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Neko [114]
3 years ago
11

Prove : cos(3x)=sin(2x)​

Physics
2 answers:
ahrayia [7]3 years ago
8 0

Explanation:

cos3x=sin2x

cos3x=cos(2π−2x)

⇒3x=2nπ+2π−2x

⇒5x=2nπ+2π

⇒x=52nπ+10π

brainliest plzz

ss7ja [257]3 years ago
6 0

Answer:

this is the answer do it help

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The following three hot samples have the same temperature. The same amount of heat is removed from each sample. Which one experi
vlabodo [156]

Answer:

Smallest drop: Water

Largest drop: Dirt

Explanation:

The heat needed to change the temperature of a sample is:

Q=cm\Delta T (1)

with Q the heat (added(+) or removed(-)), c specific heat, m the mass and \Delta T the change in temperature of the sample. So, if we solve (1) for

Sample A:

\Delta T=-\frac{Q}{cm} =\frac{Q}{4186*4.0}

\Delta T=-\frac{Q}{16744}

Sample B:

\Delta T=-\frac{Q}{cm} =\frac{Q}{2700*2.0}

\Delta T=-\frac{Q}{5400}

Sample C:

\Delta T=-\frac{Q}{cm} =\frac{Q}{1050*9.0}

\Delta T=-\frac{Q}{9450}

Note that the numbers 16744, 5400, 9450 are in the denominator of the expression -\frac{Q}{cm} that gives the drop on temperature. so, if Q is the same for the three samples the smallest denominator gives the largest drop and vice versa.

So, the smallest drop is Sample A and the largest is Sample C.

(Important: The minus sign of \Delta T implies the temperature is dropping)

8 0
3 years ago
If V is the original volume, V' is the new volume, T is the original Kelvin temperature, and T' is the new Kelvin temperature, h
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Answer:

\frac{V}{T} = \frac{V'}{T'}

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To write an expression for Charles's law;

Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Charles law is given by the formula;

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4 0
3 years ago
A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters
Artist 52 [7]

Answer:

Part a)

F = 7.76 N

Part b)

\theta = -137.7 degree

Part c)

\theta = -127.7 degree

Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

x = -19 + t - 3t^3

y = 20 + 7t - 9t^2

Part a)

differentiate x and y two times with respect to time to find the acceleration

a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)

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a_y = \frac{d^2}{dt^2}(20 + 7t - 9t^2)

a_y = \frac{d}{dt}(0 +7 - 18t)

a_y = -18

Now the acceleration of the object is given as

\vec a = (-18t)\hat i + (-18)\hat j

at t= 1.1 s we have

\vec a = -19.8 \hat i - 18 \hat j

now the net force of the object is given as

\vec F = m\vec a

\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)

\vec F = -5.74 \hat i - 5.22 \hat j

now magnitude of the force will be

F = \sqrt{5.74^2 + 5.22^2} = 7.76 N

Part b)

Direction of the force is given as

tan\theta = \frac{F_y}{F_x}

tan\theta = \frac{-5.22}{-5.74}

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Part c)

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v_x = \frac{dx}[dt}

v_x = (0 +1 - 9t^2)

v_y = \frac{dy}{dt}

v_y = (0 +7 - 18t)

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\vec v = -9.89\hat i - 12.8 \hat j

now the direction of the velocity is given as

\theta = tan^{-1}(\frac{v_y}{v_x})

\theta = tan^{-1}(\frac{-12.8}{-9.89})

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B. evaporation
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