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3 years ago

How does a bicycle rider's energy of motion changes as the gain speed while riding down a hill?

1 answer:

8 0

It is b. It gains momentum when going down. Think of it like a sled. When you go up the hill, it goes slower, but when you go down the hill, you gain speed(momentum) and it increases

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Gravity is affected by...

goldfiish [28.3K]

Mass and distance
force /pull of gravity decreases with the increase in separation between the two bodies
the amount of gravity an object possesses is proportional to the mass of that object.

4 0

3 years ago

A ball is thrown vertically upward, which is the positive direction. A little later, it returns to its point of release. The bal

Aleks [24]

Answer:

The initial velocity of the ball is <u>39.2 m/s in the upward direction.</u>

Explanation:

Given:

Upward direction is positive. So, downward direction is negative.

Tota time the ball remains in air (t) = 8.0 s

Net displacement of the ball (S) = Final position - Initial position = 0 m

Acceleration of the ball is due to gravity. So, $a=g=-9.8\ m/s^2$ (Acting down)

Now, let the initial velocity be 'u' m/s.

From Newton's equation of motion, we have:

$S=ut+\frac{1}{2}at^2$

Plug in the given values and solve for 'u'. This gives,

$0=8u-0.5\times 9.8\times 8^2\\\\8u=4.9\times 64\\\\u=\frac{4.9\times 64}{8}\\\\u=4.9\times 8=39.2\ m/s$

Therefore, the initial velocity of the ball is 39.2 m/s in the upward direction.

3 0

3 years ago

How do you calculate speed

Kryger [21]

Answer:

distance(d)=.....

time(t)=....

speed =?

we know that

speed =distance /time

8 0

3 years ago

Read 2 more answers

A source charge of 3 µC generates an electric field of 2.86 × 105 N/C at the location of a test charge. Using k = 8.99 × 109N.m^

Nataliya [291]

Variables:

Source charge, Q = 3 micro C = 3 * 10^ - 6 C

E = electric field = 2.86 * 10 ^5 N/C

K = 8.99 * 10^9 N * m^2 / C

d = distance = ?

Formula:

E = K * Q / (d^2) => d^2 = K * Q / E

=> d^2 = 8.99 * 10^9 N * m^2 / C * 3 * 10^ -6 C / (2.86 * 10^ 5 N/C)

d^2 = 9.43 * 10 ^ -2 m^2

=> d = 3.07 * 10^-1 m

Answer: 0.307 m

Note: it is a long distance due to the Electric field is very low

7 0

2 years ago

Read 2 more answers

A ball is whirled on the end of a string in a horizontal circle of radius R at speed v. By which one of the following means can

Vera_Pavlovna [14]

Explanation:

When an object moves in a circular path, it will have circular acceleration. Its magnitude of acceleration is given by :

$a=\omega^2R$

Since, $\omega=\dfrac{2\pi }{T}R$

$a=(\dfrac{2\pi}{T})^2R$

T is the time period

R is the radius of the circular path

To increase the centripetal acceleration bu a factor of 1.5 or 3/2, radius of circle must be increase by a factor of 6 and T is increased by a factor of 2 such that,

R'=6R and T'=2T

So,

$a'=(\dfrac{2\pi}{T'})^2R'$

$a'=(\dfrac{2\pi}{(2T)})^2(6R)$

$a'=\dfrac{6}{4}(\dfrac{2\pi}{T})^2R$

$a'=\dfrac{3}{2}(\dfrac{2\pi}{T})^2R$

Hence, this is the required solution.

4 0

2 years ago