Using the Equation:
v² = vi² + 2 · a · s → Eq.1
where,
v = final velocity
vi = initial velocity
a = acceleration
s = distance
<span><span>We know that vi = 0 because the ball was at rest initially.
</span><span>
Therefore,
Solving Eq.1 for acceleration,
</span></span> v² = vi² + 2 · a · s
v² = 0 + 2 · a · s
v² = 2 · a · s
Rearranging for a,
a = v ²/2·<span>s
Substituting the values,
a = 46</span>²/2×1<span>
a = 1058 m/s</span>²
<span>Now applying Newton's 2nd law of motion,
</span>
<span>F = ma
= 0.145</span>×<span>1058
F = 153.4 N</span>
W = mg = 350 newton
m = W/g = 350/9.8 = 35.71 kg
on mars
W = mg = 134 newton
g = W/m = 134/35.71 = 3.75 meters/second2
Answer:
his results in the final angle after the collision of 37.2 degrees basically what we did there is turn the vector into a right triangle. We use sohcahtoa to solve for the angle. Being.
Explanation:
Answer:
Take-off velocity = v = 81.39[m/s]
Explanation:
We can calculate the takeoff speed easily, using the following kinematic equation.
where:
a = acceleration = 4[m/s^2]
x = distance = 750[m]
vi = initial velocity = 25 [m/s]
vf = final velocity
![v_{f}=\sqrt{(25)^{2}+(2*4*750) } \\v_{f}=81.39[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3D%5Csqrt%7B%2825%29%5E%7B2%7D%2B%282%2A4%2A750%29%20%7D%20%5C%5Cv_%7Bf%7D%3D81.39%5Bm%2Fs%5D)
Answer:
Given
Frequency (f) = 3Hz
Wavelength = 9 m
Speed = ?
Explanation:
we know
Speed = wavelength * frequency
= 9*3
= 27 m/ s
