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2 years ago

2. An 873 kg dragster accelerates at a rate of 44.6 m/s during a race.

Physics

1 answer:

Ira Lisetskai [31]2 years ago

8 0

Answer:

Explanation:

According to newton's second law:

Force = mass * acceleration

Given

Mass = 873kg

acceleration = 44.66m/s²

Magnitude of the force is expressed as;

F = ma

F = 873 * 44.6

F = 38,673.9N

<em>Hence the magnitude of the net force exerted on the dragster during this time is 38,673.9N</em>

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amid [387]

Answer:

2 amps

Explanation:

Given data

Power = 460W

voltage= 230V

Required

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Recall P= IV

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3 years ago

Consider a block of mass m attached to two springs, one on the left with spring constant k1 and one on the right with spring con

Andrew [12]

Answer:

$T= 2\pi\times sqrt(m/(k1+k2))$

Explanation:

When the block is displaced by x units

F= spring force

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$F =-k_1x - k_2x$

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$-k_1x - k_2x =ma$

$-k_1x - k_2x = mx''$

a= x" ( differentiating x w.r.t time twice)

$x''+(k_1/m + k_2/m) x=0$

this the standard form of equation of oscillation spring mass system

This is the differential equation, x'' means that double differentiation of x , i.e, x'' is acceleration

since, Period $T=2\pi\sqrt{\frac{m}{K_{eq.}} }$

therefore,

$T= 2\pi\times sqrt(m/(k1+k2))$

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2 years ago

A 2.02 nF capacitor with an initial charge of 4.55 µC is discharged through a 1.22 kΩ resistor. (a) Calculate the current in the

Mandarinka [93]

Answer:

a) 0.048A

b) 0.18µC

c) 1.85A

Explanation:

The discharged current of the capacitor as a function of time is given by:

$i=\frac{q_o}{RC}*e^{-\frac{t}{\tau}}\\where:\\\tau=RC\\$

$\tau=1.22*10^3*2.02*10^{-9}\\\tau=2.46*10^{-6}s$

$i=\frac{4.55\µC}{2.46\µs}*e^{-\frac{9\µs}{2.46\µs}}\\\\i=0.048A$

$q=q_o*e^{-\frac{t}{\tau}}$

$q=4.55\µC*e^{-\frac{8\µs}{2.46\µs}}\\q=0.18\µC$

c) the maximum current occurs when t=0

$i=\frac{4.55\µC}{2.46\µs}*e^{-\frac{0\µs}{2.46\µs}}\\\\i=1.85A$

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Answer:

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