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3 years ago

When it comes to graphs I am no genius lol

Physics

1 answer:

Lunna [17]3 years ago

5 0

Answer:

Option B

Explanation:

Displacement = Area under the velocity-time graph

So to find the displacement of a particle from 7 to 8 seconds we take the triangle and calculate its area.

$Area\ of\ a\ right\ angled\ triangle = \frac{1}{2}*(base)*(height) \\\\Area\ of\ a\ right\ angled\ triangle = \frac{1}{2}*(1)*(3) \\\\Area\ of\ a\ right\ angled\ triangle = 1.5$

So the displacement is 1.5 meters

I have an uploaded image so you can understand it better hope it helps

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Shanika is an engineer at an amusement park who is experimenting with changes to the setup for a magnetic roller coaster ride. I

Whitepunk [10]

Answer:

I believe it might be point A since the question ask what will result in the ln a largest increase in potential energy

Explanation:

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3 years ago

A car traveling at 28 m/s starts to decelerate steadily. It comes to a complete stop in 13 seconds. What is its acceleration?

Maru [420]

Acceleration = (change in speed) / (time for the change)

change in speed = (speed at the end) minus (speed at the beginning)

change in speed = (zero) minus (28 m/s) = -28 m/s

Acceleration = (-28 m/s) / (13 sec)

Acceleration = -2.15 m/s²

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3 years ago

La masa de un camión es de

Nutka1998 [239]

Answer:

45000 kg and 45 tons

Explanation:

The expression in kilograms and tons is shown below;

As we know that

1 gr is 0.001 kg

So, 45000000 = 45,000 kg

And,

1 kg = 0.001 tons

So, 45000 kg = 45 tons

Therefore the same would be considered

7 0

3 years ago

The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 5

ra1l [238]

Answer:

For situation (a)

net charge E = E₊₂ + E₋₅ + E₋₃

E = K(q/d²)

where K = 8.99e9

d = 5.7cm = 5.7e-2m

Therefore,

E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)

E₋₅(y) = K(q/d²) = (8.99e9)× ((5.0e-6)÷(5.7e-2)) = 7.88e5(+y)

E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) = 4.73e5(+x)

thus

E = E₊₂ + E₋₅ + E₋₃

= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)

= 7.88e6(x) + 7.88e6(y)

use Pythagorean theorem

I <em>E </em>I = $\sqrt{(7.89e5)^{2} + (7.89e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) net magnitude = 1.115e6 $\frac{N}{C}$ @ 45° above +x axis

for situation (b)

net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆

E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)

E₊₁(y) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(-y)

E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)

E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)

thus,

E = E₊₄ + E₊₁ + E₋₁ + E₊₆

= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)

= 7.88e5(x) + 7.88e5(y)

use Pythagorean theorem

I <em>E </em>I = $\sqrt{(7.88e5)^{2} + (7.88e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) and (b) the net magnitude = 1.242e6 $\frac{N}{C}$ @ 45° above +x axis

Explanation:

I attached a sample image, i hope that corresponds to your question

5 0

3 years ago

A circular loop of flexible iron wire has an initial circumference of 165.0cm, but its circumference is decreasing at a constant

vovangra [49]

Answer:

emf induced in the loop, at the instant when 9.0s have passed = 1.576 * 10 ⁻² V.

Direction is counter clockwise.

Explanation:

See attached pictures.

6 0

4 years ago

When it comes to graphs I am no genius lol​

When it comes to graphs I am no genius lol