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spayn [35]
3 years ago
12

An experiment is designed to test what color of light will activate a photoelectric cell the best. The photocell is set in a cir

cuit that "clicks" in response to current. The faster the current, the more clicks per minute. In this experiment, the number of clicks in one minute is recorded for each color of light shining on the photocell. To change the color of light, a different color of cellophane is placed over the same flashlight and the flashlight is then located a specific distance from the photocell. In the above experiment, which factor is the independent variable? the number of clicks the color of the light the original source of the light the photocell
Physics
2 answers:
chubhunter [2.5K]3 years ago
8 0
The photocell<span>-- The click rate depends upon the filter selected.</span>
Aloiza [94]3 years ago
4 0

Answer:

No, the answer is the color of the light.

Explanation:

The number of flicks is determined by the color of the light, and the photocell is the control.

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If two metal balls each have a charge of -10^-6 C and the repulsive force between them is 1 N, how far apart are they? (Coulomb'
Vitek1552 [10]

Answer:

The distance between the charges is 9.5 cm

(c) is correct option

Explanation:

Given that,

Charge q= 10^{-6}\ C

Force F = 1 N

We need to calculate the distance between the charges

Using Coulomb's formula

F = \dfrac{kq_{1}q_{2}}{r^2}

Where, q = charge

r = distance

F = force

Put the value into the formula

1=\dfrac{9.0\times10^{9}\times(-10^{-6})^2}{r^2}

r=\sqrt{9\times10^{9}\times(-10^{-6})^2}

r=0.095\ m

r= 9.5\ cm

Hence, The distance between the charges is 9.5 cm

4 0
3 years ago
Suppose the charges attracting each other in the above situation have an equal magnitude. Find the charge.
LiRa [457]

The force between the two point charge when they are separated by 18 cm is 3 N

<h3>How do I determine the force when they are 18 cm apart?</h3>

Coulomb's law states as follow:

F = Kq₁q₂ / r²

Cross multiply

Fr² = Kq₁q₂

Kq₁q₂ => constant

F₁r₁² = F₂r₂²

Where

  • F₁ and F₂ are the initial and new force
  • r₁ and r₂ are the initial and new distance apart

With the above formula, we can obtain the force between the two point charge when they are 18 cm apart. Details below:

  • Initial distance apart (r₁) = 6 cm
  • Initial force of attraction (F₁) = 27 N
  • New distance apart (r₂) = 18 cm
  • New force of attraction (F₂) =?

F₁r₁² = F₂r₂²

27 × 6² = F₂ × 18²

972 = F₂ × 324

Divide both side by 324

F₂ = 927 / 324

F₂ = 3 N

Thus, the force when they are 18 cm apart is 3 N

Learn more about force:

brainly.com/question/28569085

#SPJ1

7 0
2 years ago
A runner has a temperature of 40°c and is giving off heat at the rate of 50cal/s (a) What is the rate of heat loss in watts? (b)
Andrew [12]

Answer:

(a)  209 Watt

(b) 4482.8 seconds

Explanation:

(a) P = 50×4.18

Where P = rate of heat loss in watt

    P = 209 Watt

Applying,

Q = cm(t₁-t₂)................ Equation 1

Where Q = amount of heat given off, c = specific heat capacity capacity of human, m = mass of the person, t₁ and t₂ = initial and final temperature.

From the question,

Given: m = 90 kg, t₁ = 40°C, t₂ = 37°C

Constant: c = 3470 J/kg.K

Substtut these values into equation 1

Q = 90×3470(40-37)

Q = 936900 J

But,

P = Q/t.............. Equation 2

Where t = time

t = Q/P............ Equation 3

Given: P = 209 Watt, Q = 936900

Substitute into equation 3

t = 936900/209

t = 4482.8 seconds

5 0
3 years ago
DISPLACEMENT WITH CONSTANT ACCELEration. Given information: displacement=64m, acceleration=9.81m\s and time interval=3s. calcula
balandron [24]
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3 0
3 years ago
A mass of 1 slug is suspended from a spring whose spring constant is 9 lb/ft. The mass is initially released from a point 1 foot
FromTheMoon [43]

Answer:

t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number

Explanation:

To solve the problem/ we first write the differential equation governing the motion. So,

m\frac{d^{2} x}{dt^{2} } = -kx \\ m\frac{d^{2} x}{dt^{2} } + kx = 0\\\frac{d^{2} x}{dt^{2} } + \frac{k}{m} x = 0

with m = 1 slug and k = 9 lb/ft, the equation becomes

\frac{d^{2} x}{dt^{2} } + \frac{9}{1} x = 0\\\frac{d^{2} x}{dt^{2} } + 9 x = 0

The characteristic equation is

D² + 9 = 0

D = ±√-9 = ±3i

The general solution of the above equation is thus

x(t) = c₁cos3t + c₂sin3t

Now, our initial conditions are

x(0) = -1 ft and x'(0) = -√3 ft/s

differentiating x(t), we have

x'(t) = -3c₁sin3t + 3c₂cos3t

So,

x(0) = c₁cos(3 × 0) + c₂sin(3 × 0)

x(0) = c₁cos(0) + c₂sin(0)

x(0) = c₁ × (1) + c₂ × 0

x(0) = c₁ + 0

x(0) = c₁ = -1

Also,

x'(0) = -3c₁sin(3 × 0) + 3c₂cos(3 × 0)

x'(0) = -3c₁sin(0) + 3c₂cos(0)

x'(0) = -3c₁ × 0 + 3c₂ × 1

x'(0) = 0 + 3c₂

x'(0) = 3c₂ = -√3

c₂ = -√3/3

So,

x(t) = -cos3t - (√3/3)sin3t

Now, we convert x(t) into the form x(t) = Asin(ωt + Φ)

where A = √c₁² + c₂² = √[(-1)² + (-√3/3)²] = √(1 + 1/3) = √4/3 = 2/√3 = 2√3/3 and Ф = tan⁻¹(c₁/c₂) = tan⁻¹(-1/-√3/3) = tan⁻¹(3/√3) = tan⁻¹(√3) = π/3.

Since tanФ > 0, Ф is in the third quadrant. So, Ф = π/3 + π = 4π/3

x(t) = (2√3/3)sin(3t + 4π/3)

So, the velocity  v(t) = x'(t) = (2√3)cos(3t + 4π/3)

We now find the times when v(t) = 3 ft/s

So (2√3)cos(3t + 4π/3) = 3

cos(3t + 4π/3) = 3/2√3

cos(3t + 4π/3) = √3/2

(3t + 4π/3) = cos⁻¹(√3/2)

3t + 4π/3 = ±π/6 + 2kπ    where k is an integer

3t  = ±π/6 + 2kπ - 4π/3

t  = ±π/18 + 2kπ/3 - 4π/9

t = π/18 + 2kπ/3 - 4π/9 or -π/18 + 2kπ/3 - 4π/9

t = π/18 - 4π/9 + 2kπ/3  or -π/18 - 4π/9 + 2kπ/3

t = -7π/18 + 2kπ/3 or -π/2 + 2kπ/3

Since t is not less than 0, the values of k ≤ 0 are not included

So when k = 1,

t = 5π/18 and π/6. So,

t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number

6 0
3 years ago
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