Answer:
<em>The range is 35.35 m</em>
Explanation:
<u>Projectile Motion</u>
It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.
Being vo the initial speed of the object, θ the initial launch angle, and
the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:

The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:



The range is 35.35 m
Answer:
The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Explanation:
Given that,
Amplitude = 0.08190 m
Frequency = 2.29 Hz
Wavelength = 1.87 m
(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave
Using formula of distance

Where, d = distance
A = amplitude
Put the value into the formula


Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Explanation:
so sorry
don't know but please mark me as brainliest please
A wave is a disturbance that moves along a medium from one end to the other. If one watches an ocean wave moving along the medium (the ocean water), one can observe that the crest of the wave is moving from one location to another over a given interval of time. The crest is observed to cover distance. The speed of an object refers to how fast an object is moving and is usually expressed as the distance traveled per time of travel. In the case of a wave, the speed is the distance traveled by a given point on the wave (such as a crest) in a given interval of time. In equation form,
Answer:
t = 1.05 s
Explanation:
Given,
The distance between your vehicle and car, 100 ft
The constant speed of your vehicle, u = 95 ft/s
Since, the velocity is constant, a =0
If the car stopped suddenly, time left for you to hit the brake, t = ?
Using the second equation of motion,
S = ut + ½ at²
Substituting the given values in the equation
100 = 95 x t
t = 100/95
= 1.05 s
Hence, the time left for you to hit the brakes and stop before rear ending them, t = 1.05 s