All categories

3 years ago

50 points!!! Kinetics

Physics

2 answers:

enyata [817]3 years ago

7 0

Answer:

h = 98m

Explanation:

The ball follows a projectile motion.

It it's at the same height at t = 2 and t =10,

It was at a max height at:

t = (2+10)/2

= 12/2

= 6s

Conventional: upwards positive

s = ut + ½at²

s = ut + ½(-9.8)t²

s = ut - 4.9t²

s is equal at t = 2 and t = 10

2u - 4.9(2)² = 10u - 4.9(10)²

2u - 19.6 = 10u - 490

8u = 470.4

u = 58.8

at t = 2

s = 58.8(2) - 4.9(2)²

s = 98 m

iris [78.8K]3 years ago

3 0

Answer:

98 m √

Explanation:

How about s = Vo * t + ½at² ?

s = h = Vo * 2s - 4.9m/s² * (2s)² = 2Vo - 19.6

and

h = Vo * 10s - 4.9m/s² * (10s)² = 10Vo - 490

Subtract 2nd from first:

0 = -8Vo + 470.4

Vo = 58.8 m/s

h = 58.8m/s * 2s - 4.9m/s² * (2s)² = 98 m

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When a potential difference of 154 V is applied to the plates of a parallel-plate capacitor, the plates carry a surface charge d

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Answer:

$4.7\mu m$

Explanation:

We are given that

Potential difference=V=154 V

Surface charge density= $\sigma=29nC/m^2=29\times 10^{-5} C/m^2$

Using $1 nC/cm^2=10^{-5} C/m^2$

We know that

$C=\frac{\epsilon_0A}{d}=\frac{Q}{V}=\frac{\sigma A}{V}$

$d=\frac{\epsilon_0V}{\sigma}$

Where

$\epsilon_0=8.85\times 10^{-12}C^2/Nm^2$

Using the formula

$d=\frac{8.85\times 10^{-12}\times 154}{29\times 10^{-5}}$

$d=4.7\times 10^{-6} m=4.7\mu m$

Using $1\mu m=10^{-6} m$

Hence, the spacing between the plates= $4.7\mu m$

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If 478 watts of power are used in 14 seconds,how much work was done

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Answer:

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Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00

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Answer:

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$1979.99974\ N/C$

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R = Radius = 4 cm

Electric field is given by

$E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C$

Volume charge density is given by

$\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3$

The volume charge density for the sphere is $2.62898\times 10^{-6}\ C/m^3$

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The magnitude of the electric field is $1979.99974\ N/C$

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