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2 years ago

8

Dr. Haviland is looking at mental rotation data but he forgot to label his chart. The data shows reaction times for 700 ms, 1400

ms and 1200 ms. What are the MOST likely corresponding degrees of rotation that will match his data?

1 answer:

ira [324]2 years ago

6 0

Answer:

Explanation:

(C) 180 degrees, 90 degrees, 120 degrees.

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Can somebody help me please? 10 point!

Vlada [557]

The correct answer is the last one:
A and C are different elements, while D is an isotope of C.

In fact, A and C are different elements, because they have a different number of protons in the nucleus (A has 3 protons, while C has 4 protons). Instead, D and C are the same element (they both have 4 protons in the nucleus), but they are different isotopes since they have a different number of neutrons (D has 4 neutrons while C has 3 neutrons)

7 0

2 years ago

Light of wavelength 436.1 nm falls on two slits spaced 0.31 mm apart. What is the required distance from the slits to the screen

kakasveta [241]

Answer:

The correct answer is "4.26 m".

Explanation:

Given:

Wavelength,

$\lambda = 436.1 \ nm$

or,

$=436.1\times 10^{-9} \ m$

Distance,

$d = 0.31 \ mm$

or,

$=0.31\times 10^{-3} \ m$

Distance between the 1st and 2nd dark fringes,

$(y_2-y_1) = 6\times 10^{-3} \ m$

As we know,

⇒ $\frac{d}{L} (y_2-y_1) = \lambda$

or,

⇒ $L=\frac{d(y_2-y_1)}{\lambda}$

By substituting the values, we get

$=\frac{0.31\times 6\times 10^{-6}}{436.1\times 10^{-9}}$

$=\frac{0.31\times 6\times 10^3}{436.1}$

$=\frac{1860}{436.1}$

$=4.26 \ m$

3 0

3 years ago

What is the speed of light in meters per second?

MAXImum [283]

The answer is approximately 2.998e+8

3 0

3 years ago

if pete (mass =90.00kg) weighs himself and finds that he weighs 30.0 pounds, how far away from the earth is he?

Svet_ta [14]

Here we know that mass of the person is 90 kg

His weight is given as 30 lbf

so here we can convert it into Newton as we know that

1 lbf = 4.45 N

Now from above conversion

$30 lbf = 30 \times 4.45 = 133.45 N$

now we can use this to find the gravity at this height

$mg = 133.45$

$90\times g = 133.45$

$g = 1.49 m/s^2$

now we know that with height gravity varies as

$g' = g(\frac{R}{R+H})^2$

$1.49 = 9.8(\frac{6.37\times 10^6}{6.37\times 10^6 + h})^2$

$h = 1.0 \times 10^7 m$

so above is the height from the surface of earth

4 0

2 years ago

The acceleration due to gravity is 9.81 m/s2, towards the Earth. Rain falling from an altitude of 9,000 m would fall for about 1

Anna [14]

Answer:

the final speed of the rain is 541 m/s.

Explanation:

Given;

acceleration due to gravity, g = 9.81 m/s²

height of fall of the rain, h = 9,000 m

time of the rain fall, t = 1.5 minutes = 90 s

Determine the initial velocity of the rain, as follows;

$h = ut + \frac{1}{2} gt^2\\\\9000 = 90u + \frac{1}{2} (9.8)(90)^2\\\\9000 = 90u + 39690\\\\90u = -30690\\\\u = \frac{-30690}{90} \\\\u = -341 \ m/s$

The final speed of the rain is calculated as;

$v^2 = u^2 + 2gh\\\\v^2 = (-341)^2 + 2(9.8\times 9000)\\\\v^2 = 292681\\\\v = \sqrt{292681} \\\\v = 541 \ m/s$

Therefore, the final speed of the rain is 541 m/s.

3 0

3 years ago

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