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Reptile [31]
3 years ago
9

Which graph uses bars to show data that are broken into intervals?

Physics
1 answer:
AnnyKZ [126]3 years ago
5 0

Answer:

A. scatter plot?

Explanation:

I dont really know if I'm right... sorry.

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Will Mark BRAINLIEST!!!!!!!!!!!!!!!!!!!
vazorg [7]

3.

a)

r = distance of each mass in each hand from center = 0.6 m

m = mass of each mass in each hand = 2 kg

v = linear speed = 1.1 m/s

L = combined angular momentum of the masses = ?

Combined angular momentum of the masses is given as

L = 2 m v r

L = 2 (2) (1.1) (0.6)

L = 2.64 kg m²/s


b)

v' = linear speed when she pulls her arms = ?

r' = distance of each mass from center after she pulls her arms = 0.15 m

Using conservation of momentum , angular momentum remains same, hence

L = 2 m v' r'

2.64 = 2 (2) (0.15) v'

v' = 4.4 m/s


4 0
4 years ago
At a certain distance from the center of the Earth, a 0.4-kg object has a weight of 2.0 N. (a) Find this distance. (b) If the ob
Alika [10]

Answer:

a) The distance of the object from the center of the Earth is 8.92x10⁶ m.

b) The initial acceleration of the object is 5 m/s².

Explanation:

a) The distance can be found using the equation of gravitational force:

F = \frac{GMm}{r^{2}}

Where:

G: is the gravitational constant = 6.67x10⁻¹¹ Nm²/kg²

M: is the Earth's mass =  5.97x10²⁴ kg  

m: is the object's mass = 0.4 kg

F: is the force or the weight = 2.0 N    

r: is the distance =?

The distance is:

r = \sqrt{\frac{GMm}{F}} = \sqrt{\frac{6.67 \cdot 10^{-11} Nm^{2}/kg^{2}*5.97 \cdot 10^{24} kg*0.4 kg}{2.0 N}} = 8.92 \cdot 10^{6} m      

Hence, the distance of the object from the center of the Earth is 8.92x10⁶ m.

         

b) The initial acceleration of the object can be calculated knowing the weight:              

W = ma                                                  

Where:            

W: is the weight = 2 N

a: is the initial acceleration =?          

a = \frac{W}{m} = \frac{2 N}{0.4 kg} = 5 m/s^{2}

Therefore, the initial acceleration of the object is 5 m/s².

           

I hope it helps you!    

4 0
3 years ago
A spherical raindrop 2.5 mm in diameter falls through a vertical distance of 3900 m. Take the cross-sectional area of a raindrop
Karolina [17]

Answer:

276.62 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s² (positive downward and negative upward)

Equation of motion

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 3900+0^2}\\\Rightarrow v=276.62\ m/s

<u>Neglecting air drag</u> the velocity of the spherical drop would be 276.62 m/s

5 0
4 years ago
What is the equilibrium constant of pure water at 25C
Tcecarenko [31]

Answer:

1.0 x 10-14.

Explanation:

We then replace the term on the right side of this equation with a constant known as the water dissociation equilibrium constant, Kw. In pure water, at 25C, the [H3O+] and [OH-] ion concentrations are 1.0 x 10-7 M. The value of Kw at 25C is therefore

6 0
4 years ago
Why is it important to consider experimental error in all the empirical results presented?
NemiM [27]
Because when your running any experiment there will always be an experimental error so always be ready for it so that you can correct that error in your experiment
3 0
3 years ago
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