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bulgar [2K]
2 years ago
5

A ball on a string moves around a complete circle, once a second, on a frictionless, horizontal table. The tension in the string

is measured to be 6.0N . What would the tension be if the ball went around in only half a second?
Mathematics
1 answer:
ki77a [65]2 years ago
5 0
The tension in the string balances out, and thus equals the centripetal force of the ball 

T = mv^2/r 
<span>
if it only takes half the time to finish one orbit it has to be moving at twice the original speed. </span>
<span>
And since v is squared T will increase by 4 </span>
<span>
T' = m(2v)^2/r </span>
<span>
T' = 4mv^/r = 4T </span>
<span>
T' = 24.0 N

I hope my answer has come to your help. Have a nice day ahead and may God bless you always!
</span>
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a rectangle pyramid fits exactly on top of a rectangular prism. the prism has a length of 18 cm, a width of 6 cm, and a height o
irina1246 [14]
To find the volume of the prism you are going to use the formula  V=WxLxH.
When you plug in your numbers you should get V= 18x6x9 which ends up equaling 972.

Now that you have that you need to find the volume of the pyramid using the formula V= L x W x H /3 and when you plug that in it looks like
V= 18 x 6 x 15 /3

Upon simplifying that you get 540.

Now you have the volume of your rectangular prism and your rectangular pyramid add them together and you should get 1512



I recommend drawing it out... it helps a lot:)
4 0
1 year ago
Point G is on segment FH such that it partitions segment FH into a ratio of 2:3.Point F is located at (5, 7) and point H is loca
Margarita [4]

Answer:

G = (9.4, 9,4)

Step-by-step explanation:

The ratio is applied in the x-distance and the y-distance. The ratio is 2:3 so you have to divide the distances by 5 and 2/5 correspond to FG and 3/5 to GH

x-distance:

x2 - x1 = 16 - 5 = 11

11/5 = 2.2

y-distance:

y2 - y1 = 13 - 7 = 6

6/5 = 1.2

Point G = Point F + (2.2*2, 1.2*2)

Point G = (5, 7) + (4.4, 2.4)

= (9.4, 9.4)

8 0
2 years ago
A force of 500.0 is represented graphically with its tail at the origin and the tip pointed in a direction 30.0° above the posit
trapecia [35]

Answer:

F^{'}=(250\sqrt{3},250 })

Step-by-step explanation:

We have F´ =500 and \alpha=30º, so x and y components:

F´ = (F_{x} , F_{y}) this is

F_{x} = F^{'} *Cos\alpha

F_{y} = F^{'} *Sin\alpha  

F_{x} = F`*Cos\alpha =500*Cos30=500*\frac{\sqrt{3} }{2} =250\sqrt{3}

F_{y}=F*Sin\alpha  =500*Sin30=500*\frac{1}{2}=250;

Finally

F' = (250\sqrt{3} , 250)

3 0
2 years ago
If mc022-1.j pg and mc022-2.j pg, what is the value of (f – g)(144)? –84 –60 0 48
Mashutka [201]

Answer:

0

Step-by-step explanation:

f(x) = √(x) + 12

g(x) = 2√(x)

(f-g)(x) = √(x) + 12 - 2√(x)

(f-g)(x) = 12 - √(x)

if x = 144

(f-g)(144) = 12 - √(144) = 12 - 12 = 0

5 0
2 years ago
Read 2 more answers
Q. You are working on an air conditioning system. A roll of cylindrical copper tubing has an outside diameter of 7/8 inch and an
diamong [38]

The refrigerant can hold 1.92 ft³

Step-by-step explanation:

The formula to apply is that of volume

v=π *r²*h where v is volume, r is radius of cylindrical object and h is height   of tubing

Finding volume of using outside  diameter

v=π*7/16*7/16*12

v=49/64 *3/1 *3.14

v=7.22 ft³

Volume using inside diameter

v=π*r²*h

v=3.14*3/8*3/8*12

v=5.30 ft³

Volume the refrigerant  can hold is;

7.22-5.30 = 1.92 ft³

Learn More

Volume of a cylindrical object:brainly.com/question/2665971

Keywords :conditioning,system,roll,copper tubing,refrigerant

#LearnwithBrainly

5 0
2 years ago
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