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3 years ago

What is the acceleration of a car that moves at a steady velocity of 100 km/h for 100 seconds? Explain your answer.

Physics

1 answer:

RUDIKE [14]3 years ago

3 0

Answer:

a = 0 m/s²

Explanation:

given,

car moving at steady velocity = 100 Km/h

1 km/h = 0.278 m/s

100 Km/h = 27.8 m/s

time of acceleration = 100 s

acceleration is equal to change in velocity per unit time.

$a=\dfrac{dv}{dt}$

change in velocity of the car is 27.8 - 27.8 = 0

$a=\dfrac{0}{100}$

a = 0 m/s²

If the car is moving with steady velocity then acceleration of the car is zero.

Hence, the acceleration of the car is equal to a = 0 m/s²

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On what factors does critical velocity depend on

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Explanation:

The critical velocity is that velocity of liquid flow, up to which its flow is streamlined (laminar)& above which its flow becomes turbulent. It's denoted by Vc & it depends upon: Coefficient of viscosity of liquid (η) Density of liquid. Radius of the tube.

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3 years ago

An 67-kg jogger is heading due east at a speed of 2.3 m/s. A 70-kg jogger is heading 61 ° north of east at a speed of 1.3 m/s. F

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Answer

given,

mass of jogger = 67 kg

speed in east direction = 2.3 m/s

mass of jogger 2 = 70 Kg

speed = 1.3 m/s in 61 ° north of east.

jogger one

$P_1 = m_1 v_1 \hat{i}$

$P_1 = 67 \times 2.3\hat{i}$

$P_1 = 154.1 \hat{i}$

$P_2 = m_2 v_2 \hat{i} +m_2 v_2 \hat{j}$

$P_2 = 70\times v cos \theta \hat{i} +70\times v sin \theta \hat{j}$

$P_2 = 70\times 1.3 cos 61^0 \hat{i} +70\times 1.3 sin 61^0\hat{j}$

$P_2 = 44.12\hat{i} +79.59\hat{j}$

now

P = P₁ + P₂

$P = 198.22 \hat{i} +79.59 \hat{j}$

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$P = \sqrt{198.22^2 + 79.59^2}$

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3 years ago

At a distance of 0.208 cm from the center of a charged conducting sphere with radius 0.100cm, the electric field is 485 n/c . wh

Anna35 [415]

We have the equation for electric field E = kQ/ $d^{2}$

Where k is a constant, Q is the charge of source and d is the distance from center.

In this case E is inversely proportional to $d^{2}$

So, $\frac{E_{1} }{E_{2}} = \frac{d_{2}^{2}}{d_{1}^{2}}$

$E_{1}$ = 485 N/C

$d_{1}$ = 0.208 cm

$d_{2}$ = 0.620 cm

$E_{2}$ = ?

$\frac{485 }{E_{2}} = \frac{0.620^{2}}{0.208^{2}}$

$E_{2}$ = $\frac{485*0.208^{2}}{0.628^{2}}$

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3 years ago

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A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she

lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

$f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L}$

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}} }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = 1.3225 \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz = 740.6 \ Hz$

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, $L_{new}$ = 2/3·L

The value of the fundamental frequency will then be given as follows;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \times \dfrac{2 \times L}{3} } =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz = 840 \ Hz$

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

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3 years ago