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Ira Lisetskai [31]
3 years ago
11

Which of the following is not a concern regarding waste

Physics
1 answer:
juin [17]3 years ago
4 0
I believe it is noise pollution

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A 15.0-μF capacitor is charged by a 130.0-V power supply, then disconnected from the power and connected in series with a 0.280-
SVETLANKA909090 [29]

The resonant frequency of a circuit is the frequency \omega_0 at which the equivalent impedance of a circuit is purely real (the imaginary part is null).

Mathematically this frequency is described as

f = \frac{1}{2\pi}(\sqrt{\frac{1}{LC}})

Where

L = Inductance

C = Capacitance

Our values are given as

C = 15*10^{-6}\mu F

L = 0.280*10^{-3}mH

Replacing we have,

f = \frac{1}{2\pi}(\sqrt{\frac{1}{LC}})

f = \frac{1}{2\pi}(\sqrt{\frac{1}{(15*10^{-6})(0.280*10^{-3})}})

f= 2455.81Hz

From this relationship we can also appreciate that the resonance frequency infers the maximum related transfer in the system and that therefore given an input a maximum output is obtained.

For this particular case, the smaller the capacitance and inductance values, the higher the frequency obtained is likely to be.

7 0
3 years ago
A 5.0-m long, 12-kg uniform ladder rests against a smooth vertical wall with the bottom of the ladder 3.0 m from the wall. the c
Semmy [17]

First establish the summation of the forces acting int the ladder

Forces in the x direction Fx = 0 = force of friction (Ff) – normal force in the wall(n2)

Forces in the y direction Fy =0 = normal force in floor (n1) – (12*9.81) –( 60*9.81)

So n1 = 706.32 N

Since Ff = un1 = 0.28*706.32 = 197,77 N = n2

Torque balance along the bottom of the ladder = 0 = n2(4 m) – (12*9.81*2.5 m) – (60*9.81 *x m)

X = 0.844 m

5/ 3 = h/ 0.844

H = 1.4 m can the 60 kg person climb berfore the ladder will slip

7 0
3 years ago
Read 2 more answers
Noah stands 170 meters away from a steep canyon wall. He shouts and hears the echo of his
bezimeni [28]
340 ms


I got it right and I hope you do as well
6 0
3 years ago
Two blocks, stacked one on top of the other, slide on a frictionless horizontal surface. The surface between the two blocks is r
devlian [24]

Answer:

F_{net} = 31.88 N

Explanation:

When top block is just or about to slide on the lower block then we can say that the frictional force on it will be maximum static friction

So we will have

F_{net} = ma

F_s = ma

\mu_s mg = ma

a = \mu_s g

a = (0.50)(9.81)

a = 4.905 m/s^2

now for the Net force on two blocks to move together

F_{net} = (m_1 + m_2) a

F_{net} = (2.3 + 4.2)(4.905)

F_{net} = 31.88 N

4 0
3 years ago
1500 kg Peugeot car is traveling at 16.67 m s ⁻¹ and accelerates to 30.56 m s ⁻¹ for 2 minutes. Calculate the impulse of the car
Elis [28]

Answer:

2084 kg*m/s

Explanation:

Impulse is change in momentum

Mathematically;

Impulse, J= F*t=mΔv   where

F= ma = 1500 * { 30.56 - 16.67}/2*60

F= 1500 *0.11575

F=  174 N

J=F*t

J= 174*120*0.1

J= 2084 kg*m/s

5 0
2 years ago
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