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3 years ago

TC

Physics

1 answer:

frosja888 [35]3 years ago

6 0

Answer:

0.08 sin 3nt +

metre. Then calculate-

a time period

(d) Time period

(

Minitial phase

wõisplacement

sec.

Explanation:

thats the answer

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You are traveling at 55 mi/h along the x-axis relative to a straight, level road and pass a car that is traveling at 45 mi/h. Th

Alik [6]

Answer:

10 mph

Explanation:

55 - 45 = 10 mph

4 0

2 years ago

Two copper rods are separated by a small gap at B. Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the f

gulaghasi [49]

Answer:

A)3.8196 * $10^{9}$ Newtons B) 2.153 * $10^{9}$ Newtons

Explanation:

Force = Pressure * Area.

Pressure is given as 120GPa

Area = Cross Sectional Area of Rod = Area Of A circle = π $R^{2}$ .

Area Expansivity β = $\frac{Change in Area}{Original Area * Temperature Rise}$

Area Expansivity β = 2α

α = 16.9x10-6/ ° C.

Radius of Rod AB = Half 0f Diameter, 200mm = $\frac{0.2m}{2}$ = 0.1 m

Radius of Rod BC = Half Of Diameter, 150mm = $\frac{0.15}{2}$ = 0.075 m.

Note: To convert from millimeter to meter you divide by 1000.

Area of Rod AB = π * $0.1^{2}$ = 0.0314 $m^{2}$

Area of Rod BC = π * $0.075^{2}$ = 0.0177 $m^{2}$ .

Change in Area = 2α * Original Area * Temperature Rise. Therefore for

Rod AB = 2 * 16.9* $10^{-6}$ * 0.0314 * 30 = 3.183 * $10^{-5}$ $m^{2}$ .

For Rod BC we have = 2 * 16.9* $10^{-6}$ * 0.0177 * 30 = 1.794∈-5 $m^{2}$ .

The forces on each rods will be given by.

Force AB = Pressure * Area of Rod AB

= 120GPa * 3.183 * $10^{-5}$ gives 3.8196 * $10^{9}$ Newtons.

Force BC = Pressure * Area of Rod BC

= 120GPa * 1.794 * $10^{-5}$ gives 2.153 * $10^{9}$ Newtons

7 0

3 years ago

b. A force 100N makes an angle of Ø with the x axis and has a y component of 30 N. Find both the x and y component of the force

horsena [70]

Answer:

Explanation:

The y component of the force is 100 sinØ . But given that y component is 30N

so 100 sinØ = 30

sinØ = 0.3

Ø = 17.5°.

X component of force = 100 cosØ

= 100 cos17.5

= 95.35 N .

Y component of force = 30 N .

Angle Ø = 17.5°.

6 0

3 years ago

ANSWER THE FOLLOWING QUESTION AND MAKE SURE TO GIVE A FULL DESCRIPTION TO HOW YOU GOT YOUR ANSWER (EX. make sure to use blank ru

vazorg [7]

Here is how I did it

6 0

3 years ago

during a lunar mission, it is necessary to increase the speed of a spacecraft by 2.2 m/s when it is moving at 400 m/s relative t

wel

The initial mass fraction of the spacecraft that must be burned and ejected to achieve an increase in speed is 0,00219 m/s

<h3>What fraction of the initial mass of the spacecraft?</h3>

Increase the speed: Vf-Vi = 2.2 m/s

Speed of aircraft: Vr = 400 m/s

Speed of ejected products: Vrel = 1000 m/s

The answer is:

$V_f - V_i = V_{rel} log_{e} (\frac{mi}{mf})\\\\2.2 = 1000 log_{e} (\frac{mi}{mf})\\\\ log_{e} (\frac{mi}{mf} ) = \frac{2.2}{1000} \\\\ log_{e} (\frac{mi}{mf} ) = 0.0022$

$\frac{mi}{mf} = e^{0,0022} \\\\\frac{mf}{mi} = e^{-0,0022} \\\\\frac{mi-mf}{mi} = 1 - e^{-0,0022} = 0,00219$

So, the initial mass fraction of the spacecraft that must be burned and ejected to achieve an increase in speed is 0,00219 m/s

Learn more about spaceship speed fraction brainly.com/question/28256735

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3 0

1 year ago