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2 years ago

8

DEFINE slope for v/t =

1 answer:

kari74 [83]2 years ago

6 0

Answer:

For example, if the acceleration is zero, then the velocity-time graph is a horizontal line (i.e., the slope is zero). If the acceleration is positive, then the line is an upward sloping line (i.e., the slope is positive). ... If the acceleration is great, then the line slopes up steeply (i.e., the slope is great).

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What is the potential difference needed to achieve a current of 16A with a resistance of

Fiesta28 [93]

Hi there!

We can use the following equation:
$\large\boxed{V = I R}$

V = potential difference (? V)
I = Current (16A)
R = Resistance (192Ω)

Plug in the givens and solve:

$F = 16 \times 192 = \boxed{3072 V}$

6 0

2 years ago

Use the figure below to answe

vovikov84 [41]

Answer:

B. Q

A. P

Explanation:

I assume the figure is the one included in my answer.

The pivot (also called the fulcrum) is point Q.

The effort needed to keep the bar balanced is force P.

7 0

3 years ago

A tetrahedron has an equilateral triangle base with 25.0-cm-long edges and three equilateral triangle sides. The base is paralle

KATRIN_1 [288]

Answer:

electric flux through the three side = 2.35 N m²/C

Explanation:

given,

equilateral triangle of base = 25 cm

electric field strength = 260 N/C

Area of triangle = $\dfrac{\sqrt{3}}{4}a^2$

= $\dfrac{\sqrt{3}}{4} 0.25^2$

= 0.0271 m³

electric flux = E. A

= 260 × 0.0271

= 7.046 N m²/C

since, tetrahedron does not enclose any charge so, net flux through tetrahedron is zero.

electric flux through the three side = (electric flux through base)/3

= $\dfrac{7.046}{3}$

electric flux through the three side = 2.35 N m²/C

4 0

3 years ago

What sport does not require a high level of fitness

Sveta_85 [38]

Rock climbing. Free diving. Sky diving. Dog sledding.

8 0

1 year ago

neptune is an average distance of 4.5×10^12m from the sun. Estimate the length of the Neptunian year.

Vikentia [17]

As per Kepler's third law we know that

$\frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3}$

now here we know that

$T_1$ = year of Neptune

$T_2$ = year of Earth

$R_1$ = distance of Neptune from Sun

$R_2$ = Distance of Earth from Sun

so now we will have

$\frac{T_1^2}{1} = \frac{(4.5 \times 10^{12})^3}{(1.5 \times 10^11)^3}$

$T_1^2 = 27000$

$T_1 = 164.3 years$

so length of year of Neptune is 164.3 years

6 0

3 years ago