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3 years ago

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A truck accelerates 2 m/s2 when it is empty. If the truck is filled so that it has twice the mass and the same amount of net for

ce is applied, how much will the truck accelerate?
8 m/s2
1 m/s2
4 m/s2
2 m/s2

2 answers:

mina [271]3 years ago

6 0

This is a conceptual problem, I’ll try to upload a picture:

hram777 [196]3 years ago

5 0

Answer :

a = 1 m/s^2

Explanation:

Given information

Net force is a constant

The mass of the truck is $m_{o}$

Initial acceleration $a_{o}$ = 2 m/ $s^{2}$

The mass of the truck is increased twice as much

$F_{net}= m_{o} a_{o}$ = ma

$F_{net}= m_{o} a_{o}$ = $2m_{o}$ a

a= $\frac{a_{o} } {2}$

a = 2/2

a = 1 m/ $s^{2}$

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Answer:

The inductor contains $N = 523962.32$ loops

Explanation:

From the question we are told that

The capacitance of the capacitor is $C = 286nF = 286 * 10^{-9} \ F$

The resonance frequency is $f = 18.0 kHz = 18*10^{3} Hz$

The diameter is $d = 1.1 mm = \frac{1.1 }{1000} = 0.00011 \ m$

The of the air-core inductor is $l = 12 \ m$

The permeability of free space is $\mu_o = 4 \pi *10^{-7} \ T \cdot m/A$

Generally the inductance of this air-core inductor is mathematically represented as

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This inductance can also be mathematically represented as

$L = \frac{1}{w^2}$

Where $w$ is the angular speed mathematically given as

$w = 2 \pi f$

So

$L = \frac{1}{4 \pi ^2 f^2}$

Now equating the both formulas for inductance

$\frac{\mu_o * N^2 \pi d^2}{4 l} = \frac{1}{4 \pi ^2 f^2}$

making N the subject of the formula

$N = \sqrt{\frac{1}{(2 \pi f)^2} * \frac{4 * l }{\mu_o * \pi d^2 C} }$

$N = \frac{1}{2 \pi f} * \frac{2}{d} * \sqrt{\frac{l}{\pi * \mu_o * C} }$

Substituting value

$N = \frac{1}{ 3.142 * 18*10^{3} * 0.00011 } \sqrt{\frac{12}{ 3.142 * 4 \pi *10^{-7}* 286 *10^{-9}} }$

$N = 523962.32$ loops

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3 years ago

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Answer:

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3 years ago

A running student has half the kinetic energy that his brother has. The student speeds up by 1 m/s, at which point he has the sa

hoa [83]

Answer:

V = (√2) + 1) m/s

Explanation:

Let the mass and speed of the running student be M and V respectively.

We are told that when he speeds up by 1 m/s, he has the same kinetic energy as his brother.

Thus, his speed at which he mow has the same kinetic energy as his brother is (V + 1) m/s

Now, we are told that the mass of the student is twice as large as that of his brother. Thus, his brother's mass is; M/2

Since kinetic energy is given by the formula K.E = ½mv²

Therefore, since we want to find the original speed of both students and that the initial condition says that the running student had half the kinetic energy of the brother, we now initial condition as;

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Since he has sped up by 1 m/s, and has a kinetic energy now equal to that of his brother, we have;

(½M(V + 1)²) = (½(M/2)V²) - - - - (Eq2)

Dividing eq 1 by eq 2 gives;

V²/(V + 1)²= 1/2

Taking square root of both sides gives;

V/(V + 1) = 1/√2

Cross multiply to give;

(√2)V = V + 1

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V = 1/((√2) - 1)

Simplifying this using surfs gives;

V = [1/((√2) - 1)] × ((√2) + 1))/((√2) + 1))

V = ((√2) + 1))/1

V = (√2) + 1) m/s

8 0

3 years ago