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3 years ago

9

If mass1 of a student is 70 kg and mass of the Jupiter is 1.901 x 10^27 kg 1

1 answer:

tamaranim1 [39]3 years ago

6 0

Answer:

lol girly same

Explanation:

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A man 6 feet tall walks at a rate of 6 feet per second away from a light that is 15 feet above the ground.

Tems11 [23]

Answer:(a)10 ft/s

(b)4 ft/s

Explanation:

Given

height of light $=15 feet$

height of man $=6 feet$

$\frac{\mathrm{d} x}{\mathrm{d} t}=6 ft/s$

From diagram

$\frac{15}{y}=\frac{6}{y-x}$

$5(y-x)=2y$

$3y=5x$

differentiate both sides

$3\times \frac{\mathrm{d} y}{\mathrm{d} t}=5\times \frac{\mathrm{d} x}{\mathrm{d} t}$

Tip of shadow is moving at the rate of

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{5}{3}\times 6=10 ft/s$

(b)rate at which length of his shadow is changing

Length of shadow is $y-x$

differentiating w.r.t time

$\frac{\mathrm{d} (y-x)}{\mathrm{d} t}=\frac{\mathrm{d} y}{\mathrm{d} t}-\frac{\mathrm{d} x}{\mathrm{d} t}$

$\frac{\mathrm{d} (y-x)}{\mathrm{d} t}=10-6=4 ft/s$

7 0

3 years ago

A woman exerts a horizontal force of 4 pounds on a box as she pushes it up a ramp that is 10 feet long and inclined at an angle

stealth61 [152]

Answer:

W = 34.64 ft-lbs

Explanation:

given,

Horizontal force = 4 lb

distance of push, d = 10 ft

angle of ramp, θ = 30°

Work done on the box = ?

We know,

W = F.d cos θ

W = 4 x 10 x cos 30°

W = 40 x 0.8660

W = 34.64 ft-lbs

Hence, work done on the box is equal to W = 34.64 ft-lbs

6 0

3 years ago

To test the performance of its tires, a car

velikii [3]

<u>Answer</u>:

The coefficient of static friction between the tires and the road is 1.987

<u>Explanation</u>:

<u>Given</u>:

Radius of the track, r = 516 m

Tangential Acceleration $a_r$ = 3.89 m/s^2

Speed,v = 32.8 m/s

<u>To Find:</u>

The coefficient of static friction between the tires and the road = ?

<u>Solution</u>:

The radial Acceleration is given by,

$a_{R = \frac{v^2}{r}$

$a_{R = \frac{(32.8)^2}{516}$

$a_{R = \frac{(1075.84)}{516}$

$a_{R = 2.085 m/s^2$

Now the total acceleration is

$\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2$

=> $= \sqrt{ (a_r)^2+(a_R)^2}$

=> $\sqrt{ (3.89 )^2+( 2.085)^2}$

=> $\sqrt{ (15.1321)+(4.347)^2}$

=> $19.4791 m/s^2$

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of static friction is

$\mu =\frac{f}{W}$

From (1) and (2)

$\mu =\frac{ma}{mg}$

$\mu =\frac{a}{g}$

Substituting the values, we get

$\mu =\frac{19.4791}{9.8}$

$\mu =1.987$

8 0

3 years ago

How do you get derived units from derived quantities

Tasya [4]

Answer:

A derived quantities is terms of the 7 base quantities via a system of quantity equations which are called SI derived units.

Explanation: there you go:)

4 0

3 years ago

The National Ambient Air Quality Standards (NAAQS) are maximum allowable levels for _____ harmful pollutants. sixteen six sixty

Harman [31]

6

I hope this helps (:

4 0

3 years ago

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