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2 years ago

5

What is the Force of Gravity of a 40 kg object on Earth?

1 answer:

AVprozaik [17]2 years ago

7 0

Gravity on Earth pulls objects to the center of the Earth about 9.8 meters per second per second. The child's mass on the moon or at a space station is 40 kg. The child's weight at these places changes because gravity is different at these locations.

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Two 1.9 kg masses are 1.1 m apart (center to center) on a frictionless table. Each has + 9.6 μC of charge.

Anni [7]

Answer:

F= 0.6 N

Explanation:

Fe(electrical force)= $k q1q2/r^2$

$k=9*10^9\\q1=-9.6 *10^-6 C\\q2= -9.6*10^-6 C\\r= 1.1m$

So,

$F=\frac{9*10^9*-9.6 *10^-6 C* -9.6*10^-6 C}{1.1^2}$

$F= 0.6 N$

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2 years ago

What happens to the atomic number of an atom when the number of neutrons in the nucleus of that atom increases? a It decreases b

kupik [55]

Answer:

It remains the same

Explanation:

It remains the same. This is because the number of protons doesn't change and the number of protons determines the atomic number.

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1 year ago

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solniwko [45]

Answer:

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2 years ago

Tell the value of the underlined digit 843,208,732,833 eight is underlined

devlian [24]

The value of the underlined 8 is, hundred billion's. Hope this helped!

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3 years ago

Read 2 more answers

The free-fall acceleration on Mars is 3.7 m/s^2. What length of pendulum has a period of 1.0 s on Earth? What length of pendulum

NemiM [27]

Answer:

Explanation:

Given

Free fall acceleration on mars $g_{m}=3.7\ m/s^2$

Time Period of pendulum on earth $T=1\ s$

Time period of Pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

for earth

$1=2\pi\cdot \sqrt{\frac{L}{9.8}}$

$L=\frac{9.8}{(2\pi )^2}$

$L=0.498\ m$

(b)For same time period on mars length is given by

$L'=\frac{g_m}{(2\pi )^2}$

$L'=\frac{3.7}{39.48}$

$L'=0.0936\ m$

$L'=9.36\ cm$

3 0

3 years ago