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3 years ago

6

A toy racer moves clockwise around a circular track of radius 3 m at a constant speed of 6 m/s. When it is at point A its veloci

ty is (6 m/s) i. When it is at point B its velocity and acceleration are ____________.

1 answer:

Llana [10]3 years ago

3 0

Answer:

2 m/s

Explanation:

divide 6 by 3

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A satellite is in a circular orbit around an unknown planet. The satellite has a speed of 1.75 104 m/s, and the radius of the or

Arisa [49]

Answer:

v = 1.32 10² m

Explanation:

In this case we are going to use the universal gravitation equation and Newton's second law

F = G m M / r²

F = m a

In this case the acceleration is centripetal

a = v² / r

The force is given by the gravitational force

G m M / r² = m v² / r

G M/r = v²

Let's calculate the mass of the planet

M = v² r / G

M = (1.75 10⁴)² 5.00 10⁶ / 6.67 10⁻¹¹

M = 2.30 10²¹ kg

With this die we clear the equation to find the orbit of the second satellite

v = √ G M / r

v = √ (6.67 10⁻¹¹ 2.30 10²¹ / 8.75 10⁶)

v = 1.32 10² m

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3 years ago

Plate Tectonics Lab Report The outcome variable (dependent variable): the outcome/dependant variable is Test variable (independe

ANTONII [103]

Answer:Test variable (independent variable): the land of the 6 locations

Outcome variable (dependent variable): the location

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2 years ago

A parallel plate capacitor is created by placing two large square conducting plates of length and width 0.1m facing each other,

borishaifa [10]

Answer:

8.854 pF

Explanation:

side of plate = 0.1 m ,

d = 1 cm = 0.01 m,

V = 5 kV = 5000 V

V' = 1 kV = 1000 V

Let K be the dielectric constant.

So, V' = V / K

K = V / V' = 5000 / 1000 = 5

C = ε0 A / d = 8.854 x 10^-12 x 0.1 x 0.1 / 0.01 = 8.854 x 10^-12 F

C = 8.854 pF

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3 years ago

A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V. What are (a) the total energy stored i

Debora [2.8K]

Answer:

(A) Total energy will be equal to $0.044\times 10^{-5}J$

(b) Energy density will be equal to $0.0175J/m^3$

Explanation:

We have given diameter of the plate d = 2 cm = 0.02 m

So area of the plate $A=\pi r^2=3.14\times 0.02^2=0.001256m^2$

Distance between the plates d = 0.50 mm = $0.50\times 10^{-3}m$

Permitivity of free space $\epsilon _0=8.85\times 10^{-12}F/m$

Potential difference V =200 volt

Capacitance between the plate is equal to $C=\frac{\epsilon _0A}{d}=\frac{8.85\times 10^{-12}\times 0.001256}{0.50\times 10^{-3}}=0.022\times 10^{-9}F$

(a) Total energy stored in the capacitor is equal to

$E=\frac{1}{2}CV^2$

$E=\frac{1}{2}\times 0.022\times 10^{-9}\times 200^2=0.044\times 10^{-5}J$

(b) Volume will be equal to $V=Ad$ , here A is area and d is distance between plates

$V=0.001256\times 0.02=2.512\times 10^{-5}m^3$

So energy density $=\frac{Energy}{volume}=\frac{0.044\times 10^{-5}}{2.512\times 10^{-5}}=0.0175J/m^3$

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A measurement must include both a number and a(an)

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<span>A measurement must include both a number and an unit of measurement. </span>

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