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2 years ago

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A glider of mass m1 on a frictionless horizontal track is connected to an object of mass m2 by a massless string. The glider acc

elerates to the right, the object accelerates downward, and the string rotates the pulley. The pulley has a moment of inertia I. What is the relationship among T1 (the tension in the horizontal part of the string), T2 (the tension in the vertical part of the string), and the weight m2g of the object

1 answer:

inysia [295]2 years ago

5 0

Answer:

m2g -> T2 -> T1

Explanation:

m2g -> T2 -> T1

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A transmission channel is made up of three sections. The first section introduces a loss of 16dB, the second an amplification (o

AlekseyPX

Answer:

$P_{out} = 0.100 W = 100 mW$

Explanation:

The attached image shows the system expressed in the question.

We can define an expression for the system.

The equivalent equation for the system would be

$G_{total} = G_{1} + G_{2} + G_{3}\\G_{total} = -16dB+20dB-10 dB = -6 dB$

so, the input signal could be expressed in dB terms

$P_{in} [dB] = 10 log_{10}(P_{in}) \\P_{in} [dB] = 10 log_{10}(0.4)\\P_{in} [dB] = -3.97 dB$ (1)

so the output signal could be expressed as.

$P_{out} = P_{in} + G_{1} + G_{2} + G_{3}\\P_{out} = -3.97 dB - 6dB = -9.97 dB$

The gain should be expressed in dB terms and power in dBm terms so

$P_{out} = -9.97 + 30 = 20.03 dBm$

using the (1) equation to find it in terms of Watts

$P_{out} = 0.100 W = 100 mW$

3 0

2 years ago

What is the magnitude of the relative angle φ

melomori [17]

Complete question is;

A ski jumper travels down a slope and leaves the ski track moving in the horizontal direction with a speed of 24 m/s. The landing incline below her falls off with a slope of θ = 59◦ . The acceleration of gravity is 9.8 m/s².

What is the magnitude of the relative angle φ with which the ski jumper hits the slope? Answer in units of ◦

Answer:

14.08°

Explanation:

The time covered will be given by the formula;

t = (2V_x•tan θ)/g

t = (2 × 24 × tan 59)/9.8

t = 8.152 s

Now, the slope of the flight path at the point of impact will be given by the formula;

tan α = V_y/V_x

We are given V_x = 24 m/s

V_y will be gotten from the formula;

v = gt

Thus;

V_y = gt

V_y = 9.8 × (8.152) = 78.89 m/s

Thus;

tan α = 78.89/24

tan α = 3.2871

α = tan^(-1) 3.2871

α = 73.08°

Thus ;

Relative angle φ = α - θ = 73.08 - 59 = 14.08°

6 0

1 year ago

The eyes of many older people have lost the ability to accommodate, and so an older person’s near point may be more than 25 cm f

tensa zangetsu [6.8K]

Answer:

Smaller refractive power

Explanation:

The refractive power of an eye is the extent to which it can converge or diverge the light rays.

Near point is the the closest point for an eye such that when an object is placed at that point the image it forms is sharp and clearly visible to the eye.

A the person ages, the ciliary muscles of the eyes weakens and as a result the lens contracts and the formation of the image takes place behind the retina instead of forming at the retina.

Thus the near point also increases and the refractive power becomes smaller.

4 0

1 year ago

100-ft-long horizontal pipeline transporting benzene develops a leak 43 ft from the high-pressure end. The diameter of the leak

Amanda [17]

Answer:

Explanation:

The mass flow rate of benzene from the leak in the pipeline containing benzene is:

$Q_m=AC_o\sqrt{2\rho g_cP_g}$

Here, $Q_m$ is the mass flow rate through the leak of the pipeline. $A$ is the area of the hole, $C_o$ is the discharge rate, $\rho$ is the fluid density, $g_c$ is the gravitational constant and $P_g$ is the constant gauge pressure within the process unit.

The diametre of the leak (d) is 0.1 in. Convert from in to ft.

$d=(0.1 in)(\frac{1ft}{12in})\\=8.33\times 10^{-3}ft$

Calculate the area (A) of the hole. The area of the hole is.

$A=\frac{\pi d^2}{4}$

Substitute 3.14 for $\pi$ and $8.33\times 10^{-3}ft$ for d and calculate A.

$A=\frac{\pi d^2}{4}\\\\\frac{(3.14)(8.33\times 10^{-3})^2}{4}\\\\5.45\times 10^{-5}ft^2$

The specific gravity of benzene is 0.8794. Specific gravity is the ratio of th density of a substance to the density of a reference substance.

Specific gravity of benzene = density of benzenee/denity of reference substance

Rewrite the expression in terms of density of benzene.

Density of benzene = specific gravity of benzene x density of reference substance

Take the reference substance as water. Density of water is $62.4\frac{Ib_m}{ft^3}$ . Calculate density of benzene.

Density of benzene = specific gravity of benzene x density of reference substance

$=(0.8794)(62.4\frac{Ib_m}{ft^3})\\\\54.9\frac{Ib_m}{ft^3}$

Calculate the pressure at the point of leak. The pressure is the average of the pressure of the high and low pressure end. Write the expression to calculate the average pressure.

Upstream x distance from upstream pressure end

$P_g$ =+DOWNSTREAM PRESSURE X DISTANCE FROM THE DOWNSTREAM PRESSURE END/ TOTAL LENGTH OF THE HORIZONTAL PIPELINE

Calculate the distance from the downstream pressure end. The distance from upstream pressure end is 43 ft. Total of the pipe is 100 ft.

Distance from the downstream pressure end = Total length of the pipe - Distance from the upstream pressure end

The distance from upstream pressure end is 43 ft. Total length of the pipe is 100 ft. Substitute the values in the equation.

Distance from the downstream pressure end = Total length of the pipe - Distance from the upstream pressure end

= 100ft - 43ft = 57 ft

Substitute 50 psig for upstream, 43 ft fr distance from the upstream pressure end, 40 psig for downstream pressure, 57 ft for distance from the downstream pressure end, and 100 ft for the total length of the horizontal pipeline and calculate $P_g$ .

Upstream x distance from upstream pressure end

$P_g$ =+DOWNSTREAM PRESSURE X DISTANCE FROM THE DOWNSTREAM PRESSURE END/ TOTAL LENGTH OF THE HORIZONTAL PIPELINE

$=\frac{(50psig\times 43ft)+(40psig \times 57ft)}{100ft}\\\\=44.3psig$

Convert the pressure from psig to $Ib_f/ft^2$

$P_g=(44.3psig)(\frac{1\frac{Ib_f}{ft^2}}{1psig})(144\frac{in^2}{ft^2})\\\\=6,379.2\frac{Ib_f}{ft^2}$

The leak is like a sharp orifice. Take the value of the discharge coefficient as 0.61.

Substitute $5.45\times 10^{-5}ft^2$ for A. 0.61 for $C_o$ , $54.9\frac{Ib_m}{ft^3}$ for $\rho$ , $32.17\frac{ft.Ib_m}{Ib_f.s^2}$ for $g_c$ , and $6,379.2\frac{Ib_f}{ft^2}$ for $P_g$ and calculate $Q_m$

$Q_m=AC_o\sqrt{2\rho g_cP_g}\\\\=(5.45\times 10^{-5}ft^2)(0.61)\sqrt{2(54.9\frac{Ib_m}{ft^3})(32.17\frac{ft.Ib_m}{Ib_f.s^2})(6,379.2\frac{Ib_f}{ft^2})}\\\\(3.3245\times 10^{-5}ft^2)\sqrt{22,533,031.21\frac{Ib^2_m}{ft^4.s^2}}\\\\=0.158\frac{Ib_m}{s}$

The mass flow rate of benzene through the leak in the pipeline is $0.158\frac{Ib_m}{s}$

8 0

2 years ago

The image shows a pendulum that is released from rest at point A. Shari tells her friend that no energy transformation occurs as

Masja [62]

Is D the right answer

6 0

1 year ago

Read 2 more answers