The distance travelled is 10 m and the velocity gained at the end of this time is 2 m/s.
<h3>Velocity of the object at the end of the time</h3>
F = mv/t
where;
- m is mass of the object
- v is velocity of the object
- t is time
Ft = mv
v = Ft/m
v = (50 x 10)/250
v = 2 m/s
<h3>Distance traveled by the object</h3>
v² = u² + 2as
where;
u is initial velocity = 0
a is acceleration
a = F/m
a = 50 N/ 250 kg
a = 0.2 m/s²
v² = 0 + 2as
s = v²/2a
s = (2²)/(2 x 0.2)
s = 10 m
Thus, the distance travelled is 10 m and the velocity gained at the end of this time is 2 m/s.
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Answer:
The resistance is 
Ω
Explanation:
<u>Data</u>
diameter: 
, 
Radius: 
Resistivity constant: 
Ω
To find Area:
To find circumference:
Now we can apply the Resistivity formula:
Ω
Ω
The time taken by the stone to hit the ground would be 5.12 seconds.
<h3>What are the three equations of motion?</h3>
There are three equations of motion given by Newton
The first equation is given as follows
v = u + at
the second equation is given as follows
S = ut + 1/2×a×t²
the third equation is given as follows
v² - u² = 2×a×s
Keep in mind that these calculations only apply to uniform acceleration.
As given in the problem, a stone is dropped from the helicopter which is ascending at the speed of 19.6 m/s
height(S) = 156.8 meters
initial velocity(u) = -19.6 m/s
acceleration(a) = 9.81 m/s²
By using the second equation of motion given by newton
S = ut + 1/2at²
S = 156.8m ,u= -19.6 m/s , a= 9.81 m/s² and t =? seconds
156.8= -19.6t + 9.81t²
t = 5.12 seconds
Thus, the time taken by the stone to hit the ground would be 5.12 seconds.
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Answer:
b) the reflected pulse returns inverted while the transmitted pulse is right side up
Explanation:
A wave pulse traveling to the right along a thin cord reaches a discontinuity where the rope becomes thicker and heavier.
Then, as far as orientation of reflected and transmitted pulses are concerned the reflected pulse returns inverted while the transmitted pulse is right side up.
Hence the correct answer is b.
Answer:
3.12 x 10^-5 m
Explanation:
Length of steel column, L = 4 m
diameter, d = 0.2 m
radius = half of diameter = 0.1 m
Young's modulus, Y = 2 x 10^11 N/m^2
Mass of truck, m = 5000 kg
Force, F = mass of truck x acceleration due to gravity
F = 5000 x 9.8 = 49000 N
Area of crossection of cable,
A = 
Let ΔL be the shrink in length of cable, then by the formula of Young's modulus
ΔL = 3.12 x 10^-5 m
Thus, the shrink in the length of cable is 3.12 x 10^-5 m.
