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3 years ago

6th grade science I mark as brainliest !

Physics

2 answers:

faltersainse [42]3 years ago

7 0

Answer:

The answer is B. molecules, atoms, elements, compund, cells, tissues, organs, organ systems, and organism

Leokris [45]3 years ago

5 0

In the 1th question the answer is B and the 2th question is A. Element.
I hope it’s helps :).

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A cube, whose edges are aligned with the , and axes, has a side length . The field is immersed in an electric field aligned with

Ivenika [448]

Complete Question

Complete Question is attached below

Answer:

$q=1.558*10^{-9}c$

Explanation:

From the question we are told that:

Side length s=1.13m

Left field strength $E_l=784.75N/m$

Right field strength $E_r=776.38 N/m$

Front field strength $E_f=725.5 N/m$

Back field strength $E_b=749.54 N/m$

Top field strength $E_t=944.95 N/m$

Bottom field strength $E_{bo}=1082.58 N/m$

Generally, the equation for Charge flux is mathematically given by

$\phi=EAcos\theta$

Where

Theta for Right,Left,Front and Back are at an angle 90

$cos 90=0$

Therefore

$\phi =0$ with respect to Right,Left,Front and Back

Generally, the equation for Charge Flux is mathematically also given by

$\phi=\frac{q}{e_o}$

Where

$Area =L*B\\\\A=1.13*1.13\\\\A=1.2769m^2$

Therefore

$Q_{net}=E_{bo}Acos\theta_{bo} +E_tAcos\theta_t$

$Q_{net}=1082.85*1.2769*cos0=944.95*1.2769cos (180)$

$Q_{net}=176N/C m^2$

Giving

$q=\phi*e_0$

$q=176N/C m^2*1.558*10^{-12}c$

$q=1.558*10^{-9}c$

4 0

3 years ago

Plzz helpp mee!!!! plzzz

Gnesinka [82]

If you add 2 miles from west then 2 miles east then it would 4 miles all together.

7 0

3 years ago

A tank whose bottom is a mirror is filled with water to a depth of 19.6 cm. A small fish floats motionless a distance of 6.40 cm

ANEK [815]

Answer:

A. 4.82 cm

B. 24.66 cm

Explanation:

The depth of water = 19.6 cm

Distance of fish = 6.40 cm

Index of refraction of water = 1.33

(A). Now use the below formula to compute the apparent depth.

$d_{app} = \frac{n_{air}}{n_{water}} \times d_{real} \\= \frac{1}{1.33} \times 6.40 \\= 4.82 cm.$

(B). the depth of the fish in the mirror.

$d_{real} = 19.6 cm + (19.6 cm – 6.40 cm) = 32.8 cm$

Now find the depth of reflection of the fish in the bottom of the tank.

$d_{app} = \frac{n_{air}}{n_{water}} \times d_{real} \\d_{app} = \frac{1}{1.33} \times 32.8 = 24.66\\$

4 0

3 years ago

Current flowing in a circuit depends on two variables. Identify these variables and their relationship to current.

Nuetrik [128]

Wow ! This question reads like it might have come from one of
Faraday or Maxwell's original laboratory notebooks.

Choice-A is the correct one, when you consider what "conductance"
means. Conductance is just 1/resistance .

So when you see
"A) Current is proportionate to the conductance of the circuit and
precisely proportional to the voltage applied across the circuit."

what it's saying is
"Current is inversely proportional to the resistance of the circuit, and
directly proportional to the voltage applied across the circuit."

If you write the equation for all those words, it looks like

I = V / R

and that's correct.

5 0

4 years ago

Read 2 more answers

Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 2.1×104 m/s when at a distance

trapecia [35]

Answer:

v_f = 6.92 x 10^(4) m/s

Explanation:

From conservation of energy,

E = (1/2)mv² - GmM/r

Where M is mass of sun

Thus,

E_i = E_f will give;

(1/2)mv_i² - GmM/(r_i) = (1/2)mv_f² - GmM/(r_f)

m will cancel out to give ;

(1/2)v_i² - GM/(r_i) = (1/2)v_f² - GM/(r_f)

Let's make v_f the subject;

v_f = √[(v_i)² + 2MG((1/r_f) - (1/r_i))]

G is Gravitational constant and has a value of 6.67 x 10^(-11) N.m²/kg²

Mass of sun is 1.9891 x 10^(30) kg

v_i = 2.1×10⁴ m/s

r_i = 2.5 × 10^(11) m

r_f = 4.9 × 10^(10) m

Plugging in all these values, we have;

v_f = √[(2.1×10⁴)² + 2(1.9891 x 10^(31)) (6.67 x 10^(-11))((1/(4.9 × 10^(10))) - (1/(2.5 × 10^(11)))] 20.408 e12

v_f = √[(441000000) + 2(1.9891 x 10^(30)) (6.67 x 10^(-11))((16.408 x 10^(-12))]

v_f = √[(441000000) + (435.38 x 10^(7))

v_f = 6.92 x 10^(4) m/s

5 0

4 years ago