Physics, 26.02.2020 18:48 lilxjayyy

An automobile with a mass of 1180 kg is traveling at a speed v = [06] m/s. a. What is its kinetic energy in SI units? b. What speed (m/s) must an 82.7-kg person move to have the same kinetic energy? c. At what speed (m/s) must a 12.1-g bullet move to have the same kinetic energy? d. What would be the speed (m/s) of the automobile if its kinetic energy were doubled?

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Answer from: jakeit432

Complete Question:

An automobile with a mass of 1180 kg is traveling at a speed v =2.51 m/s. What is its kinetic energy in SI units? What speed (m/s) must an 82.7-kg person move to have the same kinetic energy? At what speed (m/s) must is 12.1-g bullet move to have the same kinetic energy? What would be the speed (m/s) of the automobile if its kinetic energy were doubled?

a) 3717.1 J b) 9.48 m/s c) 783.8 m/s d) 3.55 m/s

Explanation:

By definition, the kinetic energy of a mass m with a speed v, is as follows:

$K = \frac{1}{2} * m *v^{2}$

if m= 1180 Kg, and v= 2.51 m/s, the kinetic energy can be calculated as follows:

$K = \frac{1}{2} * m *v^{2} = \frac{1}{2} * 1180 kg*(2.51 m/s)^{2} = 3717.1 J$

If the kinetic energy must be the same, and m= 82,7 Kg, we can write the following expression:

$K = \frac{1}{2} * m *v^{2} = \frac{1}{2} * 82.7 kg*((v)(m/s))^{2} = 3717.1 J$

We can solve the above equation as follows:

$v =\sqrt{\frac{2*K}{m} } = \sqrt{\frac{2*3717.1J}{82.7kg} } = 9.48 m/s$

If K remains the same, and m = 12.1 g = 0.0121 kg (in SI units). we can solve for v as follows:

$v =\sqrt{\frac{2*K}{m} } = \sqrt{\frac{2*3717.1J}{0.0121kg} } = 783.8 m/s$

Now, if the kinetic energy were doubled, we would have the following equation:

$K = \frac{1}{2} * m *v^{2} = \frac{1}{2} * 1180 kg*((v) m/s)^{2} = 3717.1 J * 2 = 7434.2 J$

We can solve for the new speed v as follows:

$v =\sqrt{\frac{2*K}{m} } = \sqrt{\frac{2*7434.2J}{1180kg} } = 3.55 m/s$

Answer from: Quest

when jerome paddle in flowing river towards east the net speed of jerome will increase

so the net speed of jerome will be given as

$v_j = v_1 + v_2 = 3.100 + 1.350 = 4.45 m/s$

now the speed of paul will be same as it is given as he is running on the ground

$v_p = 3.100 m/s$

so here we can say since the speed of jerome is more so he will win the race

now as we know the distance so the time difference is given as

$\delta t = \frac{d}{v_j} - \frac{d}{v_p}$

$\delta t = \frac{2080}{4.450} - \frac{2080}{3.100}$

$\delta t = -203.55 s$

so jerome will reach first and the time difference will be 203.55 s or 3.4 min

Answer from: Quest

If the heat engine is a perfect carnot cycle engine, thenefficiency = (t2-t1)/t2 where t1=temperature of cold reservoir ( ° k) = 135 t2=temperature of hot reservoir ( ° k) = 325thereforeefficiency = (t2-t1)/t2 = (325-135)/325 = 0.5846 = 58% note: this is the theoretical maximum efficiency of a perfect heat engine. real-life engines do not achieve this value.

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An automobile with a mass of 1180 kg is traveling at a speed v = [06] m/s. a. What is its kinetic e...