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3 years ago

A hydraulic system is used to lift a 20,000 N vehicle in an auto garage. If the vehicle sits on

Physics

1 answer:

ollegr [7]3 years ago

7 0

Answer:

Explanation:

Area covered by the vehicle on the piston A1 = 0.5 m2 Weight of the vehicle, F1 = 2000 kg × 9.8 m s-2 Area on which force F2 is applied, A = 0.03 m2 P1 = P2 ; F 1 A 1 F1A1 = F 2 A 2 F2A2 and F2 = F 1 A 1 F1A1 A2 ; F2 = (2000 × 9.8) 0.03 0.5 0.030.5 = 1176 NRead more on Sarthaks.com - https://www.sarthaks.com/957542/hydraulic-system-used-lift-2000-vehicle-in-an-auto-garage-if-the-vehicle-sits-on-piston-area

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A resistor with resistance R and an air-gap capacitor of capacitance C are connected in series to a battery (whose strength is "

blsea [12.9K]

Answer:

a) Q = C*emf

b) Reduction in electric field strength and electric potential

c) Initial current through the resistor = emf/R

d) The final charge = K*C*emf

Explanation:

a) The resistors and capacitors are connected in series with the battery

Using Kirchoff's voltage law, sum of all voltages in the circuit is zero

Let $V_{R}$ = Voltage dropped across the Resistor

$V_{c}$ = Voltage dropped across the capacitor

Applying KVL;

$emf - V_{R} - V_{c} = 0\\$ .........................(1)

Since the connection is in series, the same current flow through the circuit

$V_{R} = IR\\Q = CV_{c} \\V_{c} = Q/C$

Putting $V_{c}$ and $V_{R}$ into equation (1)

$emf - IR - Q/C = 0$

At the final charge, the capacitor in fully charged, and current drops to zero due to equilibrium

$I = 0A\\emf = Q/C\\Q = C* emf$

b) Current starts running through the plate because as the sheet of plastic is inserted between the plates both the electric field intensity and the electric potential reduces. The charge also reduces, then current flows

c) The current through the resistor is the current through the entire circuit ( series connection)

$I = I_{o} \exp(\frac{-t}{RC} )\\At time the initial time, t\\t = 0\\ I_{o} = \frac{emf}{R} \\$

Putting the values of t and I₀ into the formula for I written above

$I = \frac{emf}{R} \exp(0)\\I = \frac{emf}{R}$

d) NB: The initial charge on the capacitor = C * emf

The final charge will be:

$Q = K* Q_{initial} \\Q_{initial} = C *emf\\Q_{final} = KCemf$

4 0

3 years ago

1.Which car has zero acceleration?<br><br>2. which car is decelerating?

Ivanshal [37]

Answer:

1. B has no acceleration because the straight line shows that it's a constant speed not speeding up or down.

2. A because you can see the decline in speed as time goes on

5 0

3 years ago

According to Coulomb’s Law, the force between two charged objects is related to _____.

Natasha_Volkova [10]

Answer:

A.) the inverse of the square of the distance separating them

Explanation:

Coulombs law states that "the force of attraction between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them."

Mathematically, F = kq1q2/r²

Where q1 and q2 are the charges

r is the distance between the charges.

According to the law, the force between two charged objects is related to the inverse of the square of the distance separating them.

5 0

3 years ago

An ideal spring with spring constant k is hung from the ceiling. The initial length of the spring, with nothing attached to the

hram777 [196]

The mass m of the object = 5.25 kg

<h3>Further explanation</h3>

Given

k = spring constant = 3.5 N/cm

Δx= 30 cm - 15 cm = 15 cm

Required

the mass m

Solution

F=m.g

Hooke's Law

F = k.Δx

$\tt m.g=k.\Delta x\\\\m.10=3.5\times 15\\\\m=5.25~kg$

7 0

3 years ago

What is the speaker’s power output if the sound intensity level is 102 dBdB at a distance of 25 mm ? Express your answer to two

lesya [120]

Answer:

Power = 124.50 W

Explanation:

Given that:

The Sound intensity of a speaker output is 102 dB

and the distance r = 25 m

For the intensity of sound,

$\beta (dB)= 10 \ log_{10 } (\dfrac{I}{I_o})$

where;

the threshold of hearing $I_o = 10^{-12} (W/m^2)$

$\dfrac{102 }{10}= log_{10}( \dfrac{I}{10^{-12}})$

$10^{10.2} = \dfrac{I}{10^{-12}}$

$I = 10^{10.2} \times 10^{-12}$

I = 0.01585 W/m²

If we recall, we know remember that ;

Power = Intensity × A rea

Power = 0.01585 W/m² × 4 × 3.142 × (25 m)²

Power = 124.50 W

7 0

3 years ago