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3 years ago

A cat has a mass of 3 kg and runs at a speed of 6 m/s. how much kinetic energy does the cat have?

1 answer:

4 0

We have: K.E. = 1/2 mv²
Here: m = 3 Kg
v = 6 m/s

Substitute their values,
K.E. = 1/2 * 3 * 6²
K.E. = 1/2 * 3 * 36
K.E. = 54 J

In short, Your Answer would be 54 Joules

Hope this helps!

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Check the dimensional consistencies of s =vot+1/2at2

Leni [432]

Answer:

s is distance so it's dimensions become L.

Nd other side we have ut+1\2at^2.

as 1\2 is a constant it will have dimensions and apply the dimensions to other quantities.

on solving u will get L there also i,e ur LHS = RHS.

thus the equation is dimensionally consistent.

Explanation:

3 0

3 years ago

Un alambre de teléfono de 120m de largo y de 2.2mm de diámetro se estira debido a una fuerza de 380 N cual es el esfuerzo longit

sergij07 [2.7K]

Respuesta: verifique amablemente la explicación

Explicación:

Dado lo siguiente:

Longitud (L) del cable = 120 m

Diámetro (d) = 2,2 mm (2,2 / 1000) = 2,2 * 10 ^ -3 m

Fuerza (F) = 380 N

Esfuerzo longitudinal = Fuerza / Área

Área = πd² / 4 = (π * (2.2 * 10 ^ -3) ^ 2) / 4

Área = (3.142 * 4.84 * 10 ^ -6)

Área = 0.00000380132 m²

Estrés = Fuerza / Área

Estrés = 380 / 0.00000380132

Esfuerzo longitudinal = 99952128.12 = 9.9952128 * 10^7 Nm^-2

Deformación longitudinal: extensión / longitud

Extensión = 0.10 m

Longitud = 120 m

Deformación longitudinal = 0,1 m / 120 m

Deformación longitudinal = 0.0008333 = 8.33 × 10 ^ -4

6 0

2 years ago

Help with this review pls!!

o-na [289]

40 50 60 85 954 746 ity

5 0

2 years ago

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You have an initial velocity of -3.0 m/s. You then experience an acceleration of 2.5 m/s2 for 9.0s; what

Tatiana [17]

27.9 idkkkk look it up on photomath

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3 years ago

81. A 560-g squirrel with a surface area of 930cm2 falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a

Irina18 [472]

Answer:

Explanation:

Given

mass of squirrel $m=560 gm$

Surface area of squirrel $A=930 cm^2$

and the area which face $A_f=\frac{A}{2}=465 cm^2$

height of tree $h=5 m$

Coefficient of drag $C=1$

drag Force $F_d=\frac{1}{2}C\cdot \rho \cdot A\cdot v^2$

Terminal velocity is given

$F_d=mg$

$\frac{1}{2}C\cdot \rho \cdot A\cdot v^2=mg$

$v=\sqrt{\frac{2\times m\times g}{\rho \times C\times A_f}}$

$v=\sqrt{\frac{2\times 0.560\times 9.8}{1.2\times 1\times 465\times 10^{-4}}}$

$v=13.9 m/s$

(b)Mass of person $m=56 kg$

$v^2-u^2=2gh$

$u=0$

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 5}$

$v=9.89 m/s$

7 0

3 years ago

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