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Dimas [21]
2 years ago
9

15 points. give me the method.

Physics
1 answer:
AveGali [126]2 years ago
5 0

Answer:

\boxed{{160 \:  m(s)}^{ - 1} }

Explanation:

if \:the \:  frequencies \: are \to \\   f_{1} =  640Hz  \\ and \\f_{2}   = 480Hz \:  \\ but \:  \boxed{v = f \gamma }:   f =  \frac{v}{ \gamma } \\ if \:  \gamma_{1}  -  \gamma _{2}  = 1 =  \gamma  \\ f_{1}  - f_{2}  = 640 - 480 = \boxed{ 160Hz} = f \\ v = f \gamma = 160 \times 1 =  \boxed{{160 \:  m(s)}^{ - 1} }

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A balloon is expanded to the same volume as that of a human head. Do an order-of-magnitude estimate of the volume of this balloo
cestrela7 [59]

Answer:

Volume of balloon =  1000 cm^3

Explanation:

 The head of a normal person can be assumed as a sphere with radius 10 cm.

 Volume of sphere =\frac{4}{3} \pi r^3, where r is the radius.

 We have approximate radius = 10 cm.

  Approximate volume of head =\frac{4}{3} \pi r^3=\frac{4}{3} *\pi* 10^3=4188cm^3

 In the given options the closest value to the approximate volume is 1000 cm^3.

 So, volume of head = Volume of balloon =  1000 cm^3

4 0
3 years ago
A miniature quadcopter is located at x = -2.25 m and y, - 5.70 matt - 0 and moves with an average velocity having components Vv,
kupik [55]

Recall that average velocity is equal to change in position over a given time interval,

\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t}

so that the <em>x</em>-component of \vec v_{\rm ave} is

\dfrac{x_2 - (-2.25\,\mathrm m)}{1.60\,\mathrm s} = 2.70\dfrac{\rm m}{\rm s}

and its <em>y</em>-component is

\dfrac{y_2 - 5.70\,\mathrm m}{1.60\,\mathrm s} = -2.50\dfrac{\rm m}{\rm s}

Solve for x_2 and y_2, which are the <em>x</em>- and <em>y</em>-components of the copter's position vector after <em>t</em> = 1.60 s.

x_2 = -2.25\,\mathrm m + \left(2.70\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{x_2 = 2.07\,\mathrm m}

y_2 = 5.70\,\mathrm m + \left(-2.50\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{y_2 = 1.70\,\mathrm m}

Note that I'm reading the given details as

x_1 = -2.25\,\mathrm m \\\\ y_1 = -5.70\,\mathrm m \\\\ v_x = 2.70\dfrac{\rm m}{\rm s}\\\\ v_y=-2.50\dfrac{\rm m}{\rm s}

so if any of these are incorrect, you should make the appropriate adjustments to the work above.

8 0
3 years ago
A 5kg block is pulled across a table by a horizontal force of 40 N with a frictional force of 8 N opposing the motion. Draw a fo
Taya2010 [7]

Answer:

Net force on the block is 32 N.

Acceleration of the object is 6.4 m/s².

Explanation:

Let the acceleration of the object be a m/s².

Given:

Mass of the block is, m=5\ kg

Force of pull is, F=40\ N

Frictional force on the block is, f=8\ N

The free body diagram of the object is shown below.

From the figure, the net force in the forward direction is given as:

F_{net}=F-f=40-8=32\ N

Now, from Newton's second law of motion, net force is equal to the product of mass and acceleration. So,

F_{net}=ma\\32=5a\\a=\frac{32}{5}=6.4\ m/s^2

Therefore, the acceleration of the object in the forward direction is 6.4 m/s².

3 0
3 years ago
The mass luminosity relation L  M 3.5 describes the mathematical relationship between luminosity and mass for main sequence sta
ivanzaharov [21]

Answer:

(a) <u>11.3 L</u>

(b) <u>10 M</u>

Explanation:

The mass-luminosity relationship states that:

Luminosity ∝ Mass^3.5

Luminosity = (Constant)(Mass)^3.5

So, in order to find the values of luminosity or mass of different stars, we take the luminosity or mass of sun as reference. Therefore, write the equation for a star and Sun, and divide them to get:

Luminosity of a star/L = (Mass of Star/M)^3.5 ______ eqn(1)

where,

L = Luminosity of Sun

M = mass of Sun

(a)

It is given that:

Mass of Star = 2M

Therefore, eqn (1) implies that:

Luminosity of star/L = (2M/M)^3.5

Luminosity of Star = (2)^3.5 L

<u>Luminosity of Star =  11.3 L</u>

(b)

It is given that:

Luminosity of star = 3160 L

Therefore, eqn (1) implies that:

3160L/L = (Mass of Star/M)^3.5

taking ln on both sides:

ln (3160) = 3.5 ln(Mass of Star/M)

8.0583/3.5 = ln(Mass of Star/M)

Mass of Star/M = e^2.302

<u>Mass of Star = 10 M</u>

3 0
3 years ago
Calculate the mass (in SI units) of (a) a 160 lb human being; (b) a 1.9 lb cockatoo. Calculate the weight (in English units) of
kondaur [170]

Explanation:

1. Force applied on an object is given by :

F = W = mg

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g = acceleration due to gravity, g = 32 ft/s²

m=\dfrac{F}{g}

m=\dfrac{160\ lb}{32\ ft/s^2}

m = 5 kg

(b) A 1.9 lb cockatoo, F = 1.9 lb

m=\dfrac{F}{g}

m=\dfrac{1.9\ lb}{32\ ft/s^2}

m = 0.059 kg

2. (a) A 2300 kg rhinoceros, m = 2300 kg

W=2300\ kg\times 32\ ft/s^2=73600\ lb

(b) A 22 g song sparrow, m = 22 g = 0.022 kg

W=0.022\ kg\times 32\ ft/s^2=0.704\ lb

Hence, this is the required solution.

5 0
3 years ago
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