(a) The plane makes 4.3 revolutions per minute, so it makes a single revolution in
(1 min) / (4.3 rev) ≈ 0.2326 min ≈ 13.95 s ≈ 14 s
(b) The plane completes 1 revolution in about 14 s, so that in this time it travels a distance equal to the circumference of the path:
(2<em>π</em> (23 m)) / (14 s) ≈ 10.3568 m/s ≈ 10 m/s
(c) The plane accelerates toward the center of the path with magnitude
<em>a</em> = (10 m/s)² / (23 m) ≈ 4.6636 m/s² ≈ 4.7 m/s²
(d) By Newton's second law, the tension in the line is
<em>F</em> = (1.3 kg) (4.7 m/s²) ≈ 6.0627 N ≈ 6.1 N
Answer:
128 m
Explanation:
From the question given above, the following data were obtained:
Horizontal velocity (u) = 40 m/s
Height (h) = 50 m
Acceleration due to gravity (g) = 9.8 m/s²
Horizontal distance (s) =?
Next, we shall determine the time taken for the package to get to the ground.
This can be obtained as follow:
Height (h) = 50 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
50 = ½ × 9.8 × t²
50 = 4.9 × t²
Divide both side by 4.9
t² = 50 / 4.9
t² = 10.2
Take the square root of both side
t = √10.2
t = 3.2 s
Finally, we shall determine where the package lands by calculating the horizontal distance travelled by the package after being dropped from the plane. This can be obtained as follow:
Horizontal velocity (u) = 40 m/s
Time (t) = 3.2 s
Horizontal distance (s) =?
s = ut
s = 40 × 3.2
s = 128 m
Therefore, the package will land at 128 m relative to the plane
Answer:
1) 6 seconds
2) 60 m/s
Explanation:
Given:
Δy = 180 m
v₀ = 0 m/s
a = 10 m/s²
1) Find t.
Δy = v₀ t + ½ at²
180 m = (0 m/s) t + ½ (10 m/s²) t²
t = 6 s
2) Find v.
v² = v₀² + 2aΔy
v² = (0 m/s)² + 2 (10 m/s²) (180 m)
v = 60 m/s
Cody ...
Everything on this page is solved with the SAME formula !
Distance = (speed) x (time) .
Before I get into how to solve each problem, we need to notice that
this whole sheet deals with speed, NOT velocity.
'Velocity' is speed AND THE DIRECTION OF THE MOTION.
Nothing on this page ever mentions direction, so there's no velocity
anywhere on the page.
Your teacher may not be happy if you talk about this on your homework,
but that's too bad. Just don't say "velocity" in any of your answers.
Say "speed", and if the teacher complains about that, then it's time to
let the teacher have it with both barrels.
1). Speed = (distance covered) / (time to cover the distance)
2). Speed = (distance covered) / (time to cover the distance)
3). Distance = (average speed of travel) x (time traveling at that speed)
4). Time to cover the distance = (distance) / (speed)
5). Car's speed = (distance the car covered) / (time the car took)
Sprinter speed = (distance the sprinter covered) / (time the sprinter took)
Calculate the car's speed.
Calculate the sprinter's speed.
... Look at the two speeds.
Decide which one is faster.
... Subtract the slower one from the faster one.
The difference is the answer to "by how much?" .
6). Distance = (speed) x (time spent moving at that speed)
7). Average speed = (TOTAL distance covered)
divided by
(time to cover the TOTAL distance).
Answer:
The reading of the scale during the acceleration is 446.94 N
Explanation:
Given;
the reading of the scale when the elevator is at rest = your weight, w = 600 N
downward acceleration the elevator, a = 2.5 m/s²
The reading of the scale can be found by applying Newton's second law of motion;
the reading of the scale = net force acting on your body
R = mg + m(-a)
The negative sign indicates downward acceleration
R = m(g - a)
where;
R is the reading of the scale which is your apparent weight
m is the mass of your body
g is acceleration due to gravity, = 9.8 m/s²
m = w/g
m = 600 / 9.8
m = 61.225 kg
The reading of the scale is now calculated as;
R = m(g-a)
R = 61.225(9.8 - 2.5)
R = 446.94 N
Therefore, the reading of the scale during the acceleration is 446.94 N
