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3 years ago

Define Kinetic Energy:

Physics

2 answers:

mrs_skeptik [129]3 years ago

7 0

Answer:

kinetic energy is the energy an object has beause of it's motion

yawa3891 [41]3 years ago

4 0

Answer:

See explanation below

Explanation:

Kinetic energy is energy stored in a moving object.

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a ball rolls from rest down an incline with a uniform acceleration of 4m/s². what is it speed after 8 seconds

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3 years ago

A coil of conducting wire carries a current of i(t) = 14.0 sin(1.15 ✕ 103t), where i is in amperes and t is in seconds. A second

Advocard [28]

Answer:

The peak emf in second coil is 1.876 V

Explanation:

Given :

Inductance $L = 130 \times 10^{-6}$ H

The current $I(t) = 14 \sin(1.15\times 10^{3} t)$

We compare above equation with standard equation,

$I(t) = I_{o} \sin (\omega t + \phi)$

From above equation we have,

$\omega = 10^{3}$ and $\phi = 1.15$

Find the inductive resistance,

$X_{L} = \omega L$

$X_{L} = 10^{3} \times 130 \times 10^{-6}$

$X_{L} = 0.134$

The peak emf in second coil is,

$V = I_{o} X_{L}$

$V = 14 \times 0.134$

$V = 1.876$ V

Therefore, the peak emf in second coil is 1.876 V

8 0

4 years ago

Three point charges are positioned on the x axis. If the charges and corresponding positions are +32 µC at x = 0, +20 µC at x =

Troyanec [42]

Answer:

The magnitude of the electrostatic force on $\mathrm{+32\,\mu C}$ is: $\mathbf{12\;N}$

Explanation:

Consider

$q_1=+32\;\mu\text{C}$ is at position $x_1=0\;\text{m}$
$q_2=+20\;\mu\text{C}$ is at position $x_2=40\;\text{cm}$
$q_3=-60\;\mu\text{C}$ is at position $x_3=60\;\text{cm}$

The sum of all horizontal forces on $q_1$ is given as

$\sum F_x=F_{12}+F_{13}$

where

$F_{12}$ is the force exerted by $q_2$
$F_{13}$ is the force exerted by $q_3$

The force on $q_1$ exerted by $q_2$ is repulsive, so the direction of the force $F_{12}$ is to the left (negative direction). Thus

$F_{12}=-\frac{K\,|q_1|\,|q_2|}{r_{12}^2}\\F_{12}=-\frac{K\,|q_1|\,|q_2|}{(x_2\;-\;x_1)^2}\\F_{12}=\mathrm{-\frac{(8.988\times10^9\;\frac{Nm^2}{C^2})\,|+32\;\mu C|\,|+20\;\mu C|}{(40\;cm\;-\;0\;cm)^2}}\\F_{12}=\mathrm{-\frac{(8.988\times10^9\;\frac{Nm^2}{C^2})\,(32\;\mu C)\,(20\;\mu C)}{(40\;cm)^2}}\\F_{12}=\mathrm{-\frac{(8.988\times10^9\;\frac{Nm^2}{C^2})\,(32\times10^{-6}\;C)\,(20\times10^{-6}\;C)}{(0.40\;m)^2}}\\F_{12}=\mathrm{-36\;N}$

The force on $q_1$ exerted by $q_3$ is attractive, so the direction of the force $F_{13}$ is to the right (positive direction). Thus

$F_{13}=+\frac{K\,|q_1|\,|q_3|}{r_{13}^2}\\F_{13}=+\frac{K\,|q_1|\,|q_3|}{(x_3\;-\;x_1)^2}\\F_{13}=\mathrm{+\frac{(9\times10^9\;\frac{Nm^2}{C^2})\,|+32\;\mu C|\,|-60\;\mu C|}{(60\;cm\;-\;0\;cm)^2}}\\F_{13}=\mathrm{+\frac{(9\times10^9\;\frac{Nm^2}{C^2})\,(32\;\mu C)\,(60\;\mu C)}{(60\;cm)^2}}\\F_{13}=\mathrm{+\frac{(9\times10^9\;\frac{Nm^2}{C^2})\,(32\times10^{-6}\;C)\,(60\times10^{-6}\;C)}{(0.60\;m)^2}}\\F_{13}=\mathrm{+48\;N}$

Therefore

$\sum F_x=\mathrm{-36\;N+48\;N} \;\;\;\;\;\Rightarrow\;\;\;\;\; \mathbf{\sum F_x=+12\;N}$

the electrostatic force on $q_1$ is to the right and has a magnitude of $\mathbf{12\;N}$

8 0

3 years ago