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3 years ago

8

An object traveling at a constant speed but with a changing direction is accelerating.

2 answers:

Ivahew [28]3 years ago

7 0

True, it is still moving.

prohojiy [21]3 years ago

6 0

Strange as it may seem, that's true. (choice 'a'.)

"Acceleration" doesn't mean "speeding up". It means ANY change in
the speed or direction of motion. So a car with the brakes applied
and slowing down, and a point on the rim of a bicycle wheel that's
turning at a constant rate, are both accelerating.

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Write an essay about the Electromagnetic spectrum and its many uses in real life practices. Answer these questions in the essay:

vampirchik [111]

Answer:

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Explanation:

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3 years ago

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Help me please, help

Alisiya [41]

Answer:

455,000 Pa

Explanation:

PV = nRT

If n is constant:

PV / T = PV / T

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3 years ago

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A tennis player tosses a tennis ball straight up and then catches it after 2.00 s at the same height as the point of release. (a

Readme [11.4K]

Answer:

Part a)

a = -9.81 m/s/s

Part b)

v = 0

Part c)

v = 9.81 m/s

Part d)

$H = 4.905 m$

Explanation:

Part a)

During the motion of ball it will have only gravitational force on the ball

so here the acceleration of the ball is only due to gravity

so it is given as

$a = g = 9.81 m/s^2$

Part b)

As we know that ball is moving against the gravity

so here the velocity of ball will keep on decreasing as the ball moves upwards

so at the highest point of the motion of the ball the speed of ball reduce to zero

$v_f = 0$

Part c)

We know that the total time taken by the ball to come back to the initial position is T = 2 s

so in this time displacement of the ball will be zero

$\Delta y = 0 = v_y t + \frac{1}{2} at^2$

$0 = v_y (2) - \frac{1}{2}(9.81)(2^2)$

$v_y = 9.81 m/s$

Part d)

at the maximum height position we know that the final speed will be zero

so we will have

$v_f^2 - v_i^2 = 2 a d$

here we have

$0 - (9.81^2) = 2(-9.81)H$

$H = 4.905 m$

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3 years ago

A moon of mass 1×10^20kg is in a circular orbit around a planet. The planet exerts a gravitational force of 2×10^21n on the moon

vagabundo [1.1K]

Hi there!

In this instance, the centripetal force experienced by the moon is equivalent to the gravitational force.

Thus:
$\large\boxed{F_c = F_g}$

Centripetal acceleration, according to Newton's Second Law:

$\Sigma F = ma \\\\F_c = m * a_c\\\\a_c = \frac{F_c}{m}$

Therefore:

$a_c = \frac{F_g}{m_m} = \frac{2 * 10^{21}N}{1 * 10^{20}kg} = \boxed{20 N/kg}$

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2 years ago

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Artist 52 [7]

Answer: energy transformation

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2 years ago

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