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3 years ago

10

A student thinks that scientific theories are not reliable sources of information, since scientists tend to argue about the deta

ils of them. What is the flaw in the student's reasoning?

1 answer:

Nadusha1986 [10]3 years ago

7 0

A theories are information

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What are some examples of transverse waves?

Deffense [45]

Answer:

-ripples on the surface of water.

-vibrations in a guitar string.

-a Mexican wave in a sports stadium.

-electromagnetic waves – eg light waves, microwaves, radio waves.

-seismic S-waves.

Explanation:

I've done this question before

7 0

3 years ago

In a given chemical reaction, the energy of the products is kess than the energy of the reactants. Which statement is true for t

deff fn [24]

Answer: Released

Explanation: Energy is released in this reaction possibly in the form of heat thus it is an exergonic and or exothermic reaction.

5 0

3 years ago

Sixty identical drippers, each with a hole of diameter 1.00 mm, are used to water a yard. If the water in the main pipe of diame

VashaNatasha [74]

Answer:

a

$V = 5.30 *10^{-2} \ m^3$

b

$v_1 = 0.3127 \ m/s$

Explanation:

From the question we are told that

The number of identical drippers is n = 60

The diameter of each hole in each dripper is $d = 1.0 \ mm = 0.001 \ m$

The diameter of the main pipe is $d_m = 2.5 \ cm = 0.025 \ m$

The speed at which the water is flowing is $v = 3.00 \ cm/s = 0.03 \ m/s$

Generally the amount of water used in one hour = 3600 seconds is mathematically represented as

$V = A * v * 3600$

Here A is the area of the main pipe with value

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

=> $A = 3.142 * \frac{0.025^2}{4}$

=> $A = 0.0004909 \ m^2$

So

=> $V = 0.0004909 * 0.03 * 3600$

=> $V = 5.30 *10^{-2} \ m^3$

Generally the area of the drippers is mathematically represented as

$A_1= n * \pi \frac{d^2}{4}$

=> $A_1 = 60 * 3.142 * \frac{0.001 ^2}{4}$

=> $A_1 = 4.713 *10^{-5} \ m^2$

Generally from continuity equation we have that

$Av = A_1 v_1$

=> $0.0004909 * 0.03 = 4.713 *10^{-5} * v_1$

=> $v_1 = 0.3127 \ m/s$

3 0

3 years ago

A runner of mass 53.0 kg runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its cen

const2013 [10]

Answer:

0.336 rad/s

Explanation:

$\omega_1$ = Angular speed of the turntable = -0.2 rad/s

R = Radius of turntable = 2.9 m

I = Moment of inertia of turntable = $76\ kgm^2$

M = Mass of turn table = 53 kg

$v_1$ = Magnitude of the runner's velocity relative to the earth = 3.6 m/s

As the momentum in the system is conserved we have

$Mv_1R+I\omega_1=(I + MR^2)\omega_2\\\Rightarrow \omega_2=\dfrac{Mv_1R+I\omega_1}{I + MR^2}\\\Rightarrow \omega_2=\dfrac{53\times 3.6-76\times 0.2}{76+53\times 2.9^2}\\\Rightarrow \omega_2=0.336\ rad/s$

The angular velocity of the system if the runner comes to rest relative to the turntable which is the required answer is 0.336 rad/s

4 0

3 years ago

A nonconducting sphere of radius 10 cm is charged uniformly with a density of 100 nC/m3. What is the magnitude of the potential

Flura [38]

Answer:

The answer is 3.0 V

Explanation:

the formula of the potential due to a non-conductive sphere with a uniform charge at a point at a distance x from the center of the sphere is equal to:

V1 = (2*π*k*ρ*(3*r^2 - x^2))/3

where

ρ = volume charge density = 100 nC/m^3

x = distance from center = 0

r = radius = 10 cm = 0.1 m

Thus:

V1 = (2*π*k*ρ*3*r^2)/3

for x = 4 cm = 0.04 m, we have:

V4 = (2*π*k*ρ*(3*r^2 - (0.04^2))/3 = (2*π*k*ρ*(3*r^2 - 0.0016))/3

The difference between both equations:

V1 - V4 = (2*π*k*ρ*3*r^2)/3 - (2*π*k*ρ*(3*r^2 - 0.0016))/3

V1 - V4 = (2*π*k*ρ*0.0016)/3 = (2*π*9x10^9*100x10^-9*0.0016)/3 = 3.02 V = 3 V

5 0

3 years ago