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3 years ago

Which three of these statements about current and voltage in a series circuit are true ?

1 answer:

5 0

1). No. In a series circuit, the current leaving the battery is the same as the current arriving back at the battery at the other end of the circuit, and the same as the current at every point around the circuit.

2). True.
3). True
4). True

5). No. See #3.

6). No. See #2.

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A certain runner averages 4.1 m/s over a 10

Ghella [55]

Answer: 40.650406504065 or 40 minutes and 39 seconds.

Explanation:

1 k = 1000m

race = 10000m

runner time = 10000 / 4.1

runner time = 2439.0243902439024 seconds

runner time = 2439.0243902439024/60 = 40.650406504065 or 40 minutes and 39 seconds.

6 0

3 years ago

A golfer hits a golf ball upwards at an angle. We can ignore air resistance on the ball.

Natalka [10]

Answer:

Check the diagram from the photo

Explanation:

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3 years ago

A 25 kg child stands 2.5 m from the center of a frictionless merry‐go‐round, which has a 200 kg*m^2 moment of inertia and is spi

vodka [1.7K]

Answer:

Explanation:

a ) Time period T = 2 s

Angular velocity ω = 2π / T

= 2π / 2 = 3.14 rad /s

Initial moment of inertia I₁ = 200 + mr²

= 200 + 25 x 2.5²

=356.25

Final moment of inertia

I₂ = 200 + 25 X 1.5 X 1.5

= 256.25

b ) We apply law of conservation of momentum

I₁ X ω₁ = I₂ X ω₂

ω₂ = I₁ X ω₁ / I₂

Putting the values

$w_2=\frac{356.25\times3.14}{256.25}$

ω₂ = 4.365 rad s⁻¹

c ) Increase in rotational kinetic energy

=1/2 I₂ X ω₂² - 1/2 I₁ X ω₁²

.5 X 256.25 X 4.365² - .5 X 356.25 X 3.14²

= 684.95 J

This energy comes from work done against the centripetal pseudo -force.

7 0

3 years ago

A fire hose held near the ground shoots water at a speed of 6.5 m/s. At what angle(s) should the nozzle point in order that the

valina [46]

The speed of water can be split into vertical and horizontal speed components:
$v_x = 6.5 cos \theta \\ v_y = 6.5 sin \theta$

Due to the force of gravity, the y component will be parabolic. The x component will be linear:
$y(t) = -4.9t^2 + (6.5sin \theta) t \\ \\ x(t) = (6.5 cos \theta) t$
To find when the water hits the ground 2.5m away, set y= 0 and x = 2.5
$-4.9t^2 + (6.5sin \theta) t=0 \\ \\ t = \frac{6.5}{4.9} sin \theta \\ \\(6.5 cos \theta)(\frac{6.5}{4.9} sin \theta) = 2.5 \\ \\ sin \theta cos \theta = 0.29 \\ \\ sin 2\theta = 0.58 \\ \\ 2\theta = 35.4, 144.6 \\ \\ \theta = 17.7,72.3$

8 0

3 years ago

Two parallel plate capacitors 1 and 2 are identical except that capacitor 1 has charge +q on one plate and charge −q on the othe

Grace [21]

Answer:

a) the capacitance is the same for both capacitors.

b) The potential difference between the plates for the capacitor with charge +2q, is double of the one for the capacitor with charge +q.

c) The electric field magnitude between the plates for the capacitor with charge +2q, is double of the one for the capacitor with charge +q

d) The energy stored between the plates for the capacitor with charge +2q, is 4 times the value for the one with charge +q

Explanation:

a) The capacitance of a capacitor, by definition, is as follows:

$C = \frac{q}{V}$

Appying Gauss' Law to one of plates, it can be showed, that the capacitance (for a parallel plates capacitor) can be expressed as follows:

C = ε*A / d

As it can be seen, it does not depend on the charge. so we conclude that the capacitance must be the same for both capacitors, due to they are identical except for the value of the charge on the plates.

b) By definition, as we said above, the capacitance is equal to the proportion between the charge of one of the plates, and the potential difference between them.

If this proportion must remain the same, and one of the capacitors has the double of the charge than the other, the potential difference must be the double also.

c) Applying Gauss' law, to the surface of one of the plates, and assuming a constant surface charge density σ, it can be showed that the electric field can be calculated as follows:

E*A = Q/ε₀ as σ=Q/A

⇒ E = σ/ε₀

As σ is directly proportional to the charge (being the area A the same), we conclude that the electric field for the capacitor with charge +2q must be the double than the one for the capacitior with charge +q.

c) The electric potential energy, stored between plates of a capacitor, can be written as follows:

$Ue = \frac{1}{2} *\frac{q^{2}}{C}$

If the capacitance remains the same, we can conclude that the electric potential energy for the capacitor with charge +2q, as the charge is raised to the 2nd power, must be 4 times the one for the capacitor with charge +q.

4 0

3 years ago