Answer:
The correct answer is "6666.67 N".
Explanation:
The given values are:
Mass,
m = 0.100
Relative speed,
v = 4.00 x 10³
time,
t = 6.00 x 10⁻⁸
As we know,
⇒ 
On substituting the given values, we get
⇒ 
⇒ 
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A sound wave<span> in a steel rail </span>has<span> a </span>frequency of<span> 620 </span>Hz<span> and a </span>wavelength<span> of 10.5 ... Find the </span>speed<span> of </span>a wave<span> with a </span>wavelength of 5<span> m and a </span>frequency of<span> 68 </span>Hz<span>.</span>
(a) At a corresponding hill on Earth and a lesser gravity on planet Epslion, the height of the hill will cause a reduction in the initial speed of the snowboarder from 4 m/s to a value greater than zero (0).
(b) If the initial speed at the bottom of the hill is 5 m/s, the final speed at the top of the hill be greater than 3 m/s.
<h3>
Conservation of mechanical energy</h3>
The effect of height and gravity on speed on the given planet Epislon is determined by applying the principle of conservation of mechanical energy as shown below;
ΔK.E = ΔP.E
¹/₂m(v²- u²) = mg(hi - hf)
¹/₂(v²- u²) = g(0 - hf)
v² - u² = -2ghf
v² = u² - 2ghf
where;
- v is the final velocity at upper level
- u is the initial velocity
- hf is final height
- g is acceleration due to gravity
when u² = 2gh, then v² = 0,
when gravity reduces, u² > 2gh, and v² > 0
Thus, at a corresponding hill on Earth and a lesser gravity on planet Epslion, the height of the hill will cause a reduction in the initial speed of the snowboarder from 4 m/s to a value greater than zero (0).
<h3>Final speed</h3>
v² = u² - 2ghf
where;
- u is the initial speed = 5 m/s
- g is acceleration due to gravity and its less than 9.8 m/s²
- v is final speed
- hf is equal height
Since g on Epislon is less than 9.8 m/s² of Earth;
5² - 2ghf > 3 m/s
Thus, if the initial speed at the bottom of the hill is 5 m/s, the final speed at the top of the hill be greater than 3 m/s.
Learn more about conservation of mechanical energy here: brainly.com/question/6852965
Answer:
t = 1.58 s
Explanation:
given,
Speed of ranger, v = 56 km/h
v = 56 x 0.278 = 15.57 m/s
distance, d = 65 m
deceleration,a = 3 m/s²
reaction time = ?
using stopping distance formula


t is the reaction time

t = 1.58 s
hence, the reaction time of the ranger is equal to 1.58 s.