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3 years ago

14

If a group of workers can apply a force of 1,275 newtons to move a crate 26 meters, what amount of work will they have accomplis

hed

1 answer:

JulijaS [17]3 years ago

8 0

Work=applied Force x distance

= 1275 x 26
=33150 Joules

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Argon makes up 0.93% by volume of air. Calculate its solubility (in mol/L) in water at 20°C and 1.0 atm. The Henry's law constan

vladimir2022 [97]

Answer:

S= 1.40x10⁻⁵mol/L

Explanation:

The Henry's Law is given by the next expression:

$S = k_{H} \cdot p$ (1)

<em>where S: is the solubility or concentration of Ar in water, $k_{H}$ : is Henry's law constant and p: is the pressure of the Ar </em>

<u>Since the argon is 0.93%, we need to multiply the equation (1) by this percent:</u>

$S = 1.5 \cdot 10^{-3} \frac{mol}{L\cdot atm} \cdot 1.0atm \cdot \frac{0.93}{100} = 1.40 \cdot 10^{-5} \frac{mol}{L}$

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3 years ago

Discuss how friction is reduced in oce skating

belka [17]

Answer:

below

Explanation:

Ice melts, meaning it has a watery layer upon its surface. This allows things to be moving like they are on a liquid but it has the solidity of a solid. The thin metal of the ice skates also decrease the surface area meaning it exerts more force but in turn, it allows you to move faster and further reduces friction.

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2 years ago

HELP ASAP!!!! WILL GIVE BRAINLIEST!

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A planet exerts a gravitational force of magnitude 9e22 N on a star. If the planet were 2 times closer to the star (that is, if

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To solve this problem we will use the related concepts in Newtonian laws that describe the force of gravitational attraction. We will use the given value and then we will obtain the proportion of the new force depending on the Radius. From there we will observe how much the force of attraction increases in the new distance.

Planet gravitational force

$F_p = 6*10^{22}N$

$F_p = \frac{GMm}{R^2}$

$F_p = 9*10^{22}N$

Distance between planet and star

$r = \frac{R}{2}$

Gravitational force is

$F = \frac{GMm}{r^2}$

Applying the new distance,

$F = \frac{GMm}{(\frac{R}{2})^2}$

$F = 4\frac{GMm}{R^2}$

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3 years ago