Based on the calculations, the speed required for this satellite to stay in orbit is equal to 1.8 × 10³ m/s.
<u>Given the following data:</u>
- Gravitational constant = 6.67 × 10⁻¹¹ m/kg²
- Mass of Moon = 7.36 × 10²² kg
- Distance, r = 4.2 × 10⁶ m.
<h3>How to determine the speed of this satellite?</h3>
In order to determine the speed of this satellite to stay in orbit, the centripetal force acting on it must be sufficient to change its direction.
This ultimately implies that, the centripetal force must be equal to the gravitational force as shown below:
Fc = Fg
mv²/r = GmM/r²
<u>Where:</u>
- m is the mass of the satellite.
Making v the subject of formula, we have;
v = √(GM/r)
Substituting the given parameters into the formula, we have;
v = √(6.67 × 10⁻¹¹ × 7.36 × 10²²/4.2 × 10⁶)
v = √(1,168,838.095)
v = 1,081.13 m/s.
Speed, v = 1.8 × 10³ m/s.
Read more on speed here: brainly.com/question/20162935
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Answer:
52.9 N, 364.7 N
Explanation:
First of all, we need to resolve both forces along the x- and y- direction. We have:
- Force A (178 N)
- Force B (259 N)
So the x- and y- component of the total force acting on the block are:
Answer:
a) y= 3.5 10³ m, b) t = 64 s
Explanation:
a) For this exercise we use the vertical launch kinematics equation
Stage 1
y₁ = y₀ + v₀ t + ½ a t²
y₁ = 0 + 0 + ½ a₁ t²
Let's calculate
y₁ = ½ 16 10²
y₁ = 800 m
At the end of this stage it has a speed
v₁ = vo + a₁ t₁
v₁ = 0 + 16 10
v₁ = 160 m / s
Stage 2
y₂ = y₁ + v₁ (t-t₀) + ½ a₂ (t-t₀)²
y₂ = 800 + 150 5 + ½ 11 5²
y₂ = 1092.5 m
Speed is
v₂ = v₁ + a₂ t
v₂ = 160 + 11 5
v₂ = 215 m / s
The rocket continues to follow until the speed reaches zero (v₃ = 0)
v₃² = v₂² - 2 g y₃
0 = v₂² - 2g y₃
y₃ = v₂² / 2g
y₃ = 215²/2 9.8
y₃ = 2358.4 m
The total height is
y = y₃ + y₂
y = 2358.4 + 1092.5
y = 3450.9 m
y= 3.5 10³ m
b) Flight time is the time to go up plus the time to go down
Let's look for the time of stage 3
v₃ = v₂ - g t₃
v₃ = 0
t₃ = v₂ / g
t₃ = 215 / 9.8
t₃ = 21.94 s
The time to climb is
= t₁ + t₂ + t₃
t_{s} = 10+ 5+ 21.94
t_{s} = 36.94 s
The time to descend from the maximum height is
y = v₀ t - ½ g t²
When it starts to slow down it's zero
y = - ½ g t_{b}²
t_{b} = √-2y / g
t_{b} = √(- 2 (-3450.9) /9.8)
t_{b} = 26.54 s
Flight time is the rise time plus the descent date
t = t_{s} + t_{b}
t = 36.94 + 26.54
t =63.84 s
t = 64 s
Answer:
0.187 m
Explanation:
We'll begin by calculating the acceleration of the ball. This can be obtained as follow:
Mass (m) = 0.450 Kg
Force (F) = 38 N
Acceleration (a) =?
F = m × a
38 = 0.450 × a
Divide both side by 0.450
a = 38 / 0.450
a = 84.44 m/s²
Finally, we shall determine the distance. This can be obtained as follow:
Initial velocity (u) = 2.20 m/s.
Final velocity (v) = 6 m/s
Acceleration (a) = 84.44 m/s²
Distance (s) =?
v² = u² + 2as
6² = 2.2² + (2 × 84.44 × s)
36 = 4.4 + 168.88s
Collect like terms
36 – 4.84 = 168.88s
31.52 = 168.88s
Divide both side by 168.88
s = 31.52 / 168.88
s = 0.187 m
Thus, the distance is 0.187 m
One form of Ohm's Law says . . . . . Resistance = Voltage / Current .
R = V / I
R = (12 v) / (0.025 A)
R = (12 / 0.025) (V/I)
<em>R = 480 Ohms</em>
I don't know if the current in the bulb is steady, because I don't know what a car's "accumulator" is. (Floogle isn't sure either.)
If you're referring to the car's battery, then the current is quite steady, because the battery is a purely DC storage container.
If you're referring to the car's "alternator" ... the thing that generates electrical energy in a car to keep the battery charged ... then the current is pulsating DC, because that's the form of the alternator's output.
