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3 years ago

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__________ force is a fictitious force that some people believe causes you to feel as if you are being pushed outward from the c

enter of a circle while traveling in uniform circular motion.

2 answers:

11Alexandr11 [23.1K]3 years ago

6 0

Centrifugal is the answer, I hope this helped you.

jok3333 [9.3K]3 years ago

5 0

The complete sentence is:
<span>Centrifugal force is a fictitious force that some people believe causes you to feel as if you are being pushed outward from the center of a circle while traveling in uniform circular motion.

In fact, centrifugal force is an inertial force: it is not a real force, but it is due to the fact that the reference frame is rotating. The real force in a uniform circular motion is the centripetal force, which pushes towards the centre of the circle, and keeps the object in circular motion.</span>

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A ball is launched vertically with an initial speed of y˙0= 50 m/s, and its acceleration is governed by y¨=-g-cDy˙2, where the a

stira [4]

Answer:

Explanation:

Given

acceleration is given by

$a=-g-c_Dv^2$

where $\ddot{y}=a$

$\dot{y}=v$

Also acceleration is given by

$a=v\frac{\mathrm{d} v}{\mathrm{d} s}$

$ds=\frac{v}{a}dv$

$\int ds=\int \frac{v}{-g-0.001v^2}dv$

$\Rightarrow Let -g-0.001v^2=t$

$-0.001\times 2vdv=dt$

$vdv=-\frac{dt}{0.002}$

$at\ v_0=50\ m/s,\ t=-g-0.001(50)^2$

$t=-g-2.5$

at $v=0,\ t=-g$

$\int_{0}^{s}ds=\int_{-g}^{-g-2.5}\frac{-dt}{0.002t}$

$\int_{0}^{s}ds=\int^{-g}_{-g-2.5}\frac{dt}{0.002t}$

$s=\frac{1}{0.002}lnt|_{-g}^{-g-2.5}$

$s=\frac{1}{0.002}\ln (\frac{g+2.5}{g})$

$s=113.608\ m$

when air drag is neglected maximum height reached is

$h=\frac{v_0^2}{2g}$

$h=\frac{50^2}{2\times 9.8}$

$h=127.55\ m$

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3 years ago

A 2N and 6N force pull on an object to the right and a 4N force pulls to the left a 0.5kg object. What is the net force on the o

Angelina_Jolie [31]

Explanation:

F net = 2+6-4 ( 2 and 6 N are in same direction so they get added, 4N in opposite direction so it will be subtracted)

F net=4 N

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3 years ago

What is the relationship between mass and weight

sergiy2304 [10]

Mass is how heavy is it, weight is the size both are the same

5 0

3 years ago

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The electric field of a sinusoidal electromagnetic wave obeys the equation E = (375V /m) cos[(1.99× 107rad/m)x + (5.97 × 1015rad

kenny6666 [7]

Answer:

a) v = 2,9992 10⁸ m / s , b) Eo = 375 V / m , B = 1.25 10⁻⁶ T,

c) λ = 3,157 10⁻⁷ m, f = 9.50 10¹⁴ Hz , T = 1.05 10⁻¹⁵ s , UV

Explanation:

In this problem they give us the equation of the traveling wave

E = 375 cos [1.99 10⁷ x + 5.97 10¹⁵ t]

a) what the wave velocity

all waves must meet

v = λ f

In this case, because of an electromagnetic wave, the speed must be the speed of light.

k = 2π / λ

λ = 2π / k

λ = 2π / 1.99 10⁷

λ = 3,157 10⁻⁷ m

w = 2π f

f = w / 2 π

f = 5.97 10¹⁵ / 2π

f = 9.50 10¹⁴ Hz

the wave speed is

v = 3,157 10⁻⁷ 9.50 10¹⁴

v = 2,9992 10⁸ m / s

b) The electric field is

Eo = 375 V / m

to find the magnetic field we use

E / B = c

B = E / c

B = 375 / 2,9992 10⁸

B = 1.25 10⁻⁶ T

c) The period is

T = 1 / f

T = 1 / 9.50 10¹⁴

T = 1.05 10⁻¹⁵ s

the wavelength value is

λ = 3,157 10-7 m (109 nm / 1m) = 315.7 nm

this wavelength corresponds to the ultraviolet

5 0

3 years ago

The length of a certain wire is kept same while its radius is doubled. what is the new resistivity of this wire?

anastassius [24]

The text does not specify whether the resistance R of the wire must be kept the same or not: here I assume R must be kept the same.

The relationship between the resistance and the resistivity of a wire is
$\rho = \frac{AR}{L}$
where
$\rho$ is the resistivity
A is the cross-sectional area
R is the resistance
L is the wire length

the cross-sectional area is given by
$A=\pi r^2$
where r is the radius of the wire. Substituting in the previous equation ,we find
$\rho = \frac{\pi r^2 R}{L}$

For the new wire, the length L is kept the same (L'=L) while the radius is doubled (r'=2r), so the new resistivity is
$\rho' = \frac{\pi r'^2 R}{L'}= \frac{\pi (2r)^2 R}{L}=4 \frac{\pi r^2 R}{L} = 4 \rho$
Therefore, the new resistivity must be 4 times the original one.

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3 years ago

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