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3 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]3 years ago

6 0

Answer:

Explanation:

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

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8. A point of Charge +3.0 X 10^-7 Coulomb is placed 2.0 X 10^-2 from a second

Oksi-84 [34.3K]

Use the formula F = (9x10^9 Q * q) / r^2
Message me if you need help.

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4 years ago

A small metal sphere weighs 0.34 N in air and has a volume of 13 cm3 . What is the acceleration of the sphere as it falls throug

Xelga [282]

To solve this problem we will apply the concepts related to the balance of forces. Said balance will be given between buoyancy force and weight, both described as derived from Newton's second law, are given as

Buoyancy force

$F_B = V\rho g$

Here,

V = Volume

$\rho$ =Density of air

g = Acceleration due to gravity

Weight

$F_W = mg$

m = mass

g = Gravity

Our values are given as,

$\text{Weight of the sphere} = W = 0.34 N$

$\text{Volume} = V = 13 cm^3 = 13*10^{-6}m^3$

$\text{density of air} =\rho =1.29kg/m^3$

$\text{gravity}= g = 9.8 m/s^2$

Then,

$F = V\rho g$

Replacing,

$F = (13*10^{-6}m^3 )(1.29Kg/m^3)( 9.8 m/s^2) = 1.6434* 10^{-4} N$

Now net force is ,

$F_{net} = mg - F$

Mass of the sphere is

$m = \frac{W}{g} = \frac{0.34N}{9.8m/s^2} = 0.03469 kg$

Now acceleration of the sphere is

$a = \frac{F_{net}}{m}$

$a = \frac{( 0.34 N)- (1.6434* 10^{-4} N)}{0.03469 kg}$

$a = 9.822m/s^2$

Therefore the acceleration of the sphere as it falls through water is $9.822m/s^2$

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3 years ago

What is the definition of displacement

Savatey [412]

<span>the action of moving something from its place or position.</span>

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3 years ago

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Answer:

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Explanation:

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Answer:

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3 years ago