I will mark brainliest if you answer please im givingalot of points also

A marching band is practicing. The band director sees the drum major hit a drum. How far away is the director if he hears the dr

ivanzaharov [21]

Answer:

825 m

Explanation:

330 m/s

1 / 0.40 = 2.5

330 x 2.5 = 825

7 0

3 years ago

Which of the following best explains how earth is heated through radiation

lawyer [7]

Answer:

The sun touches earth during daytime and the suns rays heat our earth giving us heat. The sun heating the earth is also considered radiation.

Explanation:

6 0

3 years ago

State any two precautions taken to ensure accurate result on simple pendulum<br>

galben [10]

Explanation:

1) The bob of pendulum should be displaced with a small angle. 2) The amplitude of the oscillation of a simple pendulum should be small. 3) Fans should be switched off to reduce the air resistance. 4)The simple pendulum should be oscillate in a vertical plane only.Feb 23, 2019

6 0

3 years ago

The circumference of a sphere was measured to be

professor190 [17]

To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are

$\phi = 76cm$

$Error (dr) = 0.5cm$

The radius then would be

$\phi = 2\pi r \\76cm = 2\pi r\\r = \frac{38}{\pi} cm$

And

$\frac{d\phi}{dr} = 2\pi \\d\phi = 2\pi dr \\0.5 = 2\pi dr$

PART A ) For the Surface Area we have that,

$A = 4\pi r^2 \\A = 4\pi (\frac{38}{\pi})^2\\A = \frac{5776}{\pi}$

Deriving we have that the change in the Area is equivalent to the maximum error, therefore

$\frac{dA}{dr} = 4\pi (2r) \\dA = 4r (2\pi dr)$

Maximum error:

$dA = 4(\frac{38}{\pi})(0.5)$

$dA = \frac{76}{\pi}cm^2$

The relative error is that between the value of the Area and the maximum error, therefore:

$\frac{dA}{A} = \frac{\frac{76}{\pi}}{\frac{5776}{\pi}}$

$\frac{dA}{A} = 0.01315 = 1.31\%$

PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so

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

$V = \frac{4}{3} \pi (\frac{38}{\pi})^3$

$V = \frac{219488}{3\pi^2}$

Therefore the Maximum Error would be,

$\frac{dV}{dr} = \frac{4}{3} 3\pi r^2$

$dV = 2r^2 (2\pi dr)$

$dV = 4r^2 (\pi dr)$

Replacing the value for the radius

$dV = 4(\frac{38}{\pi})^2(0.5)$

$dV = \frac{2888}{\pi^2} cm^3$

And the relative Error

$\frac{dV}{V} = \frac{ \frac{2888}{\pi^2}}{ \frac{219488}{3\pi^2} }$

$\frac{dV}{V} = 0.03947$

$\frac{dV}{V} = 3.947\%$

3 0

3 years ago

Help please....................

Marina86 [1]

Answer:

a. arranging from least to higher pressure exerted by object B ,D A, C

b object B exert maximum pressure on surface because pressure is indirectly proportional to the surface area from the above figure we can easily conclude that object a have least surface area so, it exert more pressure on surface.

c.object C exert minimum pressure on surface because pressure is indirectly proportional to the surface area from the above figure we can easily conclude that object a have large surface area so, it exert less pressure on surface.

Explanation:

7 0

3 years ago