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2 years ago

A +6.33 uC charge q1 is attracted by a force of 0.115 N to a second charge q2 that is 1.44 m away. What is the value of q2?

Physics

1 answer:

jarptica [38.1K]2 years ago

7 0

Answer:-4.1858

Explanation: I hate Acellus

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If an astronaut has a mass of 80kg on earth, what is their mass on the moon?

stealth61 [152]

Answer:

130 N basically

Explanation:

6 0

2 years ago

In the flow past a compression corner, the upstream Mach number and pressure are 3.5 and 1 atm, respectively. Downstream of the

yulyashka [42]

Answer:

$\theta=23.7^{\circ}$

Explanation:

The ratio of pressure 2 to 1 us 5.48/1= 5.48 rounded off as 5.5.

Referring to table A.2 of modern compressible flow then $M_{\beta_1}=2.2$

Also

$M_{\beta_1}=M_1 sin \beta$ and making $sin\beta$ the subject of the formula then

$sin\beta=\frac {2.2}{3.5}\\\beta=38.94^{\circ}$

Making reference to $\theta-\beta-M$ diagram then

$\theta=23.7^{\circ}$

4 0

2 years ago

Uma partícula com carga “Q”, no vácuo, gera um potencial elétrico de 600 volts a uma distância de 6,0 m, determine o valor dessa

garik1379 [7]

Answer:

q = 400 nC

the correct answer is b

Explanation:

The expression for the electric potential of a point charge is

V = k q / r

they ask us for the electrical charge

q = V r / k

let's calculate

Q = 600 6.0 / 9 10⁹

Q = 4 10⁻⁷ C

let's reduce to nC

Q = 4 10⁻⁷ C (10⁹ nC / 1C)

q = 4 10² nC = 400 nC

the correct answer is b

Traslate

La expresión para el potencial eléctrico de una carga puntual es

V = k q/r

nos piden la carga eléctrica

q= V r /k

calculemos

Q= 600 6,0 / 9 10⁹

Q= 4 10⁻⁷ C

reduzcamos a nC

Q = 4 10⁻⁷ C(10⁹ nC/1C )

q = 4 10² nC = 400 nC

la respuesta correcta es b

8 0

2 years ago

Ok plz help this is my entrance exam question for physics in grade 9 (yes i am moving to a new school again)

vagabundo [1.1K]

Answer:

300 W

Explanation:

We know that

P = W/t

P = mgh/t

P = 40 × 10 × 3/4

P = 10 × 10 × 3

P = 300 W

4 0

2 years ago

Light of wavelength 530.00 nm is incident normally on a diffraction grating, and the first‑order maximum is observed to be 33.0∘

GarryVolchara [31]

Answer:

1028 slits/mm

Explanation:

We are given that

Wavelength of light, $\lambda=530nm=530\times 10^{-9} m$

1nm= $10^{-9} m$

$\theta=33^{\circ}$

n=1

We have to find the number of slits per mm are marked on the grating.

We know that

$dsin\theta=n\lambda$

Using the formula

$dsin33^{\circ}=1\times 530\times 10^{-9}$

$d=\frac{530\times 10^{-9}}{sin33^{\circ}}$

$d=9.731\times 10^{-7} m$

1m= $10^{3}mm$

$d=9.731\times 10^{-7}\times 10^3$ mm

$d=0.0009731mm$

Number of slits= $\frac{1}{d}$

Number of slits= $\frac{1}{0.0009731}$ /mm

Number of slits=1028/mm

Hence, 1028 slits/mm are marked on the grating.

8 0

2 years ago

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