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3 years ago

A load of 45N attached to a spring that is hanging vertically stretches the spring 0.14m. What is the spring constant?

Physics

1 answer:

Vesna [10]3 years ago

6 0

6.3 That Would be the I answer I think but Check on Google For the formula

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The amount of water vapor in the air is known as the _______.

storchak [24]

It is C the humidity

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3 years ago

Read 2 more answers

Which diagram is the best representation of gas molecules in a closed container? A. Diagram A B. Diagram B C. Diagram C D. Diagr

WITCHER [35]

Gas "floats" so if there are examples or pictures it would be the one with the most evenly spread out "dots".

8 0

2 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

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

7 0

3 years ago

A tank with a volume of 0.150 m3 contains 27.0oC helium gas at a pressure of 100 atm. How many balloons can be blown up if each

jekas [21]

Answer:

884 balloons

Explanation:

Assume ideal gas, since temperature is constant, then the product of pressure and volume is constant.

So if pressures reduces from 100 to 1.2, the new volume would be

$V_2 = \frac{P_1V_1}{P_2} = \frac{100*0.15}{1.2} = 12.5 m^2$

The spherical volume of each of the balloon of 30cm diameter (15 cm or 0.15 m in radius) is

$V_b = \frac{4}{3}\pir^3 = \frac{4}{3}\pi 0.15^3 = 0.014 m^3$

The number of balloons that 12.5 m3 can fill in is

$V_2/V_b = 12.5 / 0.014 = 884$

8 0

3 years ago

I need help PLEASE HEEEEEEEEELPP

lana66690 [7]

Answer:

a = (v2 - v1) / t

From A to B (8 - 4) m/s / 1 s = 4 m / s^2

From A to D ( 7 - 4) m/s / 5 s = .6 m / s^2

Note these equations hold for "uniform" values

They say nothing about the acceleration at intermediate points - the equation just says that his average speed increased from 4 m/s to 7 m/s during a 5 sec period

6 0

2 years ago