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2 years ago

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An electric pole shown in the figure below supports a power line that passes through it. A cable tied to the pole at B passes th

rough a hole in the struct at C and is then tied to the ground at B. There is no friction between the struct and cable. What is the force exerted by the cable onto the pole at B, if the tension in the cable is 47lb.

1 answer:

nikdorinn [45]2 years ago

3 0

We have that The force exerted by the cable at point B on the pole is

$T_1=60.189ib$

From the Diagram

We can see that A,B.C is the area showing the circumference of a cable(power line)

$\sum F_y=0\\\\$

$T_1sin51.34=47$

Generally the equation for the Tension is mathematically given as

$T_1=60.189ib$

In conclusion

The force exerted by the cable at point B on the pole is

$T_1=60.189ib$

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3 years ago

In a physics lab, Asha is given a 10.7 kg uniform rectangular plate with edge lengths 67.3 cm by 53.5 cm . Her lab instructor re

Leya [2.2K]

Answer:

$I=2.6363\ kg.m^2$

Explanation:

Given:

dimension of uniform plate, $(0.673\times 0.535)\ m^2$

mass of plate, $m=10.7\ kg$

Now we find the moment of inertia about the center of mass of the rectangular plate is given as:

$I_{cm}=\frac{1}{12} \times m(L^2+B^2)$

where:

$L=$ length of the plate

$B=$ breadth of the plate

$I_{cm}=\frac{1}{12} \times 10.7\times(0.673^2+0.535^2)$

$I_{cm}=0.6591\ kg.m^2$

We know that the center of mass of the rectangular plane is at its geometric center which is parallel to the desired axis XX' .

Now we find the distance between the center of mass and the corner:

$s=\frac{\sqrt{ (0.673^2+0.535^2)}}{2}$

$s=0.4299\ m$

Now using parallel axis theorem:

$I=I_{cm}+m.s^2$

$I=0.6591+10.7\times 0.4299^2$

$I=2.6363\ kg.m^2$

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After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 53.0 cm . The explorer finds that

larisa86 [58]

Answer:

12.4 m/s²

Explanation:

L = length of the simple pendulum = 53 cm = 0.53 m

n = Number of full swing cycles = 99.0

t = Total time taken = 128 s

T = Time period of the pendulum

g = magnitude of gravitational acceleration on the planet

Time period of the pendulum is given as

$T = \frac{t}{n}$

$T = \frac{128}{99}$

T = 1.3 sec

Time period of the pendulum is also given as

$T = 2\pi \sqrt{\frac{L}{g}}$

$1.3 = 2(3.14) \sqrt{\frac{0.53}{g}}$

g = 12.4 m/s²

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A radio station's channel, such as 100.7 fm or 92.3 fm, is actually its frequency in megahertz (mhz), where 1mhz=106 hz and 1hz=

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The frequency of the radio station is
$f=88.7 fm= 88.7 MHz = 88.7 \cdot 10^6 Hz$

For radio waves (which are electromagnetic waves), the relationship between frequency f and wavelength $\lambda$ is
$\lambda= \frac{c}{f}$
where c is the speed of light. Substituting the frequency of the radio station, we find the wavelength:
$\lambda= \frac{3 \cdot 10^8 m/s}{88.7 \cdot 10^6 Hz}=3.38 m$

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A 4.0 kg object will have a weight of approximately 14.8 N on Mars. What is the gravitational field strength on M

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