The head of a grass string trimmer has 100 g of cord wound in a light, cylindrical spool with inside diameter 3.00 cm and outsid
e diameter 18.0 cm as shown in the figure below. The cord has a linear density of 10.0 g/m. A single strand of the cord extends 16.0 cm from the outer edge of the spool. (a) When switched on, the trimmer speeds up from 0 to 2 250 rev/min in 0.230 s. What average power is delivered to the head by the trimmer motor while it is accelerating? W (b) When the trimmer is cutting grass, it spins at 1 915 rev/min and the grass exerts an average tangential force of 8.80 N on the outer end of the cord, which is still at a radial distance of 16.0 cm from the outer edge of the spool. What is the power delivered to the head under load?
2 answers:
You might be interested in
From the measured wavelength from diagram, the frequency of the sound is 6660 Hz.
<h3>What is the frequency of a wave?</h3>The frequency of a wave is the number of complete oscillation per second completed by a wave.
Frequency is related to wavelength and speed by the following formula:
- Frequency = velocity/wavelength
Velocity of sound in air = 330 m/s
The measured wavelength = 5.0 cm = 0.05 m
Frequency = 330/0.05 = 6660 Hz
Therefore, based on the measured wavelength from diagram, the frequency of the sound is 6660 Hz.
Learn more about frequency of sound at: https://brainly.in/question/15373132
#SPJ1
Answer:
He could jump 2.6 meters high.
Explanation:
Jumping a height of 1.3m requires a certain initial velocity v_0. It turns out that this scenario can be turned into an equivalent: if a person is dropped from a height of 1.3m in free fall, his velocity right before landing on the ground will be v_0. To answer this equivalent question, we use the kinematic equation:
With this result, we turn back to the original question on Earth: the person needs an initial velocity of 5 m/s to jump 1.3m high, on the Earth.
Now let's go to the other planet. It's smaller, half the radius, and its meadows are distinctly greener. Since its density is the same as one of the Earth, only its radius is half, we can argue that the gravitational acceleration g will be <em>half</em> of that of the Earth (you can verify this is true by writing down the Newton's formula for gravity, use volume of the sphere times density instead of the mass of the Earth, then see what happens to g when halving the radius). So, the question now becomes: from which height should the person be dropped in free fall so that his landing speed is 5 m/s ? Again, the kinematic equation comes in handy:
This results tells you, that on the planet X, which just half the radius of the Earth, a person will jump up to the height of 2.6 meters with same effort as on the Earth. This is exactly twice the height he jumps on Earth. It now all makes sense.
Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y = t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45 )
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² = - 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct