To start with solving this
problem, let us assume a launch angle of 45 degrees since that gives out the
maximum range for given initial speed. Also assuming that it was launched at
ground level since no initial height was given. Using g = 9.8 m/s^2, the
initial velocity is calculated using the formula:
(v sinθ)^2 = (v0 sinθ)^2
– 2 g d
where v is final
velocity = 0 at the peak, v0 is the initial velocity, d is distance = 11 m
Rearranging to find for
v0: <span>
v0 = sqrt (d * g/ sin(2 θ)) </span>
<span>v0 = 10.383 m/s</span>
Answer:
the magnitude of the force that one particle exerts on the other is 79.08 N
Explanation:
given information:
q₁ = 3.77 μC = -3.77 x 10⁻⁶ C
q₂ = 4.39 μC = 4.39 x 10⁻⁶ C
r = 4.34 cm = 4.34 x 10⁻² m
What is the magnitude of the force that one particle exerts on the other?
lFl = kq₁q₂/r²
= (9 x 10⁹) (3.77 x 10⁻⁶) (4.39 x 10⁻⁶)/(4.34 x 10⁻²)²
= 79.08 N
Answer
given,
mass of satellite = 2400 Kg
speed of the satellite = 6.67 x 10³ m/s
acceleration of satellite = ?
gravitational force of the satellite will be equal to the centripetal force
Assuming the radius of circular orbit = 8.92 x 10⁶ m
now,
F = 11970.11 N
acceleration,
a = 4.98 m/s²
Answer:0.69
Explanation:
Coefficient of kinetic friction=f/R=61.8/90=0.69
Answer:
Mass of the wooden Block is 20g.
Explanation:
The buoyant force equation will be used here
Buoyant Force= ρ*g*1/2V Here density used is of water
m*g= ρ*g*1/2V
Simplifying the above equation
2m= ρ*V Eq-1
Also we know from the question that
ρ*V = m + 0.020 Eq-2 ( Density = (Mass+20g)/Volume )
Equating Eq-1 & Eq-2 we get
2m = m+0.020
m = 0.020kg
m = 20g
