A car travels 98 km. It takes 34.92 minutes to complete the trip. What was the average speed?

Why are many adults and youth in the United states overweight?

Anna11 [10]

Answer:

I'd imagine fast food is a lot more popular than most countries over there and easier to get hold of.

Explanation:

america

6 0

2 years ago

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Help me pls, class starts soon

ratelena [41]

Answer:

Explanation:

1.)66.36

2.)11.4

3.)0.8104

4.)4158.315

i got some of them and try the rest good luck

4 0

2 years ago

On which planet can you jump about the same height as Earth? Explain your answer.

Debora [2.8K]

Answer:

Venus and Earth

Explanation:

The force of gravity depends on your mass AND the mass of the planet you stand on. Earth and Venus are about the same size, so they have about the same mass. If you go to Venus, your mass hasn't changed, and the planet mass is almost the same as earth- so the force of gravity on you (AKA your weight) will be the same.

8 0

2 years ago

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You need to push a heavy box across a rough floor, and you want to minimize the average force applied to the box during the time

7nadin3 [17]

When pushing the body it is necessary to break the frictional force generated by the floor. Once this frictional force is overcome, the body will begin to move. Ideally, if a constant velocity is maintained or close to this value, the acceleration that will be exerted will tend to be zero and therefore, by Newton's second law the value of the Force will also tend to minimum values.

Remember that this law tells us that

$F= ma$

$F= m \frac{\Delta v}{t}$

Therefore the best strategy is A. keep pushing the box forward at a steady speed

7 0

2 years ago

A spherical drop of water carrying a charge of 43 pC has a potential of 540 V at its surface (with V = 0 at infinity). (a) What

almond37 [142]

Explanation:

Given that,

Charge on a spherical drop of water is 43 pC

The potential at its surface is 540 V

(a) The electric potential on the surface is given by :

$V=\dfrac{kq}{r}$

r is the radius of the drop

$r=\dfrac{kq}{V}\\\\r=\dfrac{9\times 10^9\times 43\times 10^{-12}}{540}\\\\r=7.17\times 10^{-4}\ m$

(b) Let R is the radius of the spherical drop, when two such drops of the same charge and radius combine to form a single spherical drop. ATQ,

$\dfrac{4}{3}\pi r^3+\dfrac{4}{3}\pi r^3=\dfrac{4}{3}\pi R^3\\\\2r^3=R^3\\\\R=2^{1/3} r$

Now the charge on the new drop is 2q. New potential is given by :

$V=\dfrac{9\times 10^9\times 43\times 10^{-12}\times 2}{2^{1/3}\times 7.17\times 10^{-4}}\\\\V=856.79\ V$

Hence, the radius of the drop is $7.17\times 10^{-4}\ m$ and the potential at the surface of the new drop is 856.79 V.

5 0

2 years ago