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3 years ago

What is one way to induce an electric current

1 answer:

8 0

<span>1.an electric is induced when you move a magnet through a coil wire

2.a greater electric current is induced if you add more loops of wire</span>

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Experts, ACE, Genius... can anybody calculate for the Reactions at supports A and B please? Will give brainliest! Given: fb = 30

dybincka [34]

Answer:

Support at Cy = 1.3 x 10³ k-N

Support at Ay = 200 k-N

Explanation:

given:

fb = 300 k-N/m

fc = 100 k-N/m

D = 300 k-N

L ab = 6 m

L bc = 6 m

L cd = 6 m

To get the reaction A or C.

take summation of moment either A or C.

<em><u>Support Cy:</u></em>

∑ M at Ay = 0

(( x1 * F ) + ( D * Lab ) + ( D * L bc + D * L cd )

Cy = -------------------------------------------------------------------

( L ab + L bc )

Cy = 1.3 x 10³ k-N

<em><u>Support Ay:</u></em>

Since ∑ F = 0, A + C - F - D = 0

A = F + D - C

Ay = 200 k-N

4 0

3 years ago

Sometimes a person cannot clearly see objects close up or far away. To correct this type of vision, bifocals are often used. The

Rudik [331]

Answer:

1) P₁ = -2 D, 2) P₂ = 6 D

Explanation:

for this exercise in geometric optics let's use the equation of the constructor

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where f is the focal length, p and q are the distance to the object and the image, respectively

1) to see a distant object it must be at infinity (p = ∞)

$\frac{1}{f_1} = \frac{1}{q}$

q = f₁

2) for an object located at p = 25 cm

$\frac{1}{f_2} = \frac{1}{25} + \frac{1}{q}$

We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm

we substitute in the equations

1) f₁ = -50 cm

$\frac{1}{f_2} = \frac{1}{25} + \frac{1}{50}$

$\frac{1}{f_2}$ = 0.06

f₂ = 16.67 cm

the expression for the power of the lenses is

P = $\frac{1}{f}$

where the focal length is in meters

1) P₁ = 1/0.50

P₁ = -2 D

2) P₂ = 1 /0.16667

P₂ = 6 D

4 0

2 years ago

A ball is thrown against a wall and bounces back toward the thrower with the same speed as it had before hitting the wall. Does

lorasvet [3.4K]

Answer:

A) Although the speed is the same, the direction has changed. Therefore, the velocity has changed.

Explanation:

4 0

3 years ago

Which of the following statements are true? Select all correct responses. Choose all that apply. Choose all that apply. The tota

Ne4ueva [31]

<h3><u>Answer</u>;</h3>

-The total momentum of an isolated system is constant.

-The total momentum of any number of particles is equal to the vector sum of the momenta of the individual particles.

-The vector sum of forces acting on a particle equals the rate of change of momentum of the particle with respect to time.

<h3><u>Explanation</u>;</h3>

Momentum is a vector quantity, and therefore we need to use vector addition when summing together the momenta of the multiple bodies which make up a system.
The vector sum of forces acting on a particle is equivalent to the rate of change of momentum of the particle with respect to time. This is according to the Newton's second Law of motion. In mathematical terms, ֿF = d ֿp/dt, that is F= ma.
According to the Law of conservation of Momentum, or a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

3 0

3 years ago

A swimmer bounces almost straight up from a diving board and falls vertically feet first into a pool.she starts with a speed of

garik1379 [7]

Answer:

a) 1.20227 seconds

b) 0.98674 m

c) 7.3942875 m/s

Explanation:

t = Time taken

u = Initial velocity = 4.4 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v=u+at\\\Rightarrow 0=4.4-9.81\times t\\\Rightarrow \frac{-4.4}{-9.81}=t\\\Rightarrow t=0.44852\ s$

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=4.4\times 0.44852+\frac{1}{2}\times -9.81\times 0.44852^2\\\Rightarrow s=0.98674\ m$

b) Her highest height above the board is 0.98674 m

Total height she would fall is 0.98674+1.8 = 2.78674 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2.78674=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.78674\times 2}{9.81}}\\\Rightarrow t=0.75375\ s$

a) Her feet are in the air for 0.75375+0.44852 = 1.20227 seconds

$v=u+at\\\Rightarrow v=0+9.81\times 0.75375\\\Rightarrow v=7.3942875\ m/s$

c) Her velocity when her feet hit the water is 7.3942875 m/s

3 0

3 years ago