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3 years ago

Soil can best be described as the

1 answer:

5 0

<span>The correct option is D. Soil can best be described as the loose covering of weathered rocks and decaying organic matter. There are different types of soil, the type of soil formed depends majorly on the type of parent rock from which the soil is formed and the amount of organic matter present in the soil.</span>

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Trumpeter A holds a B-flat note on the trumpet for a long time. Person C is running towards the trumpeter at a constant velocity

Vikki [24]

You didn't mention it, but the trumpeter herself has to be standing still.

<span>Person C, the one running towards the trumpeter, hears a pitch
that is higher than B-flat. (A)

Person B, the one running away from the trumpeter, hears a pitch
that is lower than B-flat.

Person D, the one standing still the whole time, hears the B-flat.</span>

5 0

3 years ago

Read 2 more answers

How many milligrams in 1 gram. ??? HELP !!!

blagie [28]

1,000 milligrams = 1 gram
2,000 milligrams = 2 grams
3,000 milligrams = 3 grams
4,000 milligrams = 4 grams

3 0

3 years ago

Read 2 more answers

A block-and-tackle pulley hoist is suspended in a warehouse by ropes of lengths 2 m and 3 m. the hoist weighs 430 n. the ropes,

ICE Princess25 [194]

Refer to the diagram shown below.

For horizontal equilibrium,
T₃ cos38 = T₂ cos 50
0.788 T₃ = 0.6428 T₂
T₃ = 0.8157 T₂ (1)

For vertical equilibrium,
T₂ sin 50 + T₃ sin 38 = 430
0.766 T₂ + 0.6157 T₃ = 430
1.2441 T₂ + T₃ = 698.392 (2)

Substitute (1) into (2).
(1.2441 + 0.8157) T₂ = 698.392
T₂ = 339.058 N
T₃ = 0.8157(399.058) = 276.571 N

Answer:
T₂ = 339.06 N
T₃ = 276.57 N

7 0

3 years ago

A 50.0 Watt stereo emits sound waves isotropically at a wavelength of 0.700 meters. This stereo is stationary, but a person in a

photoshop1234 [79]

Answer:

a) f' = 432 Hz

b) I = 8.12*10^-4 W/m^2

Explanation:

a) To calculate the frequency of sound waves that car receives, you take into account the Doppler effect. In this case (observer moves away of the source) you have the following formula:

$f'=f(\frac{v-v_o}{v+v_s})$ (1)

where

f: frequency of the source = ?

v: speed of sound = 343 m/s

vo: speed of the observer = 40.0 m/s

vs: speed of the source = 0 m/s (stationary)

You replace the values of all parameters in the equation (1):

To calculate f' you first calculate the frequency of the sound wave, by using the following formula:

$v=\lambda f\\\\$

v: speed of sound

λ: wavelength = 0.700 m

$f=\frac{v}{\lambda}=\frac{343m/s}{0.700m}=480Hz$

Next, you replace the values of all parameters in the equation (1):

$f'=(490Hz)(\frac{343m/s-40.0m/s}{343m/s})=432Hz$

hence, the frequency perceived by the car is 432 Hz

b) To calculate the power of the sound wave, when the car is 70.0 maway from the speaker, you use the following formula:

$I=\frac{P}{4\pi r^2}$

P: power of the source = 50.0 W

r: distance to the source = 70.0 m

$I=\frac{50.0 W}{4\pi(70.0m)^2}=8.12*10^{-4}\frac{W}{m^2}$

hence, the intensity is 8.12*10^⁻4 W/m^2

3 0

3 years ago

A T-shirt cannon can shoot a 0.085 kg T-shirt at nearly 30 m/s. The T-shirt cannon has a mass of 33 kg. If the initial net momen

IgorLugansk [536]

Answer:

Approximately $0.077\; {\rm m\cdot s^{-1}}$ (assuming that external forces on the cannon are negligible.)

Explanation:

If an object of mass $m$ is moving at a velocity of $v$ , the momentum $p$ of that object would be $p = m\, v$ .

Momentum of the t-shirt:

$\begin{aligned} p(\text{t-shirt}) &= m(\text{t-shirt}) \, v(\text{t-shirt}) \\ &= 0.085\; {\rm kg} \times 30\; {\rm m \cdot s^{-1}} \\ &= 2.55 \; {\rm kg \cdot m \cdot s^{-1}} \end{aligned}$ .

If there is no external force (gravity, friction, etc.) on this cannon, the total momentum of this system should be conserved. In other words, if $p(\text{cannon})$ denote the momentum of this cannon:

$p(\text{t-shirt}) + p(\text{cannon}) = 0$ .

$p(\text{cannon}) = -p(\text{t-shirt}) = -2.55\; {\rm kg \cdot m \cdot s^{-1}}$ .

Rewrite $p = m\, v$ to obtain $v = (p / m)$ . Since the mass of this cannon is $m(\text{cannon}) = 33\; {\rm kg}$ , the velocity of this cannon would be:

$\begin{aligned} v(\text{cannon}) &= \frac{p(\text{cannon})}{m(\text{cannon})} \\ &= \frac{-2.55\; {\rm kg \cdot m \cdot s^{-1}}}{33\; {\rm kg}} \\ &\approx 0.077\; {\rm m \cdot s^{-1}}\end{aligned}$ .

8 0

1 year ago