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3 years ago

Which ball has the greater acceleration at the instant of release?

Physics

1 answer:

yuradex [85]3 years ago

6 0

They're equal.

Ignoring air resistance, any object you're not holding onto has the same, constant acceleration ... 9.8 m/s^2 down ... from the moment it leaves your hand until the moment it hits the ground.

It doesn't matter how heavy the object is, and it doesn't matter whether you throw it up, down, or sideways, or just let go and let it drop.

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what gravitational force does the moon produce on the earth is their centers are 3.88x10(8)m apart and the moon has a mass of 7.

yawa3891 [41]

Isaac Newton compared the acceleration of the moon to the acceleration of objects on earth. Believing that gravitational forces were responsible for each, Newton was able to draw an important conclusion about the dependence of gravity upon distance. This comparison led him to conclude that the force of gravitational attraction between the Earth and other objects is inversely proportional to the distance separating the earth's center from the object's center. But distance is not the only variable affecting the magnitude of a gravitational force. Consider Newton's famous equation

Read more on Brainly.com - brainly.com/question/12278932#readmore

4 0

3 years ago

An object has a displacement of 45

Anika [276]

Answer:

45° to min.

1° ----> 60 min

45° ----> ?.

45 * 60

2700 min

5 0

2 years ago

1. What is the temperature in Kelvin and Fahrenheit of silver at 51.4oC?1. What is the temperature in Kelvin and Fahrenheit of s

horsena [70]

Answer:

Temperature in Fahrenheit will be 124.52°F and in kelvin 324.4 K

Explanation:

We have given temperature in of silver in degree Celsius as $51.4^{\circ}C$

We have to convert this temperature in Fahrenheit and kelvin

For Fahrenheit

Temperature in Fahrenheit = $\frac{9}{5}\times 51.4+32=124.52^{\circ}F$

For kelvin

In kelvin = 51.4+273 = 324.4 K

So temperature in Fahrenheit will be 124.52°F and in kelvin 324.4 K

5 0

3 years ago

Pulsars are neutron stars that emit X rays and other radiation in such a way that we on Earth receive pulses of radiation from t

stira [4]

Answer:

(a) $a_{c} = 5.41\times 10^{9} m/s^{2}$

(b) $a_{t} = 2.99\times 10^{- 5} m/s^{2}$

Given:

Time period of Pulsar, $T_{P} = 33.085 ms == 33.085\times 10^{- 3} s$

Equatorial radius, R = 15 Km = 15000 m

Spinning time, $t_{s} = 9.50\times 10^{10}$

Solution:

(a) To calculate the value of the centripetal acceleration, $a_{c}$ on the surface of the equator, the force acting is given by the centripetal force: