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2 years ago

9

The core of a star must be at temperature of _____ degrees Celsius for hydrogen fusion to take place. 10,000 100,000 1,000,000 1

0,000,000 100,000,000

1 answer:

disa [49]2 years ago

6 0

The core of a star must be at the temperature of 10,000,000 degrees Celsius for hydrogen fusion to begin.

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Martin has severe myopia, with a far point on only 17 cm. He wants to get glasses that he'll wear while using his computer whose

marusya05 [52]

Answer:

Explanation:

Far point = 17 cm . That means he can not see beyond this distance .

He wants to see at an object at 65 cm away . That means object placed at 65 has image at 17 cm by concave lens . Using lens formula

1 / v - 1 / u = 1 / f

1 / - 17 - 1 / - 65 = 1 / f

= 1 / 65 - 1 / 17

= - .0434 = 1 / f

power = - 100 / f

= - 100 x .0434

= - 4.34 D .

6 0

3 years ago

You are trying to overhear a juicy conversation, but from your distance of 20.0 m , it sounds like only an average whisper of 30

12345 [234]

Answer:

r₂ = 0.2 m

Explanation:

given,

distance = 20 m

sound of average whisper = 30 dB

distance moved closer = ?

new frequency = 80 dB

using formula

$\beta = 10 log(\dfrac{I_1}{I_0})$

I₀ = 10⁻¹² W/m²

now,

$30 = 10 log(\dfrac{I_1}{10^{-12}})$

$\dfrac{I_1}{10^{-12}}= 10^3$

$I_1= 10^{-8}\ W/m^2$

to hear the whisper sound = 80 dB

$80 = 10 log(\dfrac{I_2}{10^{-12}})$

$\dfrac{I_2}{10^{-12}}= 10^8$

$I_2= 10^{-4}\ W/m^2$

we know intensity of sound is inversely proportional to square of distances

$\dfrac{I_1}{I_2}=\dfrac{r_2^2}{r_1^2}$

$\dfrac{10^{-8}}{10^{-4}}=\dfrac{r_2^2}{20^2}$

$10^{-4}=\dfrac{r_2^2}{20^2}$

r₂ = 0.2 m

6 0

3 years ago

A horizontal pipe contains water at a pressure of 110 kPa flowing with a speed of 1.4 m/s. When the pipe narrows to one half its

Pavel [41]

Answer:

a

$v_2 = 5.6 \ m/s$

b

$P_2 = 80600 \ Pa$

Explanation:

From the question we are told that

The pressure of the water in the pipe is $P_1= 110 \ kPa = 110 *10^{3 } \ Pa$

The speed of the water is $v_1 = 1.4 \ m/s$

The original area of the pipe is $A_1 = \pi \frac{d^2 }{4}$

The new area of the pipe is $A_2 = \pi * \frac{[\frac{d}{2} ]^2}{4} = \pi * \frac{\frac{d^2}{4} }{4} = \pi \frac{d^2}{16}$

Generally the continuity equation is mathematically represented as

$A_1 * v_1 = A_2 * v_2$

Here $v_2$ is the new velocity

So

$\pi * \frac{d^2}{4} * 1.4 = \pi * \frac{d^2}{16} * v_2$

=> $\frac{d^2}{4} * 1.4 = \frac{d^2}{16} * v_2$

=> $d^2 * 1.4 = \frac{d^2}{4} * v_2$

=> $1.4 = 0.25 * v_2$

=> $v_2 = 5.6 \ m/s$

Generally given that the height of the original pipe and the narrower pipe are the same , then we will b making use of the Bernoulli's equation for constant height to calculate the pressure

This is mathematically represented as

$P_1 + \frac{1}{2} * \rho * v_1 ^2 = P_2 + \frac{1}{2} * \rho * v_2 ^2$

Here $\rho$ is the density of water with value $\rho = 1000 \ kg /m^3$

$P_2 = P_1 + \frac{1}{2} * \rho [ v_1^2 - v_2^2 ]$

=> $P_2 = 110 *10^{3} + \frac{1}{2} * 1000 * [ 1.4 ^2 - 5.6 ^2 ]$

=> $P_2 = 80600 \ Pa$

4 0

2 years ago

A net force of –8750N is used to stop of 1250.kg car travelling 25m/s. What braking distance is needed to bring the car to a hal

Karolina [17]

Answer:

d = 44.64 m

Explanation:

Given that,

Net force acting on the car, F = -8750 N

The mass of the car, m = 1250 kg

Initial speed of the car, u = 25 m/s

Final speed, v = 0 (it stops)

The formula for the net force is :

F = ma

a is acceleration of the car

$a=\dfrac{F}{m}\\\\a=\dfrac{-8750}{1250}\\\\a=-7\ m/s^2$

Let d be the breaking distance. It can be calculated using third equation of motion as :

$v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-(25)^2}{2\times (-7)}\\\\d=44.64\ m$

So, the required distance covered by the car is 44.64 m.

4 0

2 years ago

20 POINT QUESTION

Valentin [98]

Label A: sublimation
Label B: condensation
Label C: melting

8 0

2 years ago

Read 2 more answers