Answer:
It took the projectile 120 s to reach the maximum height.
Explanation:
Given;
maximum height of the projectile, s = 180 km = 180,000 m
initial speed of the projectile, u = 3 km/s = 3000 m/s
final velocity at maximum height, v = 0
Apply the following kinematic equation for average velocity of the projectile;
Therefore, it took the projectile 120 s to reach the maximum height.
Answer:
529.15 m/s
Explanation:
h = Maximum height = 70000 m
g = Acceleration due to gravity = 2 m/s²
m = Mass of sulfur
As the potential and kinetic energies are conserved
The speed with which the liquid sulfur left the volcano is 529.15 m/s
Answer:
3A is the larger of the two currents.
Explanation:
Let the currents in the two wires be I₁ and I₂
given:
Magnitude of the electric field, B = 4.0μT = 4.0×10⁻⁶T
Distance, R = 10cm = 0.1m
Ratio of the current = I₁ : I₂ = 3 : 1
Now, the magnitude of a magnetic field at a distance 'R' due to the current 'I' is given as
Where is the magnitude constant = 4π×10⁻⁷ H/m
Thus, the magnitude of a magnetic field due to I₁ will be
given,
B = B₁ - B₂ (since both the currents are in the same direction and parallel)
substituting the values of B, B₁ and B₂
we get
4.0×10⁻⁶T = -
or
4.0×10⁻⁶T =
also
⇒
substituting the values in the above equation we get
4.0×10⁻⁶T =
⇒
also
⇒
⇒
Hence, the larger of the two currents is 3A
Answer:
y = 54.9 m
Explanation:
For this exercise we can use the relationship between the work of the friction force and mechanical energy.
Let's look for work
W = -fr d
The negative sign is because Lafourcade rubs always opposes the movement
On the inclined part, of Newton's second law
Y Axis
N - W cos θ = 0
The equation for the force of friction is
fr = μ N
fr = μ mg cos θ
We replace at work
W = - μ m g cos θ d
Mechanical energy in the lower part of the embankment
Em₀ = K = ½ m v²
The mechanical energy in the highest part, where it stopped
= U = m g y
W = ΔEm = - Em₀
- μ m g d cos θ = m g y - ½ m v²
Distance d and height (y) are related by trigonometry
sin θ = y / d
y = d sin θ
- μ m g d cos θ = m g d sin θ - ½ m v²
We calculate the distance traveled
d (g syn θ + μ g cos θ) = ½ v²
d = v²/2 g (sintea + myy cos tee)
d = 9.8 12.6 2/2 9.8 (sin16 + 0.128 cos 16)
d = 1555.85 /7.8145
d = 199.1 m
Let's use trigonometry to find the height
sin 16 = y / d
y = d sin 16
y = 199.1 sin 16
y = 54.9 m