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nlexa [21]
2 years ago
12

A student is trying to calculate the density of an ice cream cone. She already knows the mass, but she needs to determine the vo

lume as well. Which of the following formulas can be used to calculate the volume of the cone?
Physics
2 answers:
vladimir2022 [97]2 years ago
5 0
V= 1/3 π r²h 
this is the formula for a cone hope this helps :) 

Ostrovityanka [42]2 years ago
5 0

Answer:

V = \frac{1}{3}\pi R^2 H

Explanation:

Volume of any type of cone is given as

V = \frac{1}{3} (base area) (height)

Now we know that base of the triangular cone is always like a circle of radius R

Now the area of the base is given as

A = \pi R^2

Now in order to find volume let say the height of the cone is H

so volume of the cone is given as

V = \frac{1}{3}\pi R^2 H

You might be interested in
What element in the periodic table is the basis of all organic molecules
Dimas [21]
Carbon is the basis of all organic molecules
3 0
3 years ago
Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 21 degrees and 61 degree
kvv77 [185]

Answer:

1

When second polarizer is removed the intensity after it passes through the stack is    

                    I_f_3 = 27.57 W/cm^2

2 When third  polarizer is removed the intensity after it passes through the stack is    

                I_f_2 = 102.24 W/cm^2

Explanation:

  From the question we are told that

       The angle of the second polarizing to the first is  \theta_2 = 21^o  

        The angle of the third  polarizing to the first is     \theta_3 = 61^o

        The unpolarized light after it pass through the polarizing stack   I_u = 60 W/cm^2

Let the initial intensity of the beam of light before polarization be I_p

Generally when the unpolarized light passes through the first polarizing filter the intensity of light that emerges is mathematically evaluated as

                     I_1 = \frac{I_p}{2}

Now according to Malus’ law the  intensity of light that would emerge from the second polarizing filter is mathematically represented as

                    I_2 = I_1 cos^2 \theta_1

                       = \frac{I_p}{2} cos ^2 \theta_1

The intensity of light that will emerge from the third filter is mathematically represented as

                  I_3 = I_2 cos^2(\theta_2 - \theta_1 )

                          I_3= \frac{I_p}{2}(cos^2 \theta_1)[cos^2(\theta_2 - \theta_1)]

making I_p the subject of the formula

                  I_p = \frac{2L_3}{(cos^2 \theta [cos^2 (\theta_2 - \theta_1)])}

    Note that I_u = I_3 as I_3 is the last emerging intensity of light after it has pass through the polarizing stack

         Substituting values

                      I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (61-21)])}

                      I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (40)])}

                           =234.622W/cm^2

When the second    is removed the third polarizer becomes the second and final polarizer so the intensity of light would be mathematically evaluated as

                      I_f_3 = \frac{I_p}{2} cos ^2 \theta_2

I_f_3 is the intensity of the light emerging from the stack

                     

substituting values

                     I_f_3 = \frac{234.622}{2} * cos^2(61)

                       I_f_3 = 27.57 W/cm^2

  When the third polarizer is removed  the  second polarizer becomes the

the final polarizer and the intensity of light emerging from the stack would be  

                  I_f_2 = \frac{I_p}{2} cos ^2 \theta_1

I_f_2 is the intensity of the light emerging from the stack

Substituting values

                  I_f_2 =  \frac{234.622}{2} cos^2 (21)

                     I_f_2 = 102.24 W/cm^2

   

7 0
2 years ago
high school physics, no need detail explain, just give the answer, but you have to make sure thank you
andrey2020 [161]

Answer:

approximately 30 degrees

Explanation:

If it takes the cannonball 2 seconds to reach the maximum height, we can use the analysis of the vertical component of the velocity and the fact that the acceleration of gravity is the one acting opposite to this initial vertical component (v_y) of the velocity. We know as well that at the top of the trajectory, the vertical component of the velocity is zero, and then the movement starts going down in it trajectory. So, the final velocity for the first part of the ascending movement is zero, giving us the following equation for the velocity under an accelerated movement (with acceleration of gravity "g" acting):

v_f=v_i-a\,t\\v_f=v_y-g\,t\\0=v_y-9.8\,*\,2\\v_y=9.8\,*\,2=19.6 \frac{m}{s}

By knowing the vertical component of the initial velocity (19.6 m/s), and the actual magnitude of the total initial velocity (40 m/s), we can calculate what angle was the initial velocity vector forming above the horizontal. We use for such the fact that the sine of the angle relates the opposite side of a right angle triangle with the hypotenuse, and solve for the angle using the arcsin function:

sin(\theta)=\frac{opp}{hyp} \\sin(\theta)=\frac{19.6}{40}\\\theta=arcsin(\frac{19.6}{40})\\\theta=29.34^o

which tells us that the closer answer shown is 30^o

7 0
3 years ago
Multiple Intelligences :
vlada-n [284]

Answer:

all of the above

PLS MARK ME AS BRAINLIAST

8 0
3 years ago
a golfer hits a golf ball with a velocity of 36.0 meters/second at an angle of 28.0. if the hang time of the golf ball is 33.4 s
Kobotan [32]
Solving this using the time, we know that range = horizontal velocity x time of flight 

since there are no horizontal forces acting on the ball, there are no horizontal accelerations and the initial horizontal velocity of 36 cos 28 will be constant throughout. If we use the correct time of flight given the launch parameters, we have 

range = 36 cos 28 x 3.44 s = 109.3 m

6 0
2 years ago
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