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taurus [48]
3 years ago
15

_____ is a closed loop of conductors through which charges flow.

Physics
2 answers:
Verdich [7]3 years ago
8 0
The answer is c. Hope that helped
katovenus [111]3 years ago
8 0
Hello,

Your correct answer would be:

A circuit <span>is a closed loop of conductors through which charges flow. </span>
You might be interested in
Convert 5.5 kilometers into millimeters.​
dimaraw [331]

Answer:

5500000 millimeters

Explanation:

1 kilometre= 1000 meter

5.5 km=5.5 * 1000

=5500

Now,

1 metre = 1000 millimetres

5500 metre=1000*5500

=5500000 mm

4 0
1 year ago
Isla’s change in velocity is 30 m/s, and Hazel has the same change in velocity. Which best explains why they would have differen
irina [24]

Answer:

B

Explanation:

....

3 0
3 years ago
Read 2 more answers
What fraction of the total energy of a SHO is kinetic when the displacement is one third the amplitude
Bas_tet [7]

Answer:

The fraction of kinetic energy to the total energy is \frac{K}{T}=\frac{8}{9}.

Explanation:

displacement is one third of the amplitude.

Let the amplitude is A.

x= A/3

The kinetic energy of the body executing SHM is

K = 0.5 mw^2(A^2 - x^2)\\\\K = 0.5 m w^2 \left ( A^2 -\frac{A^2}{9} \right )\\\\K = 0.5 mw^2\times \frac{8A^2}{9}......(1)

The total energy is

T =0.5 mw^2A^2 ..... (2)

Divide (1) by (2)

\frac{K}{T}=\frac{8}{9}

5 0
3 years ago
A projectile is launched with speed v0 at an angle of θ0 above the horizontal. Find an expression for the maximum height it reac
Step2247 [10]

Answer:

h= \frac{(v_{o})^{2} sin^{2} \theta o }{2g}

Explanation:

The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).

The equation of uniform rectilinear motion (horizontal ) for the x axis is :

x = xi + vx*t   Equation (1)

Where:  

x: horizontal position in meters (m)

xi: initial horizontal position in meters (m)

t : time (s)

vx: horizontal velocity  in m/s  

The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis  are:

y= y₀+(v₀y)*t - (1/2)*g*t² Equation (2)

vfy= v₀y -gt Equation (3)

vfy²= v₀y²-2gH Equation (4)

Where:  

y: vertical position in meters (m)  

y₀ : initial vertical position in meters (m)  

t : time in seconds (s)

v₀y: initial  vertical velocity  in m/s  

vfy: final  vertical velocity  in m/s  

g: acceleration due to gravity in m/s²

H= hight that reaches the projectile above its starting point (m)

Data

v₀  : total initial speed

θ₀ : angle of v₀ above the horizontal.

g acceleration due to gravity

Calculation of the componentes x-y of the v₀

v₀x = v₀*cos θ₀

v₀y = v₀*sin θ₀

Calculation of the maximum hight that reaches the projectile above its starting point

When the projectile reaches its maximum height (h), vy = 0:

in the Equation (4)

vfy²= v₀y²-2gh

0= v₀y²-2gh

2gh= v₀y²h= \frac{(v_{o})^{2}sin^{2} \theta o }{2g}

2gh=  (v₀*sin θ₀)²

h= \frac{v_{o}^{2} sin^{2} \theta o }{2g}

7 0
3 years ago
How much force is required to accelerate a 4 kg bowling ball from 0 m/s to 2 m/s in 1 second? what amount of energy does the bow
Vladimir [108]
Well, F = ma

and a= change in vel / change in time
        = (2-0 ) / 1  = 2 m^2/s

so, F= 4* 2 = 8 N

and W = F.S  = 8 * 2 = 16 J

hope it helped
5 0
3 years ago
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