v = x/t
v = average velocity, x = displacement, t = elapsed time
Given values:
x = 6km south, t = 60min
Plug in and solve for v:
v = 6/60
v = 0.1km/min south
Increasing the number of bulbs in a series circuit decreases the brightness of the bulbs. In a series circuit, the voltage is equally distributed among all of the bulbs. Bulbs in parallel are brighter than bulbs in series. In a parallel circuit the voltage for each bulb is the same as the voltage in the circuit.
Answer:
v₀ₓ = 63.5 m/s
v₀y = 54.2 m/s
Explanation:
First we find the net launch velocity of projectile. For that purpose, we use the formula of kinetic energy:
K.E = (0.5)(mv₀²)
where,
K.E = initial kinetic energy of projectile = 1430 J
m = mass of projectile = 0.41 kg
v₀ = launch velocity of projectile = ?
Therefore,
1430 J = (0.5)(0.41)v₀²
v₀ = √(6975.6 m²/s²)
v₀ = 83.5 m/s
Now, we find the launching angle, by using formula for maximum height of projectile:
h = v₀² Sin²θ/2g
where,
h = height of projectile = 150 m
g = 9.8 m/s²
θ = launch angle
Therefore,
150 m = (83.5 m/s)²Sin²θ/(2)(9.8 m/s²)
Sin θ = √(0.4216)
θ = Sin⁻¹ (0.6493)
θ = 40.5°
Now, we find the components of launch velocity:
x- component = v₀ₓ = v₀Cosθ = (83.5 m/s) Cos(40.5°)
<u>v₀ₓ = 63.5 m/s</u>
y- component = v₀y = v₀Sinθ = (83.5 m/s) Sin(40.5°)
<u>v₀y = 54.2 m/s</u>
Answer: Option B: 1.3×10⁵ W
Explanation:
Work Done, 
Where s is displacement in the direction of force and F is force.
where, v is the velocity.
It is given that, F = 5.75 × 10³N
v = 22 m/s
P = 5.75 × 10³N×22 m/s = 126.5 × 10³ W ≈1.3×10⁵W
Thus, the correct option is B
Force is a vector quantity
so pulling from opposite side will be negative
so
750+(-500)= 250N
C is the right answer
becauseause the man on the right applies greater force.
