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2 years ago

8

Dao makes a table to identify the variables used in the equations for centripetal acceleration.

2 answers:

LiRa [457]2 years ago

6 0

Answer : The variables used in the equations for centripetal acceleration are, radius and tangential speed.

Explanation :

Centripetal acceleration : It is defined as any object moving in a circular path will have some acceleration vector that pointed towards the center of the circle.

Formula used for centripetal acceleration is,

$a_c=\frac{v^2}{r}$

where,

$a_c$ = centripetal acceleration

r = radius of path

v = velocity or speed of an object or tangential speed

In the equation of centripetal acceleration, the radius and tangential speed are the variables.

Andru [333]2 years ago

3 0

Answer:

X: tangential speed

Y: radius

Explanation:

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A long-distance swimmer is able to swim through still water at 4.0 km/h. She wishes to try to swim from Port Angeles, Washington

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Let $\theta$ be the direction the swimmer must swim relative to east. Then her velocity relative to the water is

$\vec v_{S/W}=\left(4.0\dfrac{\rm km}{\rm h}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)$

The current has velocity vector (relative to the Earth)

$\vec v_{W/E}=\left(3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath$

The swimmer's resultant velocity (her velocity relative to the Earth) is then

$\vec v_{S/E}=\vec v_{S/W}+\vec v_{W/E}$

$\vec v_{S/E}=\left(\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath+\left(4.0\dfrac{\rm km}{\rm h}\right)\sin\theta\,\vec\jmath$

We want the resultant vector to be pointing straight north, which means its horizontal component must be 0:

$\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}=0\implies\cos\theta=-\dfrac{3.0}{4.0}\implies\theta\approx138.59^\circ$

which is approximately 41º west of north.

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1 year ago

A certain part of a flat screen TV has a thickness of 150 nanometers. How<br> many meters is this?

Bess [88]

Answer:

1.5e-7 meters

.00000015 meters

Explanation:

.000000001 meters = 1 nanometer. Multiply that by 150 and an answer is there.

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1 year ago

The electric field 1.5 cm from a very small charged object points toward the object with a magnitude of 180,000 N/C. What is the

Ray Of Light [21]

Answer:

q = 4.5 nC

Explanation:

given,

electric field of small charged object, E = 180000 N/C

distance between them, r = 1.5 cm = 0.015 m

using equation of electric field

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k = 9 x 10⁹ N.m²/C²

q is the charge of the object

$q= \dfrac{Er^2}{k}$

now,

$q= \dfrac{180000\times 0.015^2}{9\times 10^9}$

q = 4.5 x 10⁻⁹ C

q = 4.5 nC

the charge on the object is equal to 4.5 nC

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1 year ago

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The position of a particle moving along the x axis may be determined from the expression x(t) = btu + ctv, where x will be in me

KIM [24]

As per given equation we have

$x = bt^u + ct^v$

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now we can say it should be meter on right side also

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$b*T^8 = M^0L^1T^0$

$b = M^0L^1T^{-8}$

similarly for other term we have

$ct^v = M^0L^1T^0$

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$c = M^0L^1T^{-7}$

<em>so above are the dimensions of b and c</em>

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1 year ago

A glider of mass m1 on a frictionless horizontal track is connected to an object of mass m2 by a massless string. The glider acc

inysia [295]

Answer:

m2g -> T2 -> T1

Explanation:

m2g -> T2 -> T1

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2 years ago