B. The sound of the engine will get louder and the pitch higher.
Answer:
the correct answer is A, the object goes 4 times as far
Explanation:
This is a projectile launching approach. Where the parameter we are controlling is the initial speed and they ask us how far it goes from the initial one. Let's calculate the range with a speed (vo)
R1 = v₀² sin 2θ / g
Now let's double vo, the new speed is
v = 2 v₀
We calculate the scope
R2 = (2v₀)² sin 2θ / g
R2 = 4 v₀² sin 2θ / g
R2 = 4 R1
Therefore the correct answer is A, the object goes 4 times further
Answer:
1)
2)displacement
3)
Explanation:
At equilibrium position the weight of the man should be balanced by force in the spring
thus we have at equilibrium
Applying values we get
2)
When we add another identical spring we get an equivalent spring with spring constant as
Applying values we get
Thus at equilibrium we have
3) Equivalent spring constant will be as calculated earlier
R 1,2 = 27.5 + 33.0 = 60.5 Ohms
1/ R 1,2,3 = 1/ 60.5 + 1 / 22 = 82.5 / 1331
R 1, 2, 3 = 1331 / 82.5 = 16.13 Ohms
I = U / R
I = 9 V / 16.13 Ohms = 0.557 A ≈ 0.56 A
Answer: C ) 0.56 Amps
Answer:
Both objects travel the same distance.
Explanation:
The two object have the same kinetic energy.
4 x 2^2/2. = 1x 4^2/2
= 8
