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Masteriza [31]
3 years ago
7

A student is conducting a physics experiment and rolls four different-

Physics
1 answer:
joja [24]3 years ago
4 0

Answer:

<em>I think the answer is C</em>

<em />

Explanation:

Its heavy but not too heavy

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Two loudspeakers are placed next to each other and driven by the same source at 500 Hz. A listener is positioned in front of the
stellarik [79]

To solve this problem we will apply the concepts related to wavelength as the rate of change of the speed of the wave over the frequency. Mathematically this is

\lambda = \frac{v}{f}

Here,

v = Wave velocity

f = Frequency,

Replacing with our values we have that,

\lambda = \frac{340}{500}

\lambda = 0.68m

The distance to move one speaker is half this

\lambda/2 = 0.34m

Therefore the minimum distance will be 0.34m

7 0
3 years ago
The 5.5 million vinyl long-playing (LP) records sold in the United States per year pales in comparison with the 1.26 billion dig
miv72 [106K]

Answer:

decline

Explanation:

Based on the scenario being described within the question it can be said that these types of firms are in the decline stage of the product life cycle. This stage refers to when a product has already passed it's peak potential and sales begin to decline until production is ultimately halted and the product dies off. Which is exactly what is happening to the LP's since everyone has moved on to digital downloads.

4 0
3 years ago
When water is boiled under a pressure of 2.00atm, the heat of vaporization is 2.20×106J/kg and the boiling point is 120∘C. At th
Vikki [24]

Answer:

2033219.05 J

Explanation:

V = Volume

P = Pressure = 2 atm

m = Mass of water = 1 kg

L_v = Heat of vaporization = 2.2\times 10^6\ J/kg

Work done in an isobaric system is given by

W=-P\Delta V\\\Rightarrow W=-2\times 101325\times (1\times 10^{-3}-0.824)\\\Rightarrow W=166780.95\ J

Work done is 166780.95 J

Change in internal energy is given by

\Delta U=Q-W

Heat is given by

Q=mL_v\\\Rightarrow Q=1\times 2.2\times 10^6\\\Rightarrow Q=2.2\times 10^6\ J

\Delta U=2.2\times 10^6-166780.95\\\Rightarrow \Delta U=2033219.05\ J

The increase in internal energy of the water is 2033219.05 J

6 0
3 years ago
7. Answer each of the following questions if the student applied a 550N force to a 100kg box on the same surface. What would be
blagie [28]
F=ma
550=100a
a=550/100
a=5.5 m/s²
5 0
3 years ago
A ball is thrown eastward into the air from the origin (in the direction of the positive x-axis). The initial velocity is 50 i +
nikklg [1K]

Answer:

The ball lands 154.3 ft from the origin at an angle of 13.6° from the eastern direction toward the south.

Explanation:

Hi there!

The position vector of the ball is described by the following equation:

r = (x0 + v0x · t + 1/2 · ax · t², y0 + v0y · t + 1/2 · ay · t², z0 + v0z · t + 1/2 · g · t²)

Where:

r =  poisition vector of the ball at time t.

x0 = initial horizontal position.

v0x = initial horizontal velocity (eastward).

t = time.

ax = horizontal acceleration (eastward).

y0 = initial horizontal position.

v0y = initial horizontal velocity (southward).

ay = horizontal acceleration (southward)

z0 = initial vertical position.

v0z = initial vertical velocity.

g = acceleration due to gravity.

We have to find at which time the vertical component of the position vector is zero (the ball is on the ground) and then we can calculate the horizontal distance traveled by the ball at that time, using the equations of the horizontal components of the position vector.

Let´s place the origin of the system of reference at the throwing point so that x0 and y0 and z0 = 0.

y =  z0 + v0z · t + 1/2 · g · t²            (z0 = 0)

0 = 48 ft/s · t - 1/2 · 32 ft/s² · t²

0 = t (48 ft/s - 16 ft / s² · t)                 (t= 0, the origin point)

0 = 48 ft/s - 16 ft / s² · t

- 48 ft/s / -16 f/s² = t

t = 3.0 s

Now, we can calculate how much distance the ball traveled in that time.

First, let´s calculate the distance traveled in the eastward direction:

x = x0 + v0x · t + 1/2 · ax · t²              (x0 = 0, ax = 0 there is no eastward acceleration)

x = 50 ft/s · 3 s

x = 150 ft

And now let´s calculate the distance traveled in southward direction:

y = y0 + v0y · t + 1/2 · ay · t²   (y0 = 0 and v0y = 0, initially, the ball does not have a southward velocity).

y =  1/2 · ay · t²

y = 1/2 · (-8 ft/s²) · (3 s)²

y = -36 ft

Then, the final position vector will be:

r = (150 ft, -36 ft, 0)

The traveled distance is the magnitude of the position vector:

|r| = \sqrt{(150ft)^{2} + (-36ft)^{2}} = 154.3 ft

To calculate the angle, we have to use trigonometry (see attached figure):

cos angle  = adjacent side / hypotenuse

cos α = x/r

cos α = 150 ft / 154.3 ft

α = 13.6°

The ball lands 154.3 ft from the origin at an angle of 13.5° from the eastern direction toward the south.

8 0
3 years ago
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