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3 years ago

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Physics

2 answers:

krek1111 [17]3 years ago

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Answer:

ye ek rod h or electric ⚡ field h P point

Virty [35]3 years ago

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3. A model rocket is launched straight upward at 58.8 m/s.

kolezko [41]

Answer:

a). 6 seconds

b). 12 seconds

c). 176.4 meters

Explanation:

a). Equation to be applied to calculate the time taken by the rocket to reach at the peak height,

v = u - gt

where v = final velocity

u = initial velocity = 58.8 m per sec

g = gravitational pull = 9.8 m per sec²

t = duration of the flight

At the peak height,

v = 0

Therefore, 0 = 58.8 - (9.8)(t)

t = $\frac{58.8}{9.8}$

= 6 seconds

b). Total time of flight = 2(Time taken to go up)

= 2×6

= 12 sec

c). Formula to get the peak height is,

$h=ut-\frac{1}{2}gt^2$

h = (58.8)6 - $\frac{1}{2}(9.8)(6)^2$

= 352.8 - 176.4

= 176.4 meters

8 0

3 years ago

If a light is moved twice (2x) as far from a surface, the area the light covers is ___ as big.

Serga [27]

Answer:

The correct option is;

- 4x

Explanation:

From the inverse square law, as the distance from the source of a physical quantity increases, the intensity of the source is spread over an area proportional to the square of the distance of the object from the source

The inverse square law can be presented as follows;

$I = \dfrac{S}{4\times \pi \times r^2 }$

As the distance, r, increases, the surface it covers also increases by the power of 2

Therefore, where the distance increases from r to 2·r, we have;

When, I, remain constant

$I = \dfrac{4\times S}{4\times \pi \times (2\cdot r)^2 } = I = \dfrac{4\times S}{4\times 4\times \pi \times r^2 } = \dfrac{S}{4\times \pi \times r^2 }$

The surface increases to 4·S by the inverse square law

Therefore, the correct option is 4 × x.

3 0

3 years ago

Quadrupling the power output from a speaker emitting a single frequency will result in what increase in loudness (in units of dB

Leviafan [203]

Answer:

6.02 dB increase

Explanation:

Let us take the initial power from the speaker P' = P Watt

then, the final power P = 4P Watt

for a given unit area, initial intensity (power per unit area) will be

I' = P Watt/m^2

and the final quadrupled sound will produce a sound intensity of

I = 4P Watt/m^2

Increase in loudness is gotten from the relation

ΔL = $10log_{10} \frac{I}{I'}$

where

I = final sound intensity

I' = initial sound intensity

imputing values of the intensity into the equation, we have

==> $10log_{10} \frac{4P}{P}$ = $10log_{10} 4$ = 6.02 dB increase

6 0

3 years ago

A radioactive isotope has a life for 15 minutes. It has an initial count rate of 36000 Bg. What will the count be after 1 hour?

lukranit [14]

Answer:

18000 Bg

Explanation:

after 15 minutes = 36000/2

After 30 minutes = 36000/4

30 minutes × 2 = 1 hour

After 1 hour = 36000/4 × 2 = 18000 Bg

3 0

3 years ago

A pipe with cross-sectional area 2.0 m^2 is joined to a second pipe with cross-sectional area 0.5 m^2. The pipes are both comple

IRISSAK [1]

Answer:

$v_2=8\ m/s$

Explanation:

It is given that,

Area of cross section of the pipe, $A_1=2\ m^2$

Area of cross section of the another pipe, $A_2=0.5\ m^2$

Speed of water in first pipe, $v_1=2\ m/s$

To find,

Speed of the water flowing in the second pipe.

Solve,

Let $v_2$ is the speed of water flowing in the second pipe. The relation between the area of cross section and the velocity is given by the continuity equation. It is given by :

$A_1v_1=A_2v_2$

$v_2=\dfrac{A_1v_1}{A_2}$

$v_2=\dfrac{2\times 2}{0.5}$

$v_2=8\ m/s$

Therefore, the water is flowing in the second pipe at the rate of 8 m/s.

3 0

3 years ago

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