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2 years ago

10

What is the relationship in which the ratio of the manipulated variable and the responding variable is constant? A. inverse prop

ortion B. direct proportion C. slope D. interdependent

1 answer:

Dafna1 [17]2 years ago

6 0

That's a direct proportion.

You might be interested in

If the 80 ohm resistor fails, will the 50 ohm and 100 ohm resistor continue to operate? Why or why not?

sleet_krkn [62]

Answer:

No, the 50 ohm and 100 ohm resistor will not continue to operate.

Explanation:

A closed circuit is the circuit in which there is no break between the negative and the positive end of the battery.

When in this, combinational circuit the 80 ohm resistor fail then there will not any continue supply of current in the circuit due to the breakage because the electron will flow from negative end of the battery to positive end if their is no breaking in the circuit.

Therefore the 50 ohm and 100 ohm circuit will not continue to operate because of the breaking of the circuit and current will not flow.

7 0

3 years ago

011 10.0 points

Sedbober [7]

<h2>The temperature of the air is 66.8° C</h2>

Explanation:

From the Newton's velocity of sound relationship , the velocity of sound is directly proportional to the square root of temperature .

In this case The velocity of sound = frequency x wavelength

= 798 x 0.48 = 383 m/sec

Suppose the temperature at this time = T K

Thus 383 ∝ $\sqrt{T}$ I

The velocity of sound is 329 m/s at 273 K ( given )

Thus 329 ∝ $\sqrt{273}$ II

Dividing I by II , we have

$\frac{383}{329}$ = $\sqrt{\frac{T}{273} }$

or $\frac{T}{273}$ = 1.25

and T = 339.8 K = 66.8° C

4 0

3 years ago

In a thunderstorm, electric charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considere

muminat

Answer:

$2.1\cdot 10^{21}$ electrons

Explanation:

The magnitude of the electric field outside an electrically charged sphere is given by the equation

$E=\frac{kQ}{r^2}$

where

k is the Coulomb's constant

Q is the charge stored on the sphere

r is the distance (from the centre of the sphere) at which the field is calculated

In this problem, the cloud is assumed to be a charged sphere, so we have:

$E_b=3.00\cdot 10^6 N/C$ is the maximum electric field strength tolerated by the air before breakdown occurs

$r=1.00 km = 1000 m$ is the radius of the sphere

Re-arranging the equation for Q, we find the maximum charge that can be stored on the cloud:

$Q=\frac{Er^2}{k}=\frac{(3.00\cdot 10^6)(1000)^2}{9\cdot 10^9}=333.3 C$

Assuming that the cloud is negatively charged, then

$Q=-333.3 C$

And since the charge of one electron is

$e=-1.6\cdot 10^{-19}C$

The number of excess electrons on the cloud is

$N=\frac{Q}{e}=\frac{-333.3}{-1.6\cdot 10^{-19}}=2.1\cdot 10^{21}$

5 0

2 years ago

In a charging process, 1 × 10^13 electrons are removed from one small metal sphere and placed on a second identical sphere. Init

liraira [26]

Answer:

r = 0.303m

= 30.3cm

Explanation:

Given that,

The number of electrons transferred from one sphere to the other,

n = 1 ×10 ¹³e le c t r o n s

The electrostatic potential energy between the spheres,

U = − 0.061 J

The charge on an electron,

q = − 1.6 × 10 ⁻¹⁹C

The coulomb constant,

K = 8.98755 × 10 ⁹ N ⋅ m ² / C 2²

Due to the transfer of electrons, both spheres become equally and oppositely.

The charge gained by the sphere due to the excess of the electron is:

q ₁ = n q

= 1 ×10 ¹³ * − 1.6 × 10 ⁻¹⁹

= -1.6 × 10⁻⁶C

The charge left on the first sphere is =

q ₂ = -q₁ = 1.6 × 10⁻⁶C

The electric potential energy between two point charges is given by the following equation:

U = K q ₁q ₂/r

q ₁ and q ₂ are the two charges.

r is the distance between the charge and the point.

K = 8.98755 × 10 ⁹ N ⋅ m ² / C ²

we have:

-0.061 = (8.98755 × 10 ⁹ * (-1.6 × 10⁻⁶)²) / r

r = (18.41 × 10 ⁻³) / 0.061

r = 0.303m

= 30.3cm

4 0

3 years ago

A stone is thrown vertically upward with a speed of 12m/s from the edge of a cliff 70 m high (a) How much later it reaches the b

jonny [76]

Answer

given,

vertical speed of stone,v = 12 m/s

height of the cliff = 70 m

a) time taken by the stone to reach at the bottom of the cliff

We know that,

S = u t + 1/2 a t²

- 70 = 12 t - 0.5 x 9.8 t²

4.9 t² - 12 t - 70 = 0

solving the equation

t = 5.2 s (neglecting the negative value)

b) again using equation of motion

v = u + a t

v = 12 - 9.8 x 5.2

v = -38.96 m/s

ignoring the negative sign

magnitude of velocity is equal to 38.96 m/s

c) total distance travel by the stone

vertical distance covered by the stone

v² = u² + 2 g h

0 = 12² - 2 x 9.8 x h

h = 7.34 m

to reach the stone to the same level distance travel be doubled.

Total distance travel by the stone

H = h + h + 70

H = 7.34 x 2 + 70

H = 84.7 m.

8 0

2 years ago