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Ghella [55]
3 years ago
11

Can someone help me with this?

Mathematics
1 answer:
Gelneren [198K]3 years ago
8 0

Answer:

all the above

Step-by-step explanation:

Don't take my word for it

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A hollow shaft has a cross-sectional area 125.6cm2 and a bore of 8cm in diameter. Calculate the outer diameter of the shaft to 2
hichkok12 [17]

Answer:

The outer diameter of the shaft to 2s.f is  12.65cm

Step-by-step explanation:

We know that the expression for the area of a circular cross section is

A= \frac{\pi d^2}{4}

given data

area of cross section= 125.6cm^2

bore diameter d= 8cm

from the area of cross section we can solve for the outer diameter D

Area of cross section=\frac{\pi D^2}{4} \\\\125.6= \frac{3.142*D^2}{4}

solving for D we have

3.142*D^2= 125.6*4\\\3.142*D^2= 502.4\\D^2= \frac{502.4}{3.142} \\D^2= 159.8981\\D=\sqrt{159.8981} \\D= 12.6450\\D= 12.65cm

6 0
3 years ago
Which of the following statements is not true? A. A foot is longer than a yard. B. A mile is longer than an inch. C. A centimete
Andrews [41]

A is incorrect, a yard is three feet.

6 0
3 years ago
2688 divided by 96 with remainder
prohojiy [21]

Answer: OK, the answer is 28.

Step-by-step explanation: Also there is no remainder for this division problem, you could have the remainder if the numbers are not the same as a multiplication problem, so did you get it?

4 0
3 years ago
You are swinging a bucket in a circle at a velocity of 7.8 ft/s. The radius of the circle you are making is 1.25 ft. The acceler
Ksju [112]

Given that,

velocity, v = 7.8 ft/s

Radius, r = 1.25 ft

To find,

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b. What is the velocity if the acceleration is 25 ft/sec²?

Solution,

(a) The centripetal acceleration is given by :

a=\dfrac{v^2}{r}

Put v = 7.8 ft/s and r = 1.25 ft

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(b) Put a = 25 ft/s² to find velocity.

v=\sqrt{ar} \\\\v=\sqrt{25\times 1.25} \\\\v=5.59\ ft/s

Hence, this is the required solution.

7 0
3 years ago
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tia_tia [17]

Answer: 7x^2-8x+10

Step-by-step explanation: hope this helps gl :)

thx for the points

4 0
3 years ago
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