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Delicious77 [7]

3 years ago

15

A boy is exerting a force of 70 N at a 50-degree angle on a lawnmower. He is accelerating at 1.8 m/s2. Round the answers to the

nearest whole number.
What is the mass of the lawnmower?
What is the normal force exerted on the lawnmower?

1 answer:

Marysya12 [62]3 years ago

5 0

The lawnmower accelerates in the positive horizontal direction, so that the net horizontal force is, by Newton's second law,

(70 N) cos(-50°) = <em>m</em> (1.8 m/s²)

where <em>m</em> is the mass of the lawnmower. Solve for <em>m</em> :

<em>m</em> = ((70 N) cos(-50°)) / (1.8 m/s²)

<em>m</em> ≈ 25 kg

The lawnmower presumably doesn't get lifted off the ground, so that the net vertical force is 0. If <em>n</em> is the magnitude of the normal force, then by Newton's second law,

<em>n</em> - <em>m g</em> + (70 N) sin(-50°) = 0

<em>n</em> = <em>m g</em> + (70 N) sin(50°)

<em>n</em> = (25 kg) (9.8 m/s²) + (70 N) sin(50°)

<em>n</em> ≈ 300 N

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Calculate a rate of cooling down of air from 80 C to 5C Show calculation. Give an answer in cubic meters per minute and cfm.

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Explanation:

Given that,

Rate of cooling of air

Initial temperature= 80°C

Final temperature = 5°C

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Using newton's law of cooling

$\dfrac{dT}{dt}=c(T-T_{0})$

$\dfrac{dT}{dt}=c(\dfrac{T_{1}+T_{2}}{2}-T_{0})$

Where, $dT=T_{1}-T_{2}$

Here, $T =\dfrac{T_{1}+T_{2}}{2}$

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$\dfrac{dT}{dt}=c(\dfrac{T_{1}+T_{2}}{2}-T_{0})$

Put the value into the formula

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A cube 6.0 cm on each side is made of a metal alloy. After you drill a cylindrical hole 2.0 cm in diameter all the way through a

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To solve this problem it is necessary to apply the concepts related to Newton's second law, the definition of density and the geometric relationships that allow us to find the volume of the figures presented.

For the particular case of the Cube with equal sides its volume is determined by

$V_c = l^3$

$V_c = 6^3 = 216cm^3$

In the case of perforated material we have that its volume is given according to the cylindrical geometry, that is to say

$V_d = \pi r^2*l$

$V_d = \pi (\frac{2}{2})^2*6$

$V_d = 6\pi cm^3$

In this way the net volume would be

$\Delta V = V_c-V_d$

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$\Delta V = 197.15cm^3 = 197.15*10^{-6}m^3$

We need to find the mass, but we have the Weight and Gravity so from Newton's second Law

$F= mg$

$m = \frac{F}{g}$

$m = \frac{6.6}{9.8}$

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PART A) From the relation of density as a unit of mass and volume we have to

$\rho = \frac{m}{V}$

$\rho = \frac{0.673}{197.15*10^{-6}}$

$\rho = 3413.64kg/m^3$

PART B) To find the weight of the cube then we apply the ratio of

$W = mg$

$W = V\rho g$

$W = (216*10^{-6})(3413.64)(9.8)$

$W = 7.22N$

3 0

3 years ago