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iVinArrow [24]
3 years ago
15

Two identical objects are moving directly toward one another at the same speed v. ~v â~v m m what is the total kinetic energy of

the system of the two objects?
Physics
1 answer:
Tresset [83]3 years ago
3 0
If they are identical, then their masses are the same. The speeds are the same. KE has no direction.
Total KE = 1/2 m v^2 + 1/2 mv^2 = mv^2

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Which of the following objects is accelerating?
jenyasd209 [6]

Answer:

A

Explanation:

Constant speed (without change in direction) is not accelerating. If you are slowing down, speeding up, or changing direction, you are accelerating

4 0
3 years ago
How much money would be saved by turning off one 100.0-W lightbulb 3.0 h/day for 365 days if the
Pavlova-9 [17]

Answer:

the money that would be saved is $13.14.

Explanation:

Given;

power consumed by the light bulb, P = 100 W = 0.1 kW

time of running the bulb, t = 3 hours for 365 days = 1,095 hours

cost rate of power consumption, C = $0.12 per kWh

Energy consumed by the light bulb for the given days;

E = Pt

E = 0.1 kW  x 1,095 hr

E = 109.5 kWh

Cost of energy consumed = 109.5 kWh   x   $0.12 / kWh

                                            = $13.14

Therefore, the money that would be saved is $13.14.

3 0
2 years ago
Rearrange the formula for mechanical energy to solve for height:
docker41 [41]

Explanation:

Given formula:

            ME=\frac{1}{2}mv²+mgh

  To make height the subject of the formula, follow the following procedures;

     Subtract \frac{1}{2}mv² from both side of equation

 M.E - \frac{1}{2}mv² = \frac{1}{2}mv² - \frac{1}{2}mv²+mgh

                  This gives:

                        M.E - \frac{1}{2}mv² = mgh

Multiply both sides of the expression by \frac{1}{mg}

  ( M.E -  \frac{1}{2}mv² ) x  \frac{1}{mg} =  \frac{1}{mg} x mgh

       h = ( M.E -  \frac{1}{2}mv² ) x  \frac{1}{mg}

Learn more:

Kinetic energy brainly.com/question/6536722

#learnwithBrainly

4 0
3 years ago
Circle the letter of each expression that has four significant figures. A. 1.25 x 10^4 B. 12.51 C. 0.0125 D. 0.1255
Andreyy89

Answer:

letter B

none zero digit are significant figures

3 0
3 years ago
A cyclical heat engine, operating between temperatures of 450º C and 150º C produces 4.00 MJ of work on a heat transfer of 5.00
gogolik [260]

Answer:

(a) Heat transfer to the environment is: 1 MJ and (b) The efficiency of the engine is: 41.5%

Explanation:

Using the formula that relate heat and work from the thermodynamic theory as:W=Q=Q_{in}-Q_{out} solving to Q_out we get:Q_{out}=Q_{in}-W=5(MJ)-4(MJ)=1(MJ) this is the heat out of the cycle or engine, so it will be heat transfer to the environment. The thermal efficiency of a Carnot cycle gives us: n=1-\frac{T_{Low} }{T_{High}} where T_Low is the lowest cycle temperature and T_High the highest, we need to remember that a Carnot cycle depends only on the absolute temperatures, if you remember the convertion of K=°C+273.15 so T_Low=150+273.15=423.15 K and T_High=450+273.15=723.15K and replacing the values in the equation we get:n=1-\frac{423.15}{723.15} =0.415=41.5%

5 0
3 years ago
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