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3 years ago

JUST PLZ HELP!!! Why does the lightbulb in the right electrical circuit turn on but not the one on the left?

Physics

2 answers:

makkiz [27]3 years ago

8 0

Because,

In left image pin is not touch to the wire.

In right image pin is touch to the wire.

Hope it helps you.....

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Mark be Brainliest.....

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tekilochka [14]3 years ago

4 0

In the open circuit the current cannot flow from one end of the power source to the other. Because of this there is no current flow, and therefore the light does not turn on.

*This is not my answer*

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3 years ago

Your friend, who is in a field 100 meters away from you, kicks a ball towards you with an initial velocity of 16 m/s. Assuming t

LekaFEV [45]

Answer:

Time, t = 5.355 seconds

Explanation:

Given the following data;

Distance = 100 m

Initial velocity = 16 m/s

Deceleration = 1 m/s²

To find the time, we would use the second equation of motion;

But since the ball is decelerating, it's acceleration would be negative.

S = ut + ½at²

Where;

S represents the displacement or height measured in meters.

u represents the initial velocity measured in meters per seconds.

t represents the time measured in seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

100 = 16t - 0.5t²

200 = 32t - t²

t² + 32t - 200 = 0

Solving the quadratic equation using the quadratic formula;

The quadratic equation formula is;

$x = \frac {-b \; \pm \sqrt {b^{2} - 4ac}}{2a}$

Substituting into the equation, we have;

$x = \frac {-32 \; \pm \sqrt {32^{2} - 4*1*(-200)}}{2*1}$

$x = \frac {-32\pm \sqrt {1024 - (-800)}}{2}$

$x = \frac {-32 \pm \sqrt {1024 + 800}}{2}$

$x = \frac {-32 \pm \sqrt {1824}}{2}$

$x = \frac {-32 \pm 42.71}{2}$

$x_{1} = \frac {-32 + 42.71}{2}$

$x_{1} = \frac {10.71}{2}$

x1 = 5.355

We do not need the negative value of x, so we proceed.

Therefore, time = 5.355 seconds

3 0

3 years ago

If the range of a projectile's trajectory is six times larger than the height of the trajectory, then what was the angle of laun

zvonat [6]

Answer:

H = 1/2 g t^2 where t is time to fall a height H

H = 1/8 g T^2 where T is total time in air (2 t = T)

R = V T cos θ horizontal range

3/4 g T^2 = V T cos θ 6 H = R given in problem

cos θ = 3 g T / (4 V) (I)

Now t = V sin θ / g time for projectile to fall from max height

T = 2 V sin θ / g

T / V = 2 sin θ / g

cos θ = 3 g / 4 (T / V) from (I)

cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ

tan θ = 2/3

θ = 33.7 deg

As a check- let V = 100 m/s

Vx = 100 cos 33.7 = 83,2

Vy = 100 sin 33,7 = 55.5

T = 2 * 55.5 / 9.8 = 11.3 sec

H = 1/2 * 9.8 * (11.3 / 2)^2 = 156

R = 83.2 * 11.3 = 932

R / H = 932 / 156 = 5.97 6 within rounding

3 0

2 years ago