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2 years ago

In an inverse relationship, when one variable increases, the other___

Physics

1 answer:

eimsori [14]2 years ago

4 0

Answer:

In an inverse relationship, when one variable increases, the other variable decreases.

Explanation:

Hope this helps! ^^

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Kathy 82 kg performer standing on a diving board at the carnival dive straight down into a small pool of water. Just before stri

mixas84 [53]

Solution :

Given weight of Kathy = 82 kg

Her speed before striking the water, $V_o$ = 5.50 m/s

Her speed after entering the water, $V_f$ = 1.1 m/s

Time = 1.65 s

Using equation of impulse,

$dP = F \times dT$

Here, F = the force ,

dT = time interval over which the force is applied for

= 1.65 s

dP = change in momentum

dP = m x dV

$= m \times [V_f - V_o]$

= 82 x (1.1 - 5.5)

= -360 kg

∴ the net force acting will be

$F=\frac{dP}{dT}$

$F=\frac{-360}{1.65}$

= 218 N

8 0

3 years ago

In an RLC series circuit that includes a source of alternating current operating at fixed frequency and voltage, the resistance

maw [93]

Answer:

Capacitive Reactance is 4 times of resistance

Solution:

As per the question:

R = $X_{L} = j\omega L = 2\pi fL$

where

R = resistance

$X_{L} = Inductive Reactance$

f = fixed frequency

Now,

For a parallel plate capacitor, capacitance, C:

$C = \frac{\epsilon_{o}A}{x}$

where

x = separation between the parallel plates

Thus

C ∝ $\frac{1}{x}$

Now, if the distance reduces to one-third:

Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.

Also,

$Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$

Also,

Z ∝ I

Therefore,

$\frac{Z}{I} = \frac{Z'}{I'}$

$\frac{\sqrt{R^{2} + (R - X_{C})^{2}}}{3I} = \frac{\sqrt{R^{2} + (R - \frac{X_{C}}{3})^{2}}}{I}$

${R^{2} + (R - X_{C})^{2}} = 9({R^{2} + (R - \frac{X_{C}}{3})^{2}})$

${R^{2} + R^{2} + X_{C}^{2} - 2RX_{C} = 9({R^{2} + R^{2} + \frac{X_{C}^{2}}{9} - 2RX_{C})$

Solving the above eqn:

$X_{C} = 4R$

6 0

3 years ago

Imagine that an electron in an excited state in a nitrogen molecule decays to its ground state, emitting a photon with a frequen

mash [69]

Since energy cannot be created nor destroyed, the change in energy of the electron must be equal to the energy of the emitted photon.

The energy of the emitted photon is given by:
$E=hf$
where
h is the Planck constant
f is the photon frequency
Substituting $f=8.88 \cdot 10^{14}Hz$ , we find
$E=hf=(6.6 \cdot 10^{-34} Js)(8.88 \cdot 10^{14} Hz)=5.86 \cdot 10^{-19} J$

This is the energy given to the emitted photon; it means this is also equal to the energy lost by the electron in the transition, so the variation of energy of the electron will have a negative sign (because the electron is losing energy by decaying from an excited state, with higher energy, to the ground state, with lower energy)
$\Delta E= -5.86 \cdot 10^{-19} J$

6 0

3 years ago

Read 2 more answers

Helpppppppp please !!!!

agasfer [191]

Answer:

second one is correct that is right

7 0

3 years ago

A rocket sled accelerates from 10 m/s to 40 m/s in 2 seconds. What is the average acceleration of the sled?

Dafna11 [192]

Answer: