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3 years ago

1 Verify the identity. Show your work. cot θ ∙ sec θ = csc θ

1 answer:

6 0

To verify the identity, we can make use of the basic trigonometric identities:
cot θ = cos θ / sin θ
sec θ = 1 / cos θ
csc θ = 1 / sin θ

Using these identities:
cot θ ∙ sec θ = (cos θ / sin θ ) ( 1 / cos θ)

We can cancel out cos θ, leaving us with
cot θ ∙ sec θ = 1 / sin θ
cot θ ∙ sec θ = = csc θ

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A horizontal force of 30N is applied to a mass of 10 kg causing it to accelerate. If the coefficient of friction is 0.20 what is

Blizzard [7]

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The force of friction is equal to the product of the vertical force applied by the surface to the object in the coefficient of friction.

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In this question ,

surface vertical force = Weight of the object

Thus ;

svf = ( mass ) × ( gravity acceleration )

_________________________________

If gravity acceleration is 10 :

svf = 10 × 10 = 100 N

So ;

frictional force = 100 × 0.20

frictional force = 20 N

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If gravity acceleration is 9.8 :

svf = 10 × 9.8 = 98 N

So ;

frictional force = 98 × 0.20

frictional force = 19.6 N

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5 0

2 years ago

When energy is transferred to air, what happens to the particles of air? (1 point) They move slower. They move faster. They cool

zubka84 [21]

This question is a critical question. as we all know, when energy is added to any state of water, the particles move faster. and when energy is taken away from any state of water, the particles reduce speed. same with the particles of air. when energy is added; they move faster. when energy is removed; they move slower. so the answer is they move faster

5 0

3 years ago

Read 2 more answers

A small water pump is used in an irrigation system. The pump takes water in from a river at 10oC, 100 kPa at a rate of 5 kg/s. T

sergij07 [2.7K]

Answer:

0.98kW

Explanation:

The conservation of energy is given by the following equation,

$\Delta U = Q-W$

$\dot{m}(h_1+\frac{1}{2}V_1^2+gz_1)-\dot{W} = \dot{m}(h_2+\frac{1}{2}V_2^2+gz_)$

Where

$\dot{m} =$ Mass flow

$h_1 =$ Specific Enthalpy (IN)

$h_2 =$ Specific Enthalpy (OUT)

$g =$ Gravity

$z_{1,2} =$ Heigth state (In, OUT)

$V_{1,2} =$ Velocity (In, Out)

Our values are given by,

$T_i = 10\°C$

$P_1 = 100kPa$

$\dot{m} = 5kg/s$

$z_2 = 20m$

For this problem we know that as pressure, temperature as velocity remains constant, then

$h_1 = h_2$

$V_1 = V_2$

Then we have that our equation now is,

$\dot{m}(gz_1) = \dot{m}(gz_2)+\dot{W}$

$\dot{W} = \frac{(5)(9.81)(0-20)}{1000}$

$\dot{W} = -0.98kW$

8 0

3 years ago

A mature thunderstorm will contain both updrafts and downdrafts. True or False

jasenka [17]

A mature thunderstorm will contain both updraft and downdrafts. The given statement is true.

When the cumulus cloud becomes very large, the water in it becomes large and heavy. Raindrops start to fall through the cloud when the rising air can no longer hold them up. Meanwhile, cool dry air starts to enter the cloud. Because cool air is heavier than warm air, it starts to descend in the cloud (known as a downdraft). The downdraft pulls the heavy water downward, making rain.

This cloud has become a cumulonimbus cloud because it has an updraft, a downdraft, and rain. Thunder and lightning start to occur, as well as heavy rain. The cumulonimbus is now a thunderstorm cell.

7 0

3 years ago

Read 2 more answers

A baton twirler is twirling her aluminum baton in a horizontal circle at a rate of 2.33 revolutions per second. A baton held hor

Nata [24]

Answer:

Explanation:

Given that;

horizontal circle at a rate of 2.33 revolutions per second

the magnetic field of the Earth is 0.500 gauss

the baton is 60.1 cm in length.

the magnetic field is oriented at 14.42°

we wil get the area due to rotation of radius of baton is

$\Delta A = \frac{1}{2} \Delta \theta R^2$

The formula for the induced emf is

$E = \frac{\Delta \phi}{\Delta t}$

$\phi = \texttt {magnetic flux}$

$E=\frac{\Delta (BA) }{\Delta t}$

$=B\frac{\Delta A}{\Delta t}$

B is the magnetic field strength

substitute

$\texttt {substitute}\ \frac{1}{2} \Delta \theta R^2 \ \ for \Delta A$

$E=B\frac{(\Delta \theta R^3/2)}{\Delta t} \\\\=\frac{1}{2} BR^2\omega$

The magnetic field of the earth is oriented at 14.42

$\omega =2.33\\\\L=60.1c,\\\\\theta=14.42\\\\B=0.5$

we plug in the values in the equation above

so, the induce EMF will be

$E=\frac{1}{2} \times (B\sin \theta)R^2\omega\\\\E=\frac{1}{2} \times (B\sin \theta)(\frac{L}{2} )\omega$

$=\frac{1}{2} \times0.5gauss\times\frac{0.0001T}{1gauss} \times\sin 14.42\times(\frac{60.1\times10^-^2m}{2} )^2(2.33rev/s)(\frac{2\pi rad}{1rev} )\\\\=2.5\times10^-^5\times0.2490\times0.0903\times14.63982\\\\=2.5\times10^-^5\times0.32917\\\\=8.229\times10^-^6V$

6 0

3 years ago