All categories

3 years ago

When referring to body composition we are talking about our leanness. Lean muscle burns _________ calories than fat.

Physics

1 answer:

stiv31 [10]3 years ago

6 0

Your answer is B.) More

You might be interested in

Jet aircraft maintenance crews are required to wear protective earplugs. Members of a particular crew wear earplugs that reduce

Taya2010 [7]

Answer:

So the sound intensity level they would experience without the earplugs is 110.32dB.

Explanation:

Given data

Sound intensity by factor =215

Sound intensity level =87 dB

To find

Sound intensity level they would experience without the earplugs

Solution

First we need to find the new sound intensity level

$I_{n}=215(10^{\frac{87}{10} } )\\I_{n}=1.08*10^{11})$

The dB can be calculated as:

$dB=10log(I_{n})\\$

Substitute the given values

$dB=10log(1.08*10^{11})\\dB=110.32dB$

So the sound intensity level they would experience without the earplugs is 110.32dB.

7 0

3 years ago

Can someone help<br> pls !

Gemiola [76]

A sort of electricity is a light bulb or a phone / computer charger. plants food water. the sun and rain . that’s what i’m guessing!

6 0

3 years ago

A student pushes a box with a total mass of 50 kg. What is the net force on the box if it accelerates 1.5 m/s

Harrizon [31]

We now that follow newton rules f=ma so net force equal to mass*acceleration=>f=50*1.5=75 N

3 0

3 years ago

Read 2 more answers

Wood is an example of a translucent material.<br> True<br> False

Svetradugi [14.3K]

False, wood is a solid structure that is not see through

3 0

3 years ago

A 1300 kg steel beam is supported by two ropes. (Figure

Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

the net horizontal force acting on the beam is

$R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0$

where $R_1,R_2$ are the magnitudes of the tensions in ropes 1 and 2, respectively;

the net vertical force acting on the beam is

$R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0$

where $m=1300\,\rm kg$ and $g=9.8\frac{\rm m}{\mathrm s^2}$ .

Eliminating $R_2$ , we have

$\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)$

$R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2$

$R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2$

$-R_1 \sin(50^\circ) = -\dfrac{mg}2$

$R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}$

Solve for $R_2$ .

$\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0$

$\dfrac{R_2}2 = -mg\cot(110^\circ)$

$R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}$

8 0

1 year ago