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3 years ago

7

A coiled spring with coils that are closely spaced then widely then closely then widely then closely, ending with a yellow line

labeled 1 second.
What is the frequency of this wave?

1

2

3

4

2 answers:

liberstina [14]3 years ago

4 0

Answer:

2, sorry I'm late but my answer is right, I just took the quiz

andreev551 [17]3 years ago

4 0

Answer:

2

Explanation:

gor it right on a unit test on 3d3nuity

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Describe the role of the teacher and the learner in a natural science and technology classroom

dem82 [27]

Teacher -> teach the topic
Learner -> learn the topic
Is that what you’re asking?

5 0

3 years ago

Thermopane window is constructed, using two layers of glass 4.0 mm thick, separated by an air space of 5.0 mm.

Bond [772]

To solve this problem it is necessary to apply the concepts related to rate of thermal conduction

$\frac{Q}{t} = \frac{kA\Delta T}{d}$

The letter Q represents the amount of heat transferred in a time t, k is the thermal conductivity constant for the material, A is the cross sectional area of the material transferring heat, $\Delta T$ , T is the difference in temperature between one side of the material and the other, and d is the thickness of the material.

The change made between glass and air would be determined by:

$(\frac{Q}{t})_{glass} = (\frac{Q}{t})_{air}$

$k_{glass}(\frac{A}{L})_{glass} \Delta T_{glass} = k_{air}(A/L)_{air} \Delta T_{air}$

$\Delta T_{air} = (\frac{k_{glass}}{k_{air}})(\frac{L_{air}}{L_{glass}}) \Delta T_{glass}$

$\Delta T_{air} = (\frac{0.84}{0.0234})(\frac{5}{4}) \Delta T_{glass}$

$\Delta T_{air} = 44.9 \Delta T_{glass}$

There are two layers of Glass and one layer of Air so the total temperature would be given as,

$\Delta T = \Delta T_{glass} +\Delta T_{air} +\Delta T_{glass}$

$\Delta T = 2\Delta T_{glass} +\Delta T_{air}$

$20\°C = 46.9\Delta T_{glass}$

$\Delta T_{glass} = 0.426\°C$

Finally the rate of heat flow through this windows is given as,

$\Delta {Q}{t} = k_{glass}\frac{A}{L_{glass}}\Delta T_{glass}$

$\Delta {Q}{t} = 0.84*24*10 -3*0.426$

$\Delta {Q}{t} = 179W$

Therefore the correct answer is D. 180W.

3 0

3 years ago

On the earth, when an astronaut throws a 0.250-kg stone vertically upward, it returns to his hand a time T later. On planet X he

Liula [17]

Answer:

correct is d) a ’= g / 2

Explanation:

For this exercise let's use the kinematics equations

On earth

v = v₀ - a t

a = (v₀- v) / T

On planet X

v = v₀ - a' t’

a ’= (v₀-v) / 2T

Let's substitute the land values in plot X

a’= a / 2

Now let's use Newton's second law

W = ma

m g = m a

a = g

We substitute

a ’= g / 2

So we see that on planet X the acceleration is half the acceleration of Earth's gravity

4 0

3 years ago

Use your calculator to evaluate -3.7 meter/second-13.9 meter/second 21.4 second-72 second

cluponka [151]

Answer :

(-3.7 meter/second) - (13.9 meter/second) = -17.6 meter/second

(21.4 second) - (72 second) = -50.6 second

Explanation :

(1) As we are given the expression :

(-3.7 meter/second) - (13.9 meter/second)

Now we have to evaluate this expression, we get:

⇒ -17.6 meter/second

(2) As we are given the expression :

(21.4 second) - (72 second)

Now we have to evaluate this expression, we get:

⇒ -50.6 second

6 0

3 years ago

A car travels around a level, circular track that is 750m across. What coefficient of friction is required to ensure the car can

Crank

The coefficient of friction must be 0.196

Explanation:

For a car moving on a circular track, the frictional force provides the centripetal force needed to keep the car in circular motion. Therefore, we can write:

$\mu mg = m\frac{v^2}{r}$

where the term on the left is the frictional force acting between the tires of the car and the road, while the term on the right is the centripetal force. The various terms are:

$\mu$ is the coefficient of friction between the tires and the road

m is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

v is the speed of the car

r is the radius of the curve

In this problem,

r = 750 m is the radius

$v=85 mph \cdot \frac{1609}{3600}=38.0 m/s$ is the speed

And solving for $\mu$ , we find the coefficient of friction required to keep the car in circular motion:

$\mu = \frac{v^2}{rg}=\frac{38.0^2}{(750)(9.8)}=0.196$

Learn more about circular motion:

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8 0

3 years ago