Solenoid is another word for

An 8 foot metal guy wire is attached to a broken stop sign to secure its position until repairs can be made. Attached to a stake

Angelina_Jolie [31]

Answer:

Explanation:

The guy wire is making a right angled triangle with the ground and stop sign . It makes an angle of 51 degree with the ground. In this triangle stop sign is the perpendicular and distance from the base on the ground forms the base of the triangle . Wire forms the hypotenuse.

base / hypotenuse = cos51

base = hypotenuse x cos51

= 8 x cos51

= 5.03 ft .

The distance of the stake with which guy wire was attached from the foot of the stop sign is 5.03 ft .

7 0

3 years ago

The journey of an electron through an external circuit involves a long and slow zigzag path that is characterized by losses in e

Kitty [74]

Answer:

d_{b} = 2 d_{a}

Explanation:

The electrical resistance for a cylindrical wire is described by the expression

R = ρ L / A

The area of a circle is

A = π r²

r = d / 2

A = π d²/4

We substitute

R = ρ L 4 /π d²

Let's apply this expression to our case, they indicate that the resistance of wire A is 4 times the resistance of wire B

$R_{a}$ = 4 R_{b}

We substitute

ρ 4/π $d_{a}$ ² = 4 (ρ 4/π d_{b}²)

1 / d_{a}² = 4 / d_{b}²

d_{a} = d_{b} / 2

4 0

3 years ago

What is an example of transparent

nataly862011 [7]

Answer:

When something is transparent, it means that it allows light to pass through or is see-through . For example:

1) glass

2) air

3) some plastics

7 0

3 years ago

Read 2 more answers

Two workers are sliding 290 kg crate across the floor. One worker pushes forward on the crate with a force of 430 N while the ot

m_a_m_a [10]

Answer:0.27

Explanation:

Given

One worker Pushes with force $F_1=430 N$

other Pulls it with a rope of rope $F_2=360 N$

mass of crate $m=290 kg$

both forces are horizontal and crate slides with a constant speed

Both forces are in the same direction so Friction will oppose the forces and will be equal in magnitude of sum of two forces because crate is moving with constant speed i.e. net force is zero on it

$f_r=F_1+F_2$

where $f_r$ is the friction force

$f_r=430+360$

$f_r=790 N$

$f_r=\mu N$

where $\mu$ is the coefficient of static friction

$N=mg$

$790=\mu 29\cdot 9.8$

$\mu =0.27$

6 0

3 years ago

Problem 4: A uniform flat disk of radius R and mass 2M is pivoted at point P A point mass of 1/2 M is attached to the edge of th

brilliants [131]

From the case we know that:

The moment of inertia Icm of the uniform flat disk witout the point mass is Icm = MR².
The moment of inerta with respect to point P on the disk without the point mass is Ip = 3MR².
The total moment of inertia (of the disk with the point mass with respect to point P) is I total = 5MR².

Please refer to the image below.

We know from the case, that:

m = 2M

r = R

m2 = 1/2M

distance between the center of mass to point P = p = R

Distance of the point mass to point P = d = 2R

We know that the moment of inertia for an uniform flat disk is 1/2mr². Then the moment of inertia for the uniform flat disk is:

Icm = 1/2mr²

Icm = 1/2(2M)(R²)

Icm = MR² ... (i)

Next, we will find the moment of inertia of the disk with respect to point P. We know that point P is positioned at the arc of the disk. Hence:

Ip = Icm + mp²

Ip = MR² + (2M)R²

Ip = 3MR² ... (ii)

Then, the total moment of inertia of the disk with the point mass is:

I total = Ip + I mass

I total = 3MR² + (1/2M)(2R)²

I total = 3MR² + 2MR²

I total = 5MR² ... (iii)

Learn more about Uniform Flat Disk here: brainly.com/question/14595971

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8 0

1 year ago