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strojnjashka [21]
2 years ago
14

Please help! this is timed!

Physics
2 answers:
sweet-ann [11.9K]2 years ago
7 0
Im pretty sure it is the first answer
Alik [6]2 years ago
5 0

Answer:

I'm positive that it is the first one. Sorry if it wrong.

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A certain capacitor, in series with a resistor, is being charged. At the end of 10 ms its charge is half the final value. The ti
vichka [17]

To solve this problem we will apply the expression of charge per unit of time in a capacitor with a given resistance. Mathematically said expression is given as

q(t) = e^{-\frac{t}{(R*C)}}

Here,

q = Charge

t = Time

R = Resistance

C = Capacitance

When the charge reach its half value it has passed 10ms, then the equation is,

\frac{1}{2}*q_{final} = e^{-\frac{0.01}{(R*C)}}

Ln(\frac{1}{2}) = -\frac{0.01}{RC}

- RC = \frac{0.01}{Ln(1/2)}

RC = 0.014s

We know that RC is equal to the time constant, then

T = RC  = 0.014s = 14ms

Therefore the time constant for the process is about 14ms

5 0
3 years ago
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Pavel [41]
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8 0
3 years ago
What measures the amount of displacement in a transverse wave
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Unlike a longitudinal wave, a transverse wave moves about, perpendicular to the direction of propagation. The particles in a transverse wave do not travel along the direction of propagation, but only oscillate up and down on its equilibrium position. With this, the displacement can be determined by measuring (in the case of electronic waves, using an oscilloscope or spectrum analyzer) and setting the desired units to measure the wave in.
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An empty truck traveling at 10 km/h has kinetic energy. How much kinetic energy does it have when loaded so that its mass and it
Katena32 [7]

Answer:

8 time increase in K.E.

Explanation:

Consider Mass of truck = m kg and speed = v m/s then

K.E. = 1/2 ×mv²

If mass and speed both are doubled i.e let m₀ = 2m and v₀ = 2v then

(K.E.)₀ =  1/2 ×2m(2v)²

(K.E.)₀ = 8 (1/2 × mv²) = 8 × K.E.

7 0
3 years ago
A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16 s, then the motor sto
Fed [463]

Answer:

a)   a = 34.375 m / s²,  b)    v_f = 550 m / s

Explanation:

This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.

a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed

             v_f = \frac{x-x_1}{t}

we substitute the values

             v_f = \frac{ 6600 -x_1}{4}  

The initial part of the movement is carried out with acceleration

             v_f = v₀ + a t

             x₁ = x₀ + v₀ t + ½ a t²

the rocket starts from rest v₀ = 0 with an initial height x₀ = 0

             x₁ = ½ a t²

             v_f = a t

we substitute the values

              x₁ = 1/2  a 16²

              x₁ = 128 a

              v_f = 16 a

let's write our system of equations  

               v_f = \frac{6600 - x_1}{4}

               x₁ = 128 a

               v_f = 16 a

we substitute in the first equation  

               16 a = \frac{6600 -128 a}{4}

               16 4 a = 6600 - 128 a

                a (64 + 128) = 6600

                a = 6600/192

                 a = 34.375 m / s²

b) let's find the time to reach this height

                x = ½ to t²

                t² = 2y / a

                t² = 2 5100 / 34.375

                t² = 296.72

                t = 17.2 s

We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part

               v_f = 16 a

               v_f = 16 34.375

               v_f = 550 m / s

8 0
2 years ago
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