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3 years ago

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A student measures the time it takes for two reactions to be completed reaction eight is completed and 57 seconds and reaction b

e is completed and 48 seconds what can about the rights of these reactions

1 answer:

Maslowich3 years ago

8 0

The answer to this problem is 1200

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A 200-foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevatio

egoroff_w [7]

Answer:

665 ft

Explanation:

Let d be the distance from the person to the monument. Note that d is perpendicular to the monument and would make 2 triangles with the monuments, 1 up and 1 down.

The side length of the up right-triangle knowing the other side is d and the angle of elevation is 13 degrees is

$dtan13^0 = 0.231d$

Similarly, the side length of the down right-triangle knowing the other side is d and the angle of depression is 4 degrees

$dtan4^0 = 0.07d$

Since the 2 sides length above make up the 200 foot monument, their total length is

0.231d + 0.07d = 200

0.301 d = 200

d = 200 / 0.301 = 665 ft

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3 years ago

Examples of increase in pressure due to increase in applied force<br>

Nina [5.8K]

Answer:

injecting

Explanation:

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3 years ago

1. A 100-kg crate is pulled across a warehouse floor using a rope with a force of 250 N at an angle of 45o from the horizontal.

harkovskaia [24]

Answer:

(a) The net force is 80.394 N

The acceleration of the crate is 0.804 m/s²

(b) the final velocity of the crate is 5.02 m/s

Explanation:

Given;

mass of the crate, m = 100 kg

applied force, F = 250 N

angle of inclination, θ = 45°

coefficient of friction, μ = 0.12

Applied force in y-direction, $F_y = Fsin \theta = 250sin45 = 176.78 \ N$

Applied force in x-direction, $F_x = Fcos \theta = 250cos45 = 176.78 \ N$

The normal force is calculated as;

N + Fy -W = 0

N = W - Fy

N = (100 x 9.8) - 176.78

N = 980 - 176.78 = 803.22 N

The frictional force is given by;

Fk = μN

Fk = 0.12 x 803.22

Fk = 96.386 N

(a) The net force is given by;

$F_{net} = F_x - F_k\\\\F_{net} = 176.78-96.386\\\\F_{net} = 80.394 \ N$

Apply Newton's second law of motion;

F = ma

$a = \frac{F_{net}}{m}\\\\ a = \frac{80.394}{100}\\\\ a = 0.804 \ m/s^2$

(b) the velocity of the crate after 5.0 s

$F = ma= \frac{m(v-u)}{t} \\\\Ft =m(v-u)\\\\v-u = \frac{Ft}{m}\\\\ v = \frac{Ft}{m} + u\\\\v = \frac{F_{net}*t}{m} + u\\\\v = \frac{80.394*5}{100} + 1\\\\v = 5.02 \ m/s$

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3 years ago

Weight on planet Mars

Sunny_sXe [5.5K]

Mars: 0.38

weight = mass x surface gravity

multiplying your weight on Earth by the number above will give you your weight on the surface of Mars

If you weigh 150 pounds (68 kg.) on Earth, you would weigh 57 lbs. (26 kg.) on Mars

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The mere exposure effect is

gulaghasi [49]

DescriptionThe mere-exposure effect is a psychological phenomenon by which people tend to develop a preference for things merely because they are familiar with them.

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3 years ago