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Fynjy0 [20]
3 years ago
6

Claim A: More potential energy can be stored by moving against the magnetic force of a stronger magnet.

Physics
1 answer:
stira [4]3 years ago
6 0

Answer:

stronger magnet thats the evidence

Explanation:

i think its help

brainliest pls

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100g converted to kg .
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Answer: 0.1 kilogram

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17. Calculate the amount of gravitational potential energy at the top of one 4 points hill. The mass of the coaster is 500 kg. T
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Answer: D. 292,338 J

This is the correct answer :)

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The vertical motion of mass A is defined by the relation x 5 10 sin 2t 1 15 cos 2t 1 100, where x and t are expressed in millime
nikklg [1K]

Explanation:

Given that,

The vertical motion of mass A is defined by the relation as :

x=10\ sin2t+15\ cos2t+100

At t = 1 s

x=10\ sin2+15\ cos2+100

x = 115.33 mm

(a) We know that,

Velocity, v=\dfrac{dx}{dt}

v=\dfrac{d(10\ sin2t+15\ cos2t+100)}{dt}

v=20\ cos2t-30\ sin2t

At t = 1 s

v=20\ cos2-30\ sin2

v = 18.94 mm/s

We know that,

Acceleration, a=\dfrac{dv}{dt}

a=\dfrac{d(20\ cos2-30\ sin2)}{dt}

a=-40\ cos2t-60\ cos2t

At t = 1 s

v=-40\ cos2-60\ cos2

a=-99.93\ mm/s^2

(b) For maximum velocity, \dfrac{dv}{dt}=a=0

-40\ cos2t-60\ cos2t=0

t = 45 seconds

For maximum acceleration, \dfrac{da}{dt}=0

80\ sin2t+120\ cos2t=0

t = 61.8 seconds

Hence, this is the required solution.

5 0
4 years ago
Vector a with arrow has a magnitude of 12.3 units and points due west. vector b with arrow points due north. (a) what is the mag
lesantik [10]

part a)

Vector a has magnitude 12.3 and its direction is west, while Vector b has unknown magnitude and its direction is north. This means that the two vectors form a right-angle triangle, so a and b are two sides, while a+b is the hypothenuse.

We know the magnitude of a+b, which is 14.5, so we can use the Pythagorean theorem to calculate the magnitude of b:

|b|=\sqrt{(a+b)^2-a^2}=\sqrt{(14.5)^2-(12.3)^2}=7.68


part b) The direction of the vector a+b relative to west can be found by calculating the tangent of the angle of the right-angle triangle described in the previous part; the tangent of the angle is equal to the ratio between the opposite side (b) and the adjacent side (a):

tan x=\frac{b}{a}=\frac{7.68}{12.3}=0.62

And the angle is

x=tan^{-1} (0.62)=31.8^{\circ}

with direction north-west.


part c)

This is exactly the same problem as the one we solved in part a): the only difference here is that the hypothenuse of the triangle is now given by a-b rather than a+b. In order to find a-b, we have to reverse the direction of b, which now points south. However, the calculations to get the magnitude of b are exactly the same as before, since the magnitude of (a-b) is the same as (a+b) (14.5 units), therefore the magnitude of b is still 7.68 units.


part d)

Again, this part is equivalent to part b); the only difference is that b points now south instead of north, so the vector (a-b) has direction south-west instead of north-west as before. Since the magnitude of the vectors involved are the same as part b), we still get the same angle, 31.8^{\circ}, but this time the direction is south-west instead of north-west.

5 0
4 years ago
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