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2 years ago

Plaease help me with thesse four 40 points

Physics

1 answer:

Zielflug [23.3K]2 years ago

3 0

Answer:

For the first one c is the answer

For the second one c is also the answer

For the third one is b

Explanation:

I took that

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A 1.40 mH inductor and a 1.00 µF capacitor are connected in series. The current in the circuit is described by I = 14.0 t, where

4vir4ik [10]

Answer:

Explanation:

Inductance L = 1.4 x 10⁻³ H

Capacitance C = 1 x 10⁻⁶ F

a )

current I = 14 .0 t

dI / dt = 14

voltage across inductor

= L dI / dt

= 1.4 x 10⁻³ x 14

= 19.6 x 10⁻³ V

= 19.6 mV

It does not depend upon time because it is constant at 19.6 mV.

b )

Voltage across capacitor

V = ∫ dq / C

= 1 / C ∫ I dt

= 1 / C ∫ 14 t dt

1 / C x 14 t² / 2

= 7 t² / C

= 7 t² / 1 x 10⁻⁶

c ) Let after time t energy stored in capacitor becomes equal the energy stored in capacitance

energy stored in inductor

= 1/2 L I²

energy stored in capacitor

= 1/2 CV²

After time t

1/2 L I² = 1/2 CV²

L I² = CV²

L x ( 14 t )² = C x ( 7 t² / C )²

L x 196 t² = 49 t⁴ / C

t² = CL x 196 / 49

t = 74.8 μ s

After 74.8 μ s energy stored in capacitor exceeds that of inductor.

7 0

2 years ago

Choose the statement(s) that is/are true about the ratio \frac{C_p}{C_v} C p C v for a gas? (Ii) This ratio is the same for all

Blababa [14]

Answer:

(i) false

(ii) true

(iii) true

(iv) false

Explanation:

(i) The ratio of Cp and Cv is not constant for all the gases. It is because the value of cp and Cv is different for monoatomic, diatomic and polyatomic gases.

So, this is false.

(ii) For monoatomic gas

Cp = 5R/2, Cv = 3R/2

So, thier ratio

Cp / Cv = 5 / 3 = 1.67

This statement is true.

(iii) for diatomic gases

Cp = 7R/2, Cv = 5R/2

Cp / Cv = 7 / 5 = 1.4

This statement is true.

(iv) It is false.

6 0

3 years ago

Clouds, wind, and rain are part of the _____.

Musya8 [376]

A its Stratosphere, Sorry I didn't see your answer, its bilogy I think not physics.. :)

8 0

3 years ago

You have a set of calipers that can measure thicknesses of a few inches with an uncertainty of ± 0.005 inches. You measure the t

Bond [772]

Answer:

a) x = (0.0114 ± 0.0001) in , b) the number of decks is 5

Explanation:

a) The thickness of the deck of cards (d) is measured and the thickness of a card (x) is calculated

x = d / 52

x = 0.590 / 52

x = 0.011346 in

Let's look for uncertainty

Δx = dx /dd Δd

Δx = 1/52 Δd

Δx = 1/52 0.005

Δx = 0.0001 in

The result of the calculation is

x = (0.0114 ± 0.0001) in

b) You want to reduce the error to Δx = 0.00002, the number of cards to be measured is

#_cards = n 52

The formula for thickness is

x = d / n 52

Uncertainty

Δx = 1 / n 52 Δd

n = 1/52 Δd / Δx

n = 1/52 0.005 / 0.00002

n = 4.8

Since the number of decks must be an integer the number of decks is 5

3 0

3 years ago

A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A

Y_Kistochka [10]

Answer:

The tensions in $T_{BC}$ is approximately 4,934.2 lb and the tension in $T_{BD}$ is approximately 6,035.7 lb

Explanation:

The given information are;

The angle formed by the two rope segments are;

The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°

The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°

Therefore, we have;

The tension in rope segment BC = $T_{BC}$

The tension in rope segment BD = $T_{BD}$

The tension in rope segment AB = $T_{AB}$ = Pulling force of tugboat = 1200 lb

By resolution of forces acting along the line A_F gives;

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = $T_{AB}$ = 1200 lb

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb............(1)

Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;

$T_{BC}$ × sin(26.0°) + $T_{BD}$ × sin(21.0°) = 0...........................(2)

Which gives;

$T_{BC}$ × sin(26.0°) = - $T_{BD}$ × sin(21.0°)

$T_{BC}$ = - $T_{BD}$ × sin(21.0°)/(sin(26.0°)) ≈ - $T_{BD}$ × 0.8175

Substituting the value of, $T_{BC}$ , in equation (1), gives;

- $T_{BD}$ × 0.8175 × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb

- $T_{BD}$ × 0.7348 + $T_{BD}$ ×0.9336 = 1200 lb

$T_{BD}$ ×0.1988 = 1200 lb

$T_{BD}$ ≈ 1200 lb/0.1988 = 6,035.6938 lb

$T_{BD}$ ≈ 6,035.6938 lb

$T_{BC}$ ≈ - $T_{BD}$ × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

$T_{BC}$ ≈ -4934.1733 lb

From which we have;

The tensions in $T_{BC}$ ≈ -4934.2 lb and $T_{BD}$ ≈ 6,035.7 lb.

8 0

3 years ago