, and \(\Delta T\) (change in temperature) is larger (heats up quicker), then \(c\) (specific heat capacity) is smaller. So the statement is True.

Step3: Analyze question 3

Power \(P=\frac{W}{t}\), where \(W\) (work) has unit \(J\) (Joule) and \(t\) (time) has unit \(s\) (second). Also, \(W = F\times d\) (force - displacement), \(F\) has unit \(N\) (Newton) and \(d\) has unit \(m\) (meter). So \(P=\frac{W}{t}=\frac{F\times d}{t}\). Since \(1\ W = 1\ J/s\) and \(1\ J=1\ N\cdot m\), then \(1\ W=\frac{N\cdot m}{s}\).

Step4: Analyze question 4

The gravitational potential energy formula is \(U = mgh\). If \(h\) is doubled (\(h' = 2h\)), then \(U'=mg(2h)=2mgh = 2U\).

Step5: Analyze question 5

By definition, \(1\ J=1\ N\cdot m\) (work done \(W = F\times d\), where \(F\) is force in Newtons and \(d\) is displacement in meters).

Step6: Analyze question 6

When a box slides down a ramp and stops due to friction:

• Kinetic energy changes (initially it has some kinetic energy, then it stops, so \(K = 0\) at the end), so option a is False.
• Mechanical energy (\(E = K+U\)) is not conserved because of non - conservative (frictional) force, so option b is False.
• The work done by frictional force \(W_f=-\mu_kmg\cos\theta\times s\) and work done by gravity \(W_g = mg\sin\theta\times s\). The relationship between them depends on \(\mu_k\) and \(\theta\), so option c is False.
• Gravitational potential energy is converted into heat (due to friction), so option d is True.

Step7: Analyze question 7

Since the gold is at the melting point, the energy used for melting is \(Q = mL_f\). Given \(Q = 8.0\times10^{7}\ J\) and \(L_f=1.1\times 10^{6}\ J/kg\), then \(m=\frac{Q}{L_f}=\frac{8.0\times 10^{7}}{1.1\times 10^{6}}\approx73\ kg\).

Step8: Analyze question 8

If the efficiency \(\eta = 32\%=0.32\) and the output energy \(E_{out}=12.0\ J\). Let the input energy be \(E_{in}\). Then \(\eta=\frac{E_{out}}{E_{in}}\), so \(E_{in}=\frac{E_{out}}{\eta}=\frac{12}{0.32}=37.5\ J\). The wasted energy \(E_{wasted}=E_{in}-E_{out}=37.5 - 12=25.5\ J\).

Step9: Analyze question 9

Using the gravitational potential energy formula \(U = mgh\), where \(U = 5730\ J\), \(g = 9.8\ m/s^{2}\), and \(h = 9.00\ m\). Then \(m=\frac{U}{gh}=\frac{5730}{9.8\times9}\approx65\ kg\).

Answer:

1. True
2. True
3. e. \(\frac{N\cdot m}{s}\)
4. c. Doubled
5. c. The amount of energy required to apply one Newton of force for one meter.
6. d. Some gravitational potential energy is transformed into heat energy.
7. a. \(73\ kg\)
8. c. \(25.5\ J\)
9. d. \(65\ kg\)