All categories

3 years ago

Which of the following is one reason why an experiment should have the larger sample size possible

1 answer:

3 0

An experiment should always have the largest sample size possible because it can show the most possible outcomes and can give you a more precise outcome for your data.

You might be interested in

A convex thin lens with refractive index of 1.50 has a focal length of 30cm in air. When immersed in a certain transparent liqui

GalinKa [24]

Answer:

$n_l = 1.97$

Explanation:

given data:

refractive index of lens 1.50

focal length in air is 30 cm

focal length in water is -188 cm

Focal length of lens is given as

$\frac{1}{f} =\frac{n_2 -n_1}{n_1} * \left [\frac{1}{r1} -\frac{1}{r2} \right ]$

$\frac{1}{f} =\frac{n_{g} -n_{air}}{n_{air}} * \left [\frac{1}{r1} -\frac{1}{r2} \right ]$

$\frac{1}{f} =\frac{n_{g} -1}{1} * \left [\frac{1}{r1} -\frac{1}{r2} \right ]$

focal length of lens in liquid is

$\frac{1}{f} =\frac{n_{g} -n_{l}}{n_{l}} * \left [\frac{1}{r1} -\frac{1}{r2} \right ]$

$=\frac{n_{g} -n_{l}}{n_{l}} [\frac{1}{(n_{g} - 1) f}$

rearrange fro $n_l$

$n_l = \frac{n_g f_l}{f_l+f(n_g-1)}$

$n_l = \frac{1.50*(-188)}{-188 + 30(1.50 -1)}$

$n_l = 1.97$

7 0

3 years ago

A particle oscillates harmonically x = A cos(ωt + φ0), with amplitude 9 m, angular frequency π s −1, and initial phase π 3 radia

Gekata [30.6K]

Answer:

$t = \frac{5}{12} s$

Explanation:

As we know that the equation of particle position is given as

$x = A cos(\omega t + \phi_0)$

Now the speed of the particle is given as

$v = A\omega sin(\omega t + \phi_0)$

now we know that potential energy and kinetic energy is equal

so we have

$\frac{1}{2} mv^2 = \frac{1}{2}kx^2$

so we will have

$A^2\omega^2 sin^2(\omega t + \phi_0) = A^2\omega^2 cos^2(\omega t + \phi_0)$

$tan^2(\omega t + \phi_0) = 1$

$\omega t + \phi_0 = \frac{\pi}{4} or \frac{3\pi}{4}$

$\pi t = \frac{3\pi}{4} - \frac{\pi}{3}$

$\pi t = \frac{5\pi}{12}$

$t = \frac{5}{12} s$

6 0

3 years ago

A projectile is launched with an initial velocity 60m/s at an angle 60° to the vertical. What the magnitude of it's displacement

emmasim [6.3K]

Answer:

the magnitude of the displacement after 5s is 137.31 m.

Explanation:

Given;

initial velocity of the projectile, u = 60 m/s

angle of projection, θ = 60°

time of motion, t = 5s

the vertical component of the velocity, $u_y= u\ sin \theta = 60sin(60^0)$

The magnitude of the displacement after 5s is calculated as;

$h = u_yt -\frac{1}{2} gt^2\\\\h = 60sin (60^0)\times 5 - \frac{1}{2} (9.8)(5)^2\\\\h = 259.81-122.5\\\\h = 137.31 \ m$

Therefore, the magnitude of the displacement after 5s is 137.31 m.

3 0

3 years ago

How much does it cost to operate a 550 W Toaster for 6 hours. If electrical energy cost 12.0 cents per kW.H

miss Akunina [59]

Answer:

6.6 cents a month!!!

8 0

3 years ago

An object is 40cm from concave mirror and the image is formed 60cm from mirror calculate the focal length of mirror?

Elden [556K]

Answer: the answer is 100 focal length.

Explanation: mark brainliest PLEASE

4 0

3 years ago