Answer: no it shouldn't be
Explanation:
To solve this problem, we must always remember that energy
is conserved. In this case, since he is falling down, he has highest potential
energy at the top and zero at bottom. While his kinetic energy is zero at the top
since he started from rest and highest at the bottom. We can also say that Potential
Energy lost is Kinetic Energy gained thus,
- ΔPE = ΔKE --->
one is negative since PE is losing energy
- m g (h2 – h1) = 0.5 m (v2^2 – v1^2)
Where,
m = mass of tarzan (cancel that out)
g = gravitational acceleration
h2 = height at the bottom= 0
h1 = height at top = 22 m
v2 = velocity at the bottom
v1 = velocity at top = 0 (started from rest)
Therefore substituting all values:
- 9.8 (- 22) = 0.5 (v2^2)
v2 = 20.77 m / s (ANSWER)
Answer:
the electric field is 3.91 x 10⁶ N/C
Explanation:
Given the data in the question;
Electric field at a point due to point charge is;
E = kq/r²
where k is the constant, r is the distance from centre of terminal to point where electric field is, q is the excess charge placed on the centre of terminal of Van de Graff,a generator
Now, given that r = 3.9 m, k = 9.0 x 10⁹ Nm²/C², q = 6.60 mC = 6.60 x 10⁻³ C
so we substitute into the formula
E = [(9.0 x 10⁹ Nm²/C²)( 6.60 x 10⁻³ C)] / ( 3.9 )²
E = 59400000 / 15.21
E = 3.91 x 10⁶ N/C
Therefore, the electric field is 3.91 x 10⁶ N/C
Answer:
The distance the train travels before coming to a (complete) stop = 40/81 km which is approximately 493.83 meters
Explanation:
The initial speed of the train u = 80 km/h = 22 2/9 m/s = 22.
m/s
The magnitude of the constant acceleration with which the train slows, a = 0.5 m/s²
Therefore, we have the following suitable kinematic equation of motion;
v² = u² - 2 × a × s
Where;
v = The final velocity = 0 (The train comes to a stop)
s = The distance the train travels before coming to a stop
Substituting the values gives;
0² = 22.² - 2 × 0.5 × s
2 × 0.5 × s = 22.²
s = 22.²/1 = 493 67/81 m = 40/81 km
The distance the train travels before coming to a (complete) stop = 40/81 km ≈ 493.83 m.
D. velocity
Velocity depends on speed and direction
