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timofeeve [1]
2 years ago
13

Earth's gravity attempts to change the velocity of all objects on Earth's surface toward _____ at a rate of 9.8 meters per secon

ds squared. outer space the Sun the horizon the center of the Earth
Physics
2 answers:
umka21 [38]2 years ago
8 0

The answer is to the ground.


Gravity refers to the force that holds together the universe. On Earth, the gravity attempts to change the velocity of all the objects on the Earth's surface toward the ground at a rate of 9.8 meters per second squared according to Galileo. 



dexar [7]2 years ago
4 0

Answer: The option is '<em>the center of the earth</em>.'

Explanation:

Gravity of the earth is defined as the the force of attraction exerted by the planet earth on the objects to attract them towards the center of the earth.

The value of the acceleration by which is velocity of freely falling object changes in the earth's atmosphere is 9.8 m/s^2.

Force=Mass\times 9.8 m/s^2

This force tends to attract the object towards the center of the earth.

From the given options the apt option is '<em>the center of the earth</em>.'

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kvasek [131]

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D. Quadruples

7 0
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What is the atomic mass of Jupiter?
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3 years ago
A. What is the RMS speed of Helium atoms when the temperature of the Helium gas is 343.0 K? (Possibly useful constants: the atom
kkurt [141]

Answer:

(a) 1462.38 m/s

(b) 2068.13 m/s

Explanation:

(a)

The Kinetic energy of the atom can be given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 10⁻²³ J/k

K.E = Kinetic Energy of atoms = 343 K

T = absolute temperature of atoms

The K.E is also given as:

K.E = (1/2)mv²

Comparing both equations:

(1/2)mv² = (3/2)KT

v² = 3KT/m

v = √[3KT/m]

where,

m = mass of Helium = (4 A.M.U)(1.66 X 10⁻²⁷ kg/ A.M.U) = 6.64 x 10⁻²⁷ kg

v = RMS Speed of Helium Atoms = ?

Therefore,

v = √[(3)(1.38 x 10⁻²³ J/K)(343 K)/(6.64 x 10⁻²⁷ kg)]

<u>v = 1462.38 m/s</u>

(b)

For double temperature:

T = 2 x 343 K = 686 K

all other data remains same:

v = √[(3)(1.38 x 10⁻²³ J/K)(686 K)/(6.64 x 10⁻²⁷ kg)]

<u>v = 2068.13 m/s</u>

8 0
3 years ago
A sonar pulse sent out by a boat arrives back after 4 seconds. If the speed of sound in water is 1600m/s, how deep is the water?
Varvara68 [4.7K]

Answer:

the boat would be deeped by 3200 m

Explanation:

Given that

The boat arrives back after 4 seconds

And, the speed of the sound in water is 1,600 m/s

We need to find out how much deep is the water

So,

As we know that

Distance = ( speed × time) ÷ 2

Here we divided by 2 because the boat arrives back

= (1600 × 4) ÷ 2

= 3200 m

Therefore the boat would be deeped by 3200 m

7 0
3 years ago
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