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Kazeer [188]
2 years ago
10

A sleepy student drops a calculator out of a window that's 20.7\text{ m}20.7 m20, point, 7, start text, space, m, end text off t

he ground. We can ignore air resistance.
What is the velocity of the calculator after falling for 1.8\,\text s1.8s1, point, 8, start text, s, end text?
Physics
1 answer:
Ira Lisetskai [31]2 years ago
8 0

Answer:

Explanation:

You can ignore how high the window actually is, unless the calculator falls further.  You just want to use the formula v_f = v_i + at  Let em know if that doesn't help and I can work you through it.

To check the distance use y = v_i*t + .5at^2.  Would just plug in 0*1.8 + .5(-9.8)1.8^2 = -15.876, so it doesn't fall more than 20.7 m so indeed, you don't have to worry about that.  

Again, let me know if you need further help.

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A carousel - a horizontal rotating platform - of radius r is initially at rest, and then begins to accelerate constantly until i
OLga [1]

Answer:

α = (ω²)/8π

Explanation:

The angular acceleration(α) of the carousel can be determined by using rotational kinematics:

ω² =ωo² + 2αθ

Let's make α the subject of this equation ;

ω² - ωo² = 2αθ

α = (ω² −ωo²)/2θ

Now, from the question, since initially at rest, thus, ωo = 0

Also,since 2 revolutions, thus, θ = 2 x 2π = 4π since one revolution is 2π

Plugging in the relevant values to get ;

α = (ω²)/2(4π)

α = (ω²)/8π

7 0
2 years ago
The table below shows data of sprints of animals that traveled 75 meters. At each distance marker, the animals' times were recor
mr_godi [17]

Answer:

Explanation:

Animal 1 because it takes 3s to go 25 meters 3.5s to go 50 meters and 5s to go 75 meters while the others take longer.

5 0
2 years ago
Read 2 more answers
A rod 150 cm long and of diameter 2.0 cm is subjected to an axial pull of 20 kn. if the modulus of elasticity of the material of
Scilla [17]
Given:
rod of circular cross section is subjected to uniaxial tension.
Length, L=1500 mm
radius, r = 10 mm
E=2*10^5 N/mm^2
Force, F=20 kN = 20,000 N
[note: newton (unit) in abbreviation is written in upper case, as in  N ]

From given above, area of cross section = π r^2 = 100 π =314 mm^2
(i) Stress,
σ
=force/area
= 20000 N / 314 mm^2
= 6366.2 N/mm^2
= 6370 N/mm^2 (to 3 significant figures)

(ii) Strain
ε
= ratio of extension / original length
= σ / E
= 6366.2 /(2*10^5)
= 0.03183 
= 0.0318 (to three significant figures)

(iii) elongation
= ε * L
= 0.03183*1500 mm
= 47.746 mm
= 47.7 mm  (to three significant figures)

5 0
2 years ago
For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t
Juliette [100K]

Answer:

24.348mm

Explanation:

NB: I'll be attaching pictures so as to depict missing mathematical expressions or special characters which are not easily found on keyboards

K = d / €^n

Note : d represents the greek alphabet epsilion.

K = 345 / 0.02⁰.²² = 816mPa

The true strain based upon the stress of 414mPa =

€= (€/k)^1/n = (414/816)¹/⁰.²² = 0.04576

However the true relationship between true strain and length is given by

€ = ln(Li/Lo)

Making Li the subject of formula by rearranging,

Li = Lo.e^€

Li = 520e⁰.⁰⁴⁵⁷⁶

Li = 544.348mm

The amount of elongation can be calculated from

Change in L = Li - Lo = 544.348 - 520 change in L = 24.348mm.

8 0
2 years ago
Consider two waves defined by the wave functions y1(x,t)=0.50msin(2π3.00mx+2π4.00st) and y2(x,t)=0.50msin(2π6.00mx−2π4.00st). Wh
guapka [62]

Answer:

They two waves has the same amplitude and frequency but different wavelengths.

Explanation: comparing the wave equation above with the general wave equation

y(x,t) = Asin(2Πft + 2Πx/¶)

Let ¶ be the wavelength

A is the amplitude

f is the frequency

t is the time

They two waves has the same amplitude and frequency but different wavelengths.

4 0
2 years ago
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