Ask question

Ask question

All categories

allochka39001 [22]

3 years ago

15

A uniform pole AB of weight 5w and length 8a is suspended horizontally by tho strings attached to it at C and D where AC=DB=a. A

body of weight 9w hangs vertically from the pole at E where ED
=2a.
Calculate the tension in each string.

1 answer:

luda_lava [24]3 years ago

4 0

Moment about D=0 or,T1*6a-5w*3a-9w*2a=0 T2+T1=5w+9w

You might be interested in

Two everyday examples of a starting motion

Tasya [4]

Pushing, pulling is the answer

8 0

3 years ago

A weight of 1400 pounds is suspended from two cables as shown in the figure. What is the tension in the left cable? _________ po

juin [17]

Answer:

Following are the solution to this question:

Explanation:

Law:

$\to \theta= 180^{\circ}- 50^{\circ}- 25^{\circ}$

$= 180^{\circ}- 75^{\circ}\\\\= 105^{\circ}$

$\to \frac{T_{L}}{\sin (90+50)}= \frac{T_{R}}{\sin (25+90)}=\frac{1400}{\sin (105)}$

$\to T_L=931.65 \ pounds \\\\ \to T_R=1313.59 \ pounds \\\\$

4 0

3 years ago

Assume that in the Stern-Gerlach experiment for neutral silver atoms, the magnetic field has a magnitude of B = 0.21 T. (a) What

Marat540 [252]

Answer:

Explanation

Assume that in the Stern-Gerlach experiment for neutral silver atoms, the magnetic field has a magnitude of B = 0.21 T

A. To calculate the energy difference in the magnetic moment orientation

∆E = 2μB

For example, any electron's magnetic moment is measured to be 9.284764×10^−24 J/T

Then

μ = 9.284764 × 10^-24 J/T

∆E = 2μB

∆E = 2 × 9.284764 × 10^-24 × 0.21

∆E = 3.8996 × 10^-24 J

Then, to eV

1eV = 1.602 × 10^-19J

∆E = 3.8996 × 10^-24 J × 1eV / 1.602 × 10^-19J

∆E = 2.43 × 10^-5 eV

B. Frequency?

To determine the frequency of radiation hitch would induce the transition between the two states is,

∆E = hf

Where h is plank constant

h = 6.626 × 10-34 Js

Then, f = ∆E / h

f = 3.8996 × 10^-24 / 6.626 × 10^-34

f = 5.885 × 10^9 Hz

f ≈ 5.89 GHz

C. The wavelength of the radiation

From wave equation

v = fλ

In electromagnetic, we deal with speed of light, v = c

And the speed of light in vacuum is

c = 3 × 10^8 m/s

c = fλ

λ = c / f

λ = 3 × 10^8 / 5.885 × 10^9

λ = 0.051 m

λ = 5.1 cm

λ = 51 mm

D. It belongs to the microwave

From table

Micro waves ranges from

•Wavelength 10 to 0.01cm

Then we got λ = 5.1 cm, which is in the range.

•Frequency 3GHz to 3 Thz

Then, we got f ≈ 5.89 GHz, which is in the range

•Energy 10^-5 to 0.01 eV

We got ∆E = 2.43 × 10^-5 eV, which is in the range of the microwave

The value above is in microwave range

5 0

3 years ago

A cylinder is fitted with a piston, beneath which is a spring, as in the drawing. The cylinder is open to the air at the top. Fr

Mazyrski [523]

Answer:

x = 0.0537 m or 5.37 cm

Explanation:

Given:

spring constant'k'= 4900 N/m

radius 'r' =0.029 m

Area 'A' =r²π = 0.029²π => 2.6 x $10^{-3}$ m²

Here, Pressure 'P' is given by,

Pressure = Force / Area

And we know that, for a spring :

F = kx, where k is the spring constant and x is the change in length.

P = kx/A

As P = 101325 Pa

101325 = 4900x / ( 2.6 x $10^{-3}$ )

x = 0.0537 m or 5.37 cm

6 0

3 years ago

Two particles A and B start simultaneously from a Point P with velocities 20 m/s and 30 m/s respectively. A and B move with acce

zysi [14]

Answer:

<u>20 m/s</u>

Explanation:

<u>Given</u>

u(A) = 20 m/s
u(B) = 30 m/s
acceleration equal in magnitude but opposite in direction

<u>Solving</u>

Velocity of A at Q = 30 m/s
From, P to Q, <u>Δv(A) = 30 - 20 = +10 m/s</u>
Therefore, velocity of B at Q will be decreased by 10 as it is equal in magnitude but opposite in direction to A
Δv(B) = v(B at Q) - u(B at P)
-10 m/s = v(B at Q) - 30 m/s
v(B at Q) = 30 - 10 = <u>20 m/s</u>

6 0

2 years ago