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OLEGan [10]
2 years ago
15

The Himalayas are the highest mountain in the world (positive)​. grammar question

Physics
1 answer:
andrew-mc [135]2 years ago
7 0

Answer:

positive degree:

No other peak of the Himalayas is as high as Mount everest

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A car accelerates from 10.0 m/s to 30.0 m/s at a rate of 3.00 m/s2. how far does the car travel while accelerating?
Snezhnost [94]
Answer is in attachment.

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3 years ago
Momentum is mass times .
anzhelika [568]

Answer:True

Explanation:

4 0
3 years ago
Read 2 more answers
A mass weighting 48 lbs stretches a spring 6 inches. The mass is in a medium that exerts a viscous resistance of 27 lbs when the
Mademuasel [1]

Answer:

a)

u(t)=0.499ft.e^{-\frac{144.76lb/s}{2(48lb)}t}cos(\omega t)\\\\u(t)=0.499ft.e^{-1.5t}cos(\omega t)

b)

m = 48lb

c)

b = 144.76lb

Explanation:

The general equation of a damping oscillate motion is given by:

u(t)=u_oe^{-\frac{b}{2m}t}cos(\omega t-\alpha)    (1)

uo: initial position

m: mass of the block

b: damping coefficient

w: angular frequency

α: initial phase

a. With the information given in the statement you replace the values of the parameters in (1). But first, you calculate the constant b by using the information about the viscous resistance force:

|F_{vis}|=bv\\\\b=\frac{|F_{vis}|}{v}\\\\|F_{vis}|=27lbs=27*32.17ft.lb/s^2=868.59ft.lb/s^2\\\\b=\frac{868.59}{6}lb/s=144.76lb/s

Then, you obtain by replacing in (1):

6in = 0.499 ft

u(t)=0.499ft.e^{-\frac{144.76lb/s}{2(48lb)}t}cos(\omega t)\\\\u(t)=0.499ft.e^{-1.5t}cos(\omega t)

b.

mass, m = 48lb

c.

b = 144.76 lb/s

8 0
3 years ago
Define wheel and axle 4 example of wheel and axle​
natita [175]

Answer:

A system of two co-axial cylindersof different diameters which rotate together is called wheel and axle example; the door knob , knob of the tap ,screw driver,water tap

5 0
3 years ago
The index of refraction of a transparent substance is 1.7, and the speed of light in air is 3.00 · 108 s. Calculate the speed of
Natalija [7]
Given the index of refraction, n and speed of light in the vacuum, c, we can solve for the speed of light in the transparent substance by the equation below.
                                                n = c / v
where v is our unknown.
Substituting the known values,
                                            1.7 = (3 x 10^8 m/s) / v
The value of v is equal to 1.76 x 10^8 m/s. 
5 0
3 years ago
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