The nut distance from the Bullwinkle after uniform motion is 21 m.
We need to know about the uniform motion to solve this problem. The uniform motion is an object's motion under acceleration. It should follow the rule
vt = vo + a . t
vt² = vo² + 2a . s
s = vo . t + 1/2 . a . t²
where vt is final velocity, vo is initial velocity, a is acceleration, t is time and s is displacement.
From the question above, the parameters given are
m = 0.5 kg
s = 15 m
vx = 12 m/s
vo = 0 m/s
a = g = 9.8 m/s²
Find the time taken of the nut for landing
s = vo . t + 1/2 . a . t²
15 = 0 + 1/2 . 9.8 . t²
t² = 3.06
t = 1.75 s
Find the distance of nut in horizontal direction
vx = x / t
12 = x / 1.75
x = 12 . 1.75
x = 21 m
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A group of organs working together comprises an organ system, B.
Answer:
v= 1.71 m/s
Explanation:
Given that
Distance between two successive crests = 4.0 m
λ = 4 m
T= 7 sec
T is the time between 3 waves.
3 waves = 7 sec
1 wave = 7 /3 sec
So t= 7/3 s
We know that frequency f
f= 1/t= 3/7 Hz
Lets take speed of the wave is v
v= f λ
f=frequency
λ=wavelength
v= 3/7 x 4 = 12 /7
v= 1.71 m/s
Answer:
Explanation:
Given that,
Mass m = 2kg
Initial Velocity Vi = 3m/s
Force applied. = 4N
Distance cause by the force applied d = 5m
a. Work done by the force applied?
Work done by force applied can be determined by
W = F×d•Cosx
Where x is direction is the force and displacement
The force and displacement are in the same direction, then, x =0°
W = F×d•Cosx
W = 4×5Cos0
W = 20J.
b. Final velocity
The change in kinetic energy is equal to work done
∆K.E = W
½mVf² - ½mVi² = W
½mVf² = ½mVi² + W
½ × 2 Vf² = ½ × 2 × 3² + 20
Vf² = 9+20
Vf² = 29
Vf = √29
Vf = 5.39m/s
The final velocity is 5.39m/s