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Oxana [17]
3 years ago
13

a battery is tested and it is found that the voltage Vout = 1.64V with the switch open and 1.63V with the switch closed. Find th

e values of Vbat and Rbat

Physics
1 answer:
Burka [1]3 years ago
8 0

Rate brainliest please. Thank you

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Electric Field of Ring of Charge A ring shaped charge has a radius of 0.250 meters and bears a total charge of +5.00 C What is t
Eduardwww [97]

Answer:

6.44 × 10^10 N/C

Explanation:

Electric field due to the ring on its axis is given by

E = K q r / (r^2 + x^2)^3/2

Where r be the radius of ring and x be the distance of point from the centre of ring and q be the charge on ring.

r = 0.25 m, x = 0.5 m, q = 5 C

K = 9 × 10^9 Nm^2/C^2

E = 9 × 10^9 × 5 × 0.25 / (0.0625 + 0.25)^3/2

E = 6.44 × 10^10 N/C

8 0
3 years ago
Acceleration and Force
olga55 [171]

Answer:

I'm pretty sure its 3m/s^2 for the acceleration but I don't know the force part sorry .

Explanation:

15m/s - 0m/s divided by 5 s = 3m/s

I'm no expert or anything so I could be wrong but this is the best I can give you. Sorry

6 0
2 years ago
You illuminate the grating in a spectrometer at normal incidence θi=0° with a beam of light that has a wavelength of 6562.8 Å. T
monitta

Answer:

a) θ₁ = 23.14 ° , b) θ₂ = 51.81 °

Explanation:

An address network is described by the expression

     d sin θ = m λ

Where is the distance between lines, λ is the wavelength and m is the order of the spectrum

The distance between one lines, we can find used a rule of proportions

     d = 1/600

     d = 1.67 10⁻³ mm

    d = 1-67 10⁻³ m

Let's calculate the angle

    sin θ = m λ / d

    θ  = sin⁻¹ (m λ / d)

First order

    θ₁ = sin⁻¹ (1 6.5628 10⁻⁷ / 1.67 10⁻⁶)

    θ₁ = sin⁻¹ (3.93 10⁻¹)

    θ₁ = 23.14 °

Second order

     θ₂ = sin⁻¹ (2 6.5628 10⁻⁷ / 1.67 10⁻⁶)

     θ₂ = sin⁻¹ (0.786)

     θ₂ = 51.81 °

3 0
3 years ago
Determine the stopping distances for a car with an initial speed of 88 km/h and human reaction time of 2.0 s for the following a
seropon [69]

Explanation:

Given that,

Initial speed of the car, u = 88 km/h = 24.44 m/s

Reaction time, t = 2 s

Distance covered during this time, d=24.44\times 2=48.88\ m

(a) Acceleration, a=-4\ m/s^2

We need to find the stopping distance, v = 0. It can be calculated using the third equation of motion as :

s=\dfrac{v^2-u^2}{2a}

s=\dfrac{-(24.44)^2}{2\times -4}

s = 74.66 meters

s = 74.66 + 48.88 = 123.54 meters

(b) Acceleration, a=-8\ m/s^2

s=\dfrac{v^2-u^2}{2a}

s=\dfrac{-(24.44)^2}{2\times -8}

s = 37.33 meters

s = 37.33 + 48.88 = 86.21 meters

Hence, this is the required solution.

4 0
3 years ago
Earth attracts a person with a gravitational force of 7.0 × 102 newtons. What is the magnitude of the force with which the indiv
ehidna [41]
The best option is B) <span>7.0 × 10² newtons.

</span>If Earth attracts a person with a gravitational force of <span><span>7.0 × 10² </span>newtons, 
the person attracts Earth with a gravitational force of 7.0 × 10² newtons.</span>
5 0
3 years ago
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