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4 years ago

A simple pendulum 2 m long swings through a maximum angle of 30 ? with the vertical.

1 answer:

6 0

The complete question was calculate the period T assuming the smallest amplitude.
Using the equation;
T = 2 π√(L/g)
Where T is the period in seconds, L is the length of the rod or wire in meters and g is the acceleration due to gravity.
Hence; T = 2×3.14 × √(2/9.81)
= 6.28 × 0.4515
= 2.836 seconds

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Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T

djverab [1.8K]

Answer:

$v_{2}$ will be less than $v_{1}$ and $P_{2}$ will be greater than $P_{1}$ .

Explanation:

As we know from the conservation of mass, the rate at which any amount of fluid mass ( $m_{1}$ ) is entering in a system is equal to the rate at which the same amount of fluid mass ( $m_{2}$ ) is leaving the system.

Rate of mass flow can be written as,

$m = \rho A v$

where $\rho$ is the density of the fluid, $A$ is the area through which the fluid is flowing and $v$ is the velocity of the fluid.

Now, according to the problem, as the density of the fluid does not change, we can write

$&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}$

where $A_{1}$ and $A_{2}$ are the cross-sectional areas through which the fluid is passing and $v_{1}$ and $v_{2}$ are the velocities of the fluid through the respective cross-sectional areas.

As according to the problem, $A_{2} > A_{1}$ , so from the above formula $v_{2} < v_{1}$ .

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

So, as in this case $v_{2} < v_{1}$ the pressure in the large cross-sectional area $P_{2}$ will be greater than the pressure $P_{1}$ in the small cross sectional area, i.e.,

$P_{2} > P_{1}$ .

6 0

4 years ago

<img src="https://tex.z-dn.net/?f=%5Cbold%7B%20Choose%20%5C%3A%20the%20%5C%3A%20correct%20%5C%3A%20Answer%20%5C%3A%20%3A%29%20%5

Ulleksa [173]

Answer:

B and B

Explanation:

Question 1:

The distance is a unit of length, such as miles, meters, kilometers

The correct answer for this question would be 8m

Question 2:

Speed has a unit of length per unit of time.

The correct answer for this would be meters per second

-Chetan K

3 0

3 years ago

A horizontal spring is attached to a wall at one end and a mass at the other. The mass rests on a frictionless surface. You pull

LenKa [72]

Answer:

54%

Explanation:

So, we have that the "magnitude of its displacement from equilibrium is greater than (0.66)A—''. Thus, the first step to take in answering this question is to write out the equation showing the displacement in simple harmonic motion which is = A cos w×t.

Therefore, we will have two instances t the displacement that is to say at a point 2π/w - a2 and the second point at a = a2.

Let us say that 2π/w = A, then, we have that a = A cos ^-1 (0.66)/2π. Also, we have that a2 = A/2 - A cos^- (0.66) / 2π.

The next thing to do is to calculate or determine the total length of of the required time. Thus, the total length is given as:

2a1 + ( A - 2a2) = 2A{ cos^-1 (0.66)}/ π.

Therefore, the total percentage of the period does the mass lie in these regions = 100 × {2a1 + ( A - 2a2) }/A = 2 { cos^-1 (0.66)}/ π × 100 = 54%.

Thus, the total percentage of the period does the mass lie in these regions = 54%.

6 0

3 years ago

Monochromatic light with wavelength 590 nm passes through a single slit 2. 30 ?m wide and 1. 90 m from a screen. Find the distan

lilavasa [31]

The distance between the first and second-order dark fringe is 0.441 m.

Diffraction of a single slit:

When the light wave passes through a single slit of width which is comparable to the wavelength of the light, then the light wave bends at the edges of the slit. This is called diffraction.

Note: It is assumed that the slit is 3*10^(-3) mm wide. 1 nm = 10^(-9) m and 1mm = 10^(-3) m.

The dark fringes are obtained at the position which satisfies the equation,

d*sinθ = mλ

where d is the slit width, λ is the wavelength of the wave, θ is the angle of diffraction and m denotes the order of the dark fringe.

For first order fringe (m=1), the angle of diffraction θ₁ is,

d*sinθ₁ = λ

sinθ₁ =λ/d

Substitute λ=590 nm, d=3*10^(-3) mm, and solve it.

sinθ₁ =(590 nm)/(3*10^(-3) mm)

sinθ₁ =(590*10^(-9))/ (3*10^(-3)*10^(-3) m)

sinθ₁ =0.196

θ₁ =11.03 degree

Similarly, for second-order fringe (m=2), the angle of diffraction θ₂ is,

d*sinθ₂= 2λ

sinθ₂ =2λ/d

Substitute λ=590 nm, d=3*10^(-3) mm, and solve it.

sinθ₂ =(2*590 nm)/(3*10^(-3) mm)

sinθ₂ =(2*590*10^(-9))/ (3*10^(-3)*10^(-3) m)

sinθ₂ =0.393

θ₂ =23.14 degree

From geometry, the positions x₁ and x₂ of the first and second-order dark fringe from the center of the screen are x₁=Dtanθ₁ and x₂= Dtanθ₂ where D is the distance of the screen from the slit. The distance s between the first-order and second-order dark fringe is then given by,

s=D(tanθ₂-tanθ₁)

Substitute D=1.90 m, θ₁=11.03 degree, and θ₂=23.14 degree in this equation and solve it.

s=1.90*(tan(23.14)-tan(11.03))

s=1.90*(0.427-0.195)

s=0.441 m

Learn more about diffraction here:

brainly.com/question/12290582

#SPJ4

4 0

2 years ago

In a blood pressure reading of 110/70, 110 represents ________ pressure and 70 represents ________ pressure.

kap26 [50]

Systolic pressure; Doastolic pressure.

6 0

3 years ago