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Virty [35]
1 year ago
15

What is the acceleration due to gravity on earth

Physics
2 answers:
Zanzabum1 year ago
5 0
The acceleration due to gravitation on earth is 9.806 m/s^2












Svetradugi [14.3K]1 year ago
4 0
The acceleration due to gravity on earth is 9.8 m / s^2
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How are gas particles arranged in a container?
joja [24]

Answer:

3. Tightly Packed

Explanation:

Hope this helps

I learned this

4 0
3 years ago
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In an experiment what are all the parts of an experiment that remain unchanged are called<br>​
stira [4]
The answer for this question is Control Variable because it doesn’t change throughout the experiment.
3 0
4 years ago
A 3.5-kg object placed on an inclined plane (angle 30? above the horizontal) is connected by a string going over a pulley to a 1
Blababa [14]

Answer:

a= 0.22 m/s²

Explanation:

Given that

M = 3.5 kg

θ = 30°

m = 1 kg

μ= 0.3

The force due to gravity

F₁= M g sinθ

F₁=3.5 x 10 x sin 30

F₁= 17.5 N

F₂ = m g

F₂ = 1 x 10 = 10 N

The maximum value of the friction force on the incline plane

Fr = μ M g cosθ

Fr = 0.3 x 2.5 x 10 cos30°

Fr= 6.49 N

Lets take acceleration of the system is a  m/s²

F₁ - F₂  - Fr =  (M+m) a

17.5 - 10 - 6.49 = (3.5+1)a

a= 0.22 m/s²

7 0
3 years ago
HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!<br> I NEED THE ANSWER NOW
FrozenT [24]
90 F = 43 OR 0.9F = 0.43
(F = 43 / 90 OR 0.43 / 0.9 =) 0.48 N

upwards force = downwards force
(R =) 1.2 N
5 0
3 years ago
A person of mass 55 kg swings on a rope length 4 m from rest (when the rope makes an angle of 30 degrees with the vertical) and
vovangra [49]

Answer:

θ = 19.66°

Explanation:

To determine the angle that the rope makes with the vertical for the two people, you first take into account the potential energy of the first person before he swings on the rope:

U=mgh

h: distance to the ground

g: gravitational acceleration = 9.8m/s^2

m: mass of the first person = 55 kg

In the image attache below you can notice that the height h is:

h=4-4cos(30\°)=0.53m

Then, the potential energy is:

U=(55kg)(9.8m/s^2)(0.53m)=285.67J

When the first person picks up the second person (when the rope is exactly vertical), all the potential energy becomes kinetic energy. Next, when both people reaches the maximum height h' the energy must be equal to the initial potential energy of the first person:

U'=(m_1+m_2)gh'=285.67\ J

From the previous equation you can get h':

h'=\frac{285.67J}{(55kg+70kg)(9.8m/s^2)}=0.2332m

Finally, you obtain the angle between the rope at the height h,' and the vertical, by calculating the following:

h'=4-4cos(\theta)\\\\\theta=cos^{-1}(\frac{4-h'}{4})=cos^{-1}(\frac{4-0.2332}{4})=19.66\°

hence, the angle between the rope and the vertical, when the two people are in the rope is 19.66°

8 0
3 years ago
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