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3 years ago

Body mass index (bm) is a measure of body weight relative

Physics

1 answer:

kow [346]3 years ago

8 0

BMI is a measure of body fat determined by ones Height, Weight, and Gender.

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The dimensions of a cylinder are changing, but the height is always equal to the diameter of the base of the cylinder. If the he

VashaNatasha [74]

Answer:

dV/dt = 9 cubic inches per second

Explanation:

Let the height of the cylinder is h

Diameter of cylinder = height of the cylinder = h

Radius of cylinder, r = h/2

dh/dt = 3 inches /s

Volume of cylinder is given by

$V = \pi r^{2}h$

put r = h/2 so,

$V = \pi \frac{h^{3}}{4}$

Differentiate both sides with respect to t.

$\frac{dV}{dt}=\frac{3h^{2}}{4}\times \frac{dh}{dt}$

Substitute the values, h = 2 inches, dh/dt = 3 inches / s

$\frac{dV}{dt}=\frac{3\times 2\times 2}{4}\times 3$

dV/dt = 9 cubic inches per second

Thus, the volume of cylinder increases by the rate of 9 cubic inches per second.

6 0

2 years ago

How many ohms of resistance must be present in a circuit that has 240 volts and a current of 15 amps?

Mkey [24]

THE ANSWER IS 16 ohms or however its spelled

8 0

3 years ago

Read 2 more answers

A rectangular barge, 5.2 m long and 2.4 m wide, floats in fresh water. Suppose that a 410-kg crate of auto parts is loaded onto

sesenic [268]

<h2>Answer:</h2><h2>The depth of barge float=3 cm</h2><h2>Explanation:</h2>

Length of rectangular barge=5.2 m

Width of rectangular barge=2.4m

Mass of crate=410 kg

Let h be the height of barge float

Volume of barge float= $l\times b\times h=5.2\times 2.4\times h=12.48h$

Density of water= $10^3kg/m^3$

Weight of water displaced by barge=Buoyant force=-Weight of horse

$Volume\;of\;water\times density\;of\;water\times g=410\times g$

$12.48h\times 1000=410$

$h=\frac{410}{12.48\times 1000}=0.03 m$

1 m=100 cm

$0.03 m=0.03\times 100=3$ cm

Hence, the depth of barge float=3 cm

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4 0

3 years ago

Read 2 more answers

How do the three types of boundary's work together to keep the Earth at equilibrium?

Maurinko [17]

There are three types: divergent, convergent, and transform boundaries. I hope this helps.

8 0

3 years ago

Read 2 more answers

Two 3.0 μC charges lie on the x-axis, one at the origin and the other at What is the potential (relative to infinity) due to the

Airida [17]

Complete Question:

Two 3.0µC charges lie on the x-axis, one at the origin and the other at 2.0m. A third point is located at 6.0m. What is the potential at this third point relative to infinity? (The value of k is 9.0*10^9 N.m^2/C^2)

Answer:

The potential due to these charges is 11250 V

Explanation:

Potential V is given as;

$V =\frac{Kq}{r}$

where;

K is coulomb's constant = 9x10⁹ N.m²/C²

r is the distance of the charge

q is the magnitude of the charge

The first charge located at the origin, is 6.0 m from the third charge; the potential at this point is:

$V =\frac{9X10^9 X3X10^{-6}}{6} =4500 V$

The second charge located at 2.0 m, is 4.0 m from the third charge; the potential at this point is:

$V =\frac{9X10^9 X3X10^{-6}}{4} =6750 V$

Total potential due to this charges = 4500 V + 6750 V = 11250 V

6 0

3 years ago