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4 years ago

How would we be able to detect a large asteroid if it were heading straight for earth?

1 answer:

7 0

<span>Answer: NASA in US using its super powerful telescope checks all the meteorites ,asteroids that may land on Earth. It deflects the asteroids ,meteorites from their respective path using copper ball . But it is not wholly practical as sometimes these meteorites get enter into key hole .</span>

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PLZ help asap :-/<br>............................

Misha Larkins [42]

Explanation:

$\underline{\boxed{\large{\bf{Option \; A!! }}}}$

Here,

$\rm { R_1}$ = 2Ω
$\rm { R_2}$ = 2Ω
$\rm { R_3}$ = 2Ω
$\rm { R_4}$ = 2Ω

We have to find the equivalent resistance of the circuit.

Here, $\rm { R_1}$ and $\rm { R_2}$ are connected in series, so their combined resistance will be given by,

$\longrightarrow \rm { R_{(1,2)} = R_1 + R_2} \\$

$\longrightarrow \rm { R_{(1,2)} = (2 + 2) \; Omega} \\$

$\longrightarrow \rm { R_{(1,2)} = 4 \; Omega} \\$

Now, the combined resistance of $\rm { R_1}$ and $\rm { R_2}$ is connected in parallel combination with $\rm { R_3}$ , so their combined resistance will be given by,

$\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \dfrac{1}{R_{(1,2)}} + \dfrac{1}{R_3} } \\$

$\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1}{4} + \dfrac{1}{2} \Bigg ) \;\Omega} \\$

$\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1 + 2}{4} \Bigg ) \;\Omega} \\$

$\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{3}{4} \Bigg ) \;\Omega} \\$

Reciprocating both sides,

$\longrightarrow \rm {R_{(1,2,3)}= \dfrac{4}{3} \;\Omega} \\$

Now, the combined resistance of $\rm { R_1}$ , $\rm { R_2}$ and $\rm { R_3}$ is connected in series combination with $\rm { R_4}$ . So, equivalent resistance will be given by,

$\longrightarrow \rm {R_{(1,2,3,4)}= R_{(1,2,3)} + R_4} \\$

$\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4}{3} + 2 \Bigg ) \; \Omega} \\$

$\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4 + 6}{3} \Bigg ) \; \Omega} \\$

$\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{10}{3} \Bigg ) \; \Omega} \\$

$\longrightarrow \bf {R_{(1,2,3,4)}= 3.33 \; \Omega} \\$

Henceforth, Option A is correct.

$\underline{\boxed{\large{\bf{Option \; B!! }}}}$

Here, we have to find the amount of flow of current in the circuit. By using ohm's law,

$\longrightarrow$ V = IR

$\longrightarrow$ 3 = I × 3.33

$\longrightarrow$ 3 ÷ 3.33 = I

$\longrightarrow$ 0.90 Ampere = I

Henceforth, Option B is correct.

$\tt \purple{Hope \; it \; helps \; you, Army! \heartsuit } \\$

4 0

3 years ago

What is the power of a refrigerator with voltage 110 V and<br> current 0.8 A?

tia_tia [17]

Answer: 88

Explanation:

8 0

3 years ago

Can anyone solve this for me?? I need help ASAP!!

Pie

Answer:

option A is rightttttttt

.......................

6 0

3 years ago

A certain metal with work function of Φ = 1.7 eV is illuminated in vacuum by 1.4 x 10-6 W of light with a wavelength of λ = 600

Makovka662 [10]

Answer:

1) n = 4.47*10^12 photons

2) K = 0.25 eV

Explanation:

This is a problem about the photoelectric effect.

1) In order to calculate the number of photons that impact the metal, you take into account the power of the light, which is given by:

$P=\frac{E}{t}=1.4*10^{-6}\frac{J}{s}$ (1)

Furthermore you calculate the energy of a photon with a wavelength of 600nm, by using the following formula:

$E_p=h\frac{c}{\lambda}$ (2)

c: speed of light = 3*10^8 m/s

h: Planck's constant = 6.626*10^-34 Js

λ: wavelength = 600*10^-9 m

You replace the values of the parameters in the equation (2):

$E_p=(6.626*10^{-34}Js)\frac{3*10^8m/s}{600*10^{-9}m}=3.131*10^{-19}J$

Next, you calculate the quotient between the power of the light (equation (1)) and the energy of the photon:

$n=\frac{P}{E_p}=\frac{1.4*10^{-6}J/s}{3.131*10^{-19}J}=4.47*10^{12}photons$

The number of photons is 4.47*10^12

2) The kinetic energy of the electrons emitted by the metal is given by the following formula:

$K=E_p-\Phi$ (3)

Ep: energy of the photons

Φ: work function of the metal = 1.7 eV

You first convert the energy of the photons to eV:

$E_p=3.131*10^{-19}J*\frac{6.242*10^{18}eV}{1J}=1.954eV$

You replace in the equation (3):

$K=1.95eV-1.7eV=0.25eV$

The kinetic energy of the electrons emitted by the metal is 0.25 eV

6 0

4 years ago

A 2.00 kg rock is dropped from the top of a 30.0 m high building. Calculate the ball’s momentum at the time that it strikes the

Darya [45]

Explanation:

We use the Theorem of conservation of mechanical energy for finding the velocity when it strikes the ground:

Ei = Ef

Ki + Ui = Kf + Uf

Ui = Kf

m g h = 1/2 m v^2

v = sqrt(2gh)

So the momentum will be:

p = mv = m * sqrt(2gh)

5 0

3 years ago