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3 years ago

6

Why does the cyclist have less kinetic energy at position A than at position B?

1 answer:

g100num [7]3 years ago

7 0

the answer is A. Their is less friction between the tire and the road at position A than at position B.

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PLeAsE hElp <br> What is the kinetic energy of 14 Kg traveling at a velocity of 3m/s east

harkovskaia [24]

Answer:

Please see the given attachment.

Explanation:

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Thank you.

<h2><em><u>PLEASE</u></em><em><u> </u></em><em><u>MARK</u></em><em><u> </u></em><em><u>ME</u></em><em><u> </u></em><em><u>AS </u></em><em><u>BRAINLEST</u></em><em><u>.</u></em></h2>

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3 years ago

What is the formula to find PE and KE ?

Lapatulllka [165]

PE stands for Potential Energy. It is the stored energy in an object due to its position with respect to some reference. It is expressed in Joules.

P.E = m * g * h OR P.E. = mgh

m - mass of the body
g - acceleration due to gravity
h - height attained due to the body's displacement.

K.E. stands for Kinetic Energy. It is the energy possessed by a body due to its motion.

K.E = 1/2 mv² where m = mass of the body; v = velocity with which the body is moving.

5 0

3 years ago

Water is flowing from a garden hose. A child places his thumb to cover most of those outlet, causing a thin jet of high speed wa

Oxana [17]

Answer: maximum height= 40.8m

Explanation: shown in the attachment.

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3 years ago

Monochromatic light passes through a double slit, producing interference, the distance between the slit centres is 1.2 mm and th

Alik [6]

Answer:

The wavelength of the light is $7200\ \AA$ .

Explanation:

Given that,

Distance between the slit centers d= 1.2 mm

Distance between constructive fringes $\beta= 0.3\ cm$

Distance between fringe and screen D= 5 m

We need to calculate the wavelength

Using formula of width

$\beta=\dfrac{D\lambda}{d}$

Put the value into the formula

$0.3\times10^{-2}=\dfrac{5\times\lambda}{1.2\times10^{-3}}$

$\lambda=\dfrac{0.3\times10^{-2}\times1.2\times10^{-3}}{5}$

$\lambda=7.2\times10^{-7}\ m$

$\lambda=7200\ \AA$

Hence, The wavelength of the light is $7200\ \AA$ .

8 0

3 years ago

A prismatic bar AB of length L and solid circular cross section (diameter d) is loaded by a distributed torque of constant inten

Lyrx [107]

Answer:

a) the maximum shear stress τ $_{max}$ the bar is 16T $_{max}$ /πd³

b) the angle of twist between the ends of the bar is 16tL² / πGd⁴

Explanation:

Given the data in the question, as illustrated in the image below;

d is the diameter of the prismatic bar of length AB

t is the intensity of distributed torque

(a) Determine the maximum shear stress tmax in the bar

Maximum Applied torque T_max = tL

we know that;

shear stress τ = 16T/πd³

where d is the diameter

so

τ $_{max}$ = 16T $_{max}$ /πd³

Therefore, the maximum shear stress τ $_{max}$ the bar is 16T $_{max}$ /πd³

(b) Determine the angle of twist between the ends of the bar.

let theta ( $\theta$ ) be the angle of twist

polar moment of inertia $I_p}$ = πd⁴/32

now from the second image;

lets length dx which is at distance of "x" from "B"

Torque distance x

T(x) = tx

Elemental angle twist = d $\theta$ = T(x)dx / G $I_{p}$

so

d $\theta$ = tx.dx / G(πd⁴/32)

d $\theta$ = 32tx.dx / πGd⁴

so total angle of twist $\theta$ will be;

$\theta$ = $\int\limits^L_0 \, d\theta$

$\theta$ = $\int\limits^L_0 \,$ 32tx.dx / πGd⁴

$\theta$ = 32t / πGd⁴ $\int\limits^L_0 \, xdx$

$\theta$ = 32t / πGd⁴ [ L²/2]

$\theta$ = 16tL² / πGd⁴

Therefore, the angle of twist between the ends of the bar is 16tL² / πGd⁴

7 0

3 years ago