Ms. Stafford wants to try racing with a push start. A student pushes her at 5 m/s before the rocket kicks in. The rocket still o
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Answer:
<u>Part A</u>
I = 1.4 mW/m²
<u>Part B</u>
β = 91.46 dB
Explanation:
<u>Part A</u>
Sound intensity is the power per unit area of sound waves in a direction perpendicular to that area. Sound intensity is also called acoustic intensity.
For a spherical sound wave, the sound intensity is given by;
Where;
P is the source of power in watts (W)
I is the intensity of the sound in watt per square meter (W/m2)
r is the distance r away
Given:
P = 34 W,
A = 1.0 cm²
r = 44 m
The sound intensity at the position of the microphone is calculated to be;
I = 0.0013975 W/m²
≈ I = 0.0014 W/m² = 1.4 × 10⁻³ W/m²
I = 1.4 mW/m²
The sound intensity at the position of the microphone is 1.4 mW/m².
<u>Part B</u>
Sound intensity level or acoustic intensity level is the level of the intensity of a sound relative to a reference value. It is a a logarithmic quantity. It is denoted by β and expressed in nepers, bels, or decibels.
Sound intensity level is calculated as;
β dB
Where,
β is the Sound intensity level in decibels (dB)
I is the sound intensity;
I₀ is the reference sound intensity;
By pluging-in, I₀ is 1.0 × 10⁻¹² W/m²
∴ β
β
β = 91.46 dB
The sound intensity level at the position of the microphone is 91.46 dB.
Answer:
20.4
Explanation:
To start doing this problem, first draw a free body diagram of the table. My teacher always tells us to do this, and I find that it is very helpful. I have attached a free body diagram to this answer- take a look at it.
First, let us see if Net force = MA. To do that, we need to determine whether the object is at equilibrium horizontally. For an object to be at equilibrium, it either needs to be moving at a constant velocity or not moving at all. Also, if an object is at equilibrium, there will not be any acceleration. But we know that there IS acceleration horizontally, so it cannot be in equilibrium. If it is not in equilibrium, we can use the formula ∑F= ma.
Let us determine the net force. Since the object is moving horizontally, we can ignore the weight and normal force, because they are vertical forces. The only horizontal forces we need to worry about are the applied force and force of friction.
Applied force = 1055 N (490 + 565)
Friction force= Unknown
To find the friction force, use the kinetic friction formula, Friction = μkN
μk is the coefficient, which the problem includes- it is 0.613.
N is the normal force, which we have to find.
*To find the normal force, we have to determine if the object is at equilibrium VERTICALLY. Since it has no acceleration vertically (it's not moving up/down), it is at equilibrium. Now, when an object is at equilibrium in one direction, it means that all the forces in that direction are equal. What are our vertical forces? Weight (mg) and Normal force (N). So it means that the Normal force is equal to the Weight.
Weight = mg = (40)(9.8) = 392 N
Normal force = 392 N
Now, plug it back into the formula (μkN): (0.613)(392) = 240.296 N
Friction = 240.296 N
Now that we know the friction, we can find the horizontal net force. Just subtract the friction force, 240.296 from the applied force, 1055 N
Horizontal Net Force: 814.704 N
Now that we know the net force, plug in the numbers for the formula
∑F= ma.
814.704 = (40.0)(a)
*Divide on both sides)
a = 20.3676 m/s^2
Round it to 3 significant figures, to get:
20.4
