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Dimas [21]
3 years ago
9

15 points. give me the method.

Physics
1 answer:
AveGali [126]3 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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Explain the difference between the synodic period of a planet and the sidereal period. why is the synodic period often much less
Ilia_Sergeevich [38]
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(I will explain the second question with an example, so it's easier to understand)
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3 years ago
4.
tatiyna
The answer is A. A folkway is a closely held belief by a specific group of people.
6 0
3 years ago
Read 2 more answers
A capacitor consists of two metal surfaces separated by an insulating layer. A new capacitor has no charge on either of its surf
mihalych1998 [28]
<h2>Answer:</h2><h3>(A) the positively charged surface increases and the energy stored in the capacitor increases.</h3>

When charging a capacitor transferring charge from one surface to the other, the first surface becomes negatively charged while the second surface becomes positively charged. As you transfer the charge, the voltage of the positively charged surface increases and the energy stored in the capacitor also increases. We can solve this by the definition of <em>capacitance</em><em> </em>that is <em>a  measure of the ability of a capacitor  to store energy. </em>For any capacitor, the capacitance is  a constant defined as:

C=\frac{Q}{V_{ab}}

To maintain C constant, if Q increases V also increases.

On the other hand, the potential energy U can be expressed as:

U=\frac{Q^{2}}{2C}

In conclusion, as Q increases the potential energy also increases.

5 0
3 years ago
A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle
tatiyna

Answer:

a) 17.8 m/s

b) 28.3 m

Explanation:

Given:

angle A = 53.0°

sinA = 0.8

cosA = 0.6

width of the river,d = 40.0 m,

the far bank was 15.0 m lower than the top of the ramp h = 15.0 m,

The river itself was 100 m below the ramp H = 100 m,

(a) find speed v

vertical displacement

-h= vsinA\times t-gt^2/2

putting values h=15 m, v=0.8

-15 = 0.8vt - 4.9t^2  ............. (1)

horizontal displacement d = vcosA×t = 0.6×v ×t

so v×t = d/0.6 = 40/0.6

plug it into (1) and get

-15 = 0.8\times40/0.6 - 4.9t^2

solving for t we get

t = 3.734 s

also, v = (40/0.6)/t = 40/(0.6×3.734) =  17.8 m/s

(b) If his speed was only half the value found in (a), where did he land?

v = 17.8/2 = 8.9 m/s

vertical displacement = -H =v sinA t - gt^2/2

⇒ 4.9t^2 - 8.9\times0.8t - 100 = 0

t = 5.30 s

then

d =v×cosA×t = 8.9×0.6×5.30= 28.3 m

3 0
3 years ago
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