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Salsk061 [2.6K]

3 years ago

8

What is a net force on an object that has a mass of 20.0 kg, an applied force of 100 n moving on a surface with a friction coeff

icient of 0.21?

1 answer:

sergiy2304 [10]3 years ago

5 0

The net force on the object as described is; 58.84N

Two forces acting on the object are;

The <em>applied force and the frictional force.</em>

In essence; the frictional force can be evaluated as;

Frictional force; = coefficient × Weight of object.

Frictional force = 0.21 × 20 × 9.8.

Frictional force = 41.16N

The Net force = Applied force - frictional force

Net force = 100 - 41.16N

Net Force = 58.84 N.

Read more:

brainly.com/question/94428

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This thermal energy flows as heat within the box and floor, ultimately raising the temperature of both of these objects.

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A wire of resistance R is cut into ten equal parts which are then connected in parallel. The equivalent resistance of the combin

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Answer:

<em>The equivalent resistance of the combination is R/100</em>

Explanation:

<u>Electric Resistance</u>

The electric resistance of a wire is directly proportional to its length. If a wire of resistance R is cut into 10 equal parts, then each part has a resistance of R/10.

Parallel connection of resistances: If R1, R2, R3,...., Rn are connected in parallel, the equivalent resistance is calculated as follows:

$\displaystyle \frac{1}{R_e}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+...+\frac{1}{R_n}$

If we have 10 wires of resistance R/10 each and connect them in parallel, the equivalent resistance is:

$\displaystyle \frac{1}{R_e}=\frac{1}{R/10}+\frac{1}{R/10}+\frac{1}{R/10}...+\frac{1}{R/10}$

This sum is repeated 10 times. Operating each term:

$\displaystyle \frac{1}{R_e}=\frac{10}{R}+\frac{10}{R}+\frac{10}{R}+...+\frac{10}{R}$

All the terms have the same denominator, thus:

$\displaystyle \frac{1}{R_e}=10\frac{10}{R}=\frac{100}{R}$

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The equivalent resistance of the combination is R/100

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Which form of energy is involved in weighing fruit on a spring scale?

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Water from a vertical pipe emerges as a 10-cm-diameter cylinder and falls straight down 7.5 m into a bucket. The water exits the

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The diameter of the column of the water as it hits the bucket is 4.04 cm

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By using the kinematics equation to determine the speed of the water in the bucket and applying the equation of continuity to estimate the diameter of the column, we have the following;

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$\mathbf{ 50 = 12.29 \times r_f^2}$

$\mathbf{ r_f= \sqrt{\dfrac{50}{12.29} }}$

$\mathbf{ V_f= 2.02 \ cm }$

Since diameter = 2r;

∴

The diameter of the column of the water is:

= 2(2.02) cm

= 4.04 cm

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3 years ago