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3 years ago

9

In order to control for experimenter _____ as well as confounding variables such as severity of condition, attitudes, age, etc.,

researchers assign subjects to groups by random assignment.

1 answer:

White raven [17]3 years ago

3 0

The answer to this question is the random assignment to control experimeter BIAS. This is the method or the only way in order to control for experimeter BIAS as well as confounding variables such as the severity condition, the attitudes, the age and etc., The researchers or the person who conduct subjects to groups by random assignment.

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A sphere of mass 40kg is attaracted by a second sphere of mass 15 kg when their centers are 20 cm apart with a force of 0.1mg-wt

ser-zykov [4K]

<span>gravitational force = 0.1 mg-wt. = 0.1 * 10^-6 * 9.8 N = Gm1m2/r^2
m1 = 40 kg m2 =15 kg and r = 0.2 m
Put in and find G</span>

3 0

3 years ago

A wire with a linear mass density of 1.17 g/cm moves at a constant speed on a horizontal surface and the coefficient of kinetic

stira [4]

Answer:

The value is $B = 0.2312 \ T$

The direction is into the surface

Explanation:

From the question we are told that

The mass density is $\mu =\frac{m}{L} = 1.17 \ g/cm =0.117 kg/m$

The coefficient of kinetic friction is $\mu_k = 0.250$

The current the wire carries is $I = 1.24 \ A$

Generally the magnetic force acting on the wire is mathematically represented as

$F_F = F_B$

Here $F_F$ is the frictional force which is mathematically represented as

$F_F = \mu_k * m * g$

While $F_B$ is the magnetic force which is mathematically represented as

$F_B = BILsin(\theta )$

Here $\theta =90^o$ is the angle between the direction of the force and that of the current

So

$F_B = BIL$

So

$BIL = \mu_k * m * g$

=> $B = \mu_k * \frac{m}{L} * [\frac{g}{I} ]$

=> $B = 0.25 * 0.117 * [\frac{9.8}{1.24} ]$

=> $B = 0.2312 \ T$

Apply the right hand curling rule , the thumb pointing towards that direction of the current we see that the direction of the magnetic field is into the surface as shown on the first uploaded image

8 0

3 years ago

Which has the greater acceleration

Alik [6]

Answer: Object B

The velocity of object A is depicted in the graph as a straight line (constant speed therefore no acceleration).
The graph indicates that the velocity of object B increases (the object is accelerating).

5 0

3 years ago

An isotropic point source emits light at wavelength 510 nm, at the rate of 170 W. A light detector is positioned 410 m from the

Wewaii [24]

Answer:

$\frac{dB}{dt} = 3.03 \times 10^6 T/s$

Explanation:

As we know that the power emitted by the source is given as

$P = 170 W$

now we know that

$P = \frac{N}{t} (\frac{hc}{\lambda})$

now we know that energy density is given as

$u = \frac{B^2}{2\mu_0} + \frac{\epsilon_0 E^2}{2}$

now we have

$E = B c$

$u = \frac{B^2}{2\mu_0}$

intensity is defined as

$I = \frac{P}{A}$

now we have

$\frac{I}{c} = u = \frac{B^2}{2\mu_0}$ [/tex]

now we have

$\frac{dB}{dt} = \omega B$

$\frac{dB}{dt} = \frac{2\pi c B}{\lambda}$

$\frac{dB}{dt} = \frac{2\pi c \sqrt{2\mu_0 I}}{\lambda\sqrt c}$

here we have

$I = \frac{P}{4\pi r^2}$

$I = \frac{170}{4\pi (410)^2}$

$I = 8.05 \times 10^{-5}$

now we have

$\frac{dB}{dt} = \frac{2\pi\sqrt{2\mu_0 c (8.05 \times 10^{-5})}}{(510 nm)}$

$\frac{dB}{dt} = 3.03 \times 10^6 T/s$

4 0

3 years ago

A 0.111 kg hockey puck moving at 55 m/s is caught by a 80 kg goalie at rest. With what speed does the goalie slide on the (frict

Andrews [41]

Answer: 0.076 m/s

Explanation:

Momentum is conserved:

m v = (m + M) V

(0.111 kg) (55 m/s) = (0.111 kg + 80. kg) V

V = 0.076 m/s

After catching the puck, the goalie slides at 0.076 m/s.

6 0

3 years ago