This is an example Newton's Third Law. All the kinectic energy from the moving car transferred the potential energy of the parked car. This potential is not much since the brakes are on (hopefully) and it's not in a non-moving position.
First, we convert kcal to joules:
1 kcal = 4.184 kJ
475 kcal = 1987.4 kJ
Now, calculating the change in internal energy:
ΔU = Q + W; where Q is the heat supplied to the system and W is the work done on the system.
ΔU = -500 + 1987.4
ΔU = 1487.4 kJ
The minimum stopping distance when the car is moving at 32.0 m/s is 348.3 m.
<h3>
Acceleration of the car </h3>
The acceleration of the car before stopping at the given distance is calculated as follows;
v² = u² + 2as
when the car stops, v = 0
0 = u² + 2as
0 = 15² + 2(76.5)a
0 = 225 + 153a
-a = 225/153
a = - 1.47 m/s²
<h3>Distance traveled when the speed is 32 m/s</h3>
If the same force is applied, then acceleration is constant.
v² = u² + 2as
0 = 32² + 2(-1.47)s
2.94s = 1024
s = 348.3 m
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The energy in electron volts of the photons that has the following frequencies is as follows:
- 620 THz = 2.564eV
- 3.10GHz = 1.28 × 10-⁵eV
- 46.0 MHz = 1.902 × 10-⁷eV
<h3>How to calculate energy?</h3>
The energy of a photon can be calculated using the following formula:
E = hf
Where;
- E = energy
- h = Planck's constant (6.626 × 10-³⁴ J/s)
- f = frequency
First, we convert the frequencies to hertz as follows;
- 620THz = 6.2 × 10¹⁴Hz
- 3.10GHz = 3.1 × 10⁹Hz
- 46.0MHz = 4.6 × 10⁷Hz
- E = 6.626 × 10-³⁴ × 6.2 × 10¹⁴ = 2.564eV
- E = 6.626 × 10-³⁴ × 3.1 × 10⁹ = 1.28 × 10-⁵eV
- E = 6.626 × 10-³⁴ × 4.6 × 10⁷ = 1.902 × 10-⁷eV
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Answer:
a = 0.1 s b. 10 s
Explanation:
Given that,
The frequency in circular motion, f = 10 Hz
(a) Let T is the period of itsrotation. We know that,
T = 1/f
So,
T = 1/10
= 0.1 s
(b) Frequency is number of rotations per unit time. So,

Hence, this is the required solution.