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3 years ago

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Please help on this one!!

1 answer:

inysia [295]3 years ago

3 0

The energy transformations that occur as you coast down long hill on a bicycle, including the brakes to make the bike stop at the bottom, is that at the top of the hill you have high GPE AND LOW KE, on your way down you have HIGH KE AND LOW GPE, and at the bottom you have thermal energy due to the stop of the brakes.

the law of conversation of energy and describe the energy transformations that occur as you coast down a long hill on a bicycle and then apply the brakes to make the bike stop at the bottom.

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Please help!!! 50 POINTS

taurus [48]

Answer:

The Energy Involved In A Reaction That Changes Methane Gas And Oxygen Into Carbon Dioxide And Water.

Explanation:

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3 years ago

Read 2 more answers

Two cars are heading towards each other but are 12 km apart. one car is going 70 km/hr, and the other is going 50 km/hr. how muc

vova2212 [387]

<span>12-50t=70t, t= 0.1h = 6 minutes.</span>

3 0

3 years ago

Ship A is located 4.1 km north and 2.3 km east of ship B. Ship A has a velocity of 22 km/h toward the south and Ship B has a vel

castortr0y [4]

Answer:

a) x component = -31.25 km/hr

b) y component = 46.64 km/hr

Explanation:

Given data:

A position is 4km north and 2.5 km east to B

Ship A velocity = 22 km/hr

ship B velocity = 40 km/hr

A velocity wrt to velocity of B

$\vec{V_{AB}} =\vec{V_A} - \vec{V_B}$

$\vec{V_A} = 22 km/hr$

$\vec{V_B} = 40 cos38\hat{i} + 40sin 38 \hat{j}$

$= 31.52\hat{i} + 24.62 \hat{j}$

putting respective value to get velocity of A with respect to B

$\vec{V_{AB}} = -22 \hat {j} - (31.52\hat{i} + 24.62 \hat{j})$

$\vec{V_{AB}} = -31.52\hat{i} - 46.62\hat{j}$

a) x component = -31.25 km/hr

b) y component = 46.64 km/hr

6 0

3 years ago

A train is moving at 15ms-1. It slows down to 10.5ms over a time of 4s. Calculate the distance it travels before stopping if acc

Nostrana [21]

Answer:

3WGGRGWERGRG

Explanation:

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4 0

3 years ago

A point charge of -0.70 μC is fixed to one corner of a square. An identical charge is fixed to the diagonally opposite corner. A

MakcuM [25]

Answer:

The magnitude and algebraic sign of q is $14\sqrt{2}\ \mu C$

Explanation:

Given that,

Point charge = -0.70 μC[/tex]

We need to calculate the force for all charges

The electric force at first corner

$F_{1}=\dfrac{-k0.70\times10^{-6}q}{r^2}$

The electric force at opposite corner

$F_{3}=\dfrac{-k0.70\times10^{-6}q}{r^2}$

The net force is

$F=\sqrt{F_{1}^2+F_{2}^2}$

Put the value into the formula

$F=\sqrt{(\dfrac{-k0.70\times10^{-6}q}{r^2})^2+(\dfrac{-k0.70\times10^{-6}q}{r^2})^2}$

The electric force at second corner

$F_{2}=\dfrac{-kq^2}{2r^2}$

The net force acting on either of the charges is zero.

So, $F=F'$

$\sqrt{(\dfrac{k0.70\times10^{-6}q}{r^2})^2+(\dfrac{-k0.70\times10^{-6}q}{r^2})^2}=\dfrac{kq^2}{2r^2}$

$\sqrt{2}\times\dfrac{0.70\times10^{-6}kq}{r^2}=\dfrac{kq^2}{2r^2}$

$q=14\sqrt{2}\ \mu C$

Hence, The magnitude and algebraic sign of q is $14\sqrt{2}\ \mu C$

8 0

3 years ago