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3 years ago

PLEASE HELP

Physics

1 answer:

quester [9]3 years ago

7 0

(a) the electric field generated by a single point charge is radial, and its direction is determined by the sign of the source charge. In particular, when the source charge is positive as in this example, the field lines are directed away from the charge (see attached figure).

(b) The electric field strength at distance r from the source charge is given by
$E=k_e \frac{Q}{r^2}$
where ke is the Coulomb's constant and Q is the source charge. In our problem, the distance is r=2.00 cm=0.02 m, so the electric field at that point is
$E=(8.99 \cdot 10^9 N m^2 C^{-2}) \frac{2.0 \cdot 10^{-5}C}{(0.02 m)^2}=4.5 \cdot 10^8 N/C$

The force experienced by the test charge in this electric field is equal to
$F=qE$
where $q=+5 \mu C=+5 \cdot 10^{-6}C$ is our test charge. Substituting, we find that the force is equal to
$F=qE=(5\cdot 10^{-6}C)(4.5 \cdot 10^8 N/C)=2250 N$

(c) The electric potential at a distance r from the source (positive) charge is given by
$V=k_e \frac{Q}{r}$
and the electric potential energy of a test charge q at distance r from the source charge is
$U=qV=k_e \frac{qQ}{r}$
We can see that when the test charge q is moved closer to the source Q, the distance r decreases, and so the potential energy U of the charge increases. So, the change in potential energy is positive.

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Answer:4.08 N

Explanation:

Given data

superball dropped from a height of 10 cm

Mass of ball $\left ( m\right )$ =0.05kg

time of contact $\left ( t\right )$ =34.3 $\times 10^{-3}$ s

Now we know impulse = $Force\times time\ of\ contact$ =Change in momentum

$F_{average}\times t$ = $m\left ( v-(-v)\right )$

and velocity at the bottom is given by

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$F_{average}\times 34.3\times 10^{-3}$ = $0.05\left ( 1.4-(-1.4)\right )$

$F_{average}$ =4.081N

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3 years ago

Two hypothetical discoveries in Part A deal with moons that, like Earth's moon, are relatively large compared to their planets.

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Answer:

Unusually large moons form in giant impacts, which are relatively rare events

Explanation:

Solution:

- Finding large moons comparable in size to their planets result from impacts of two astro-bodies. The probability of such an event occurring is very rare.

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A rotating object has an angular acceleration of α = 0 rad/s2. Which one or more of the following three statements is consistent

Murrr4er [49]

Answer:

A,B and C

Explanation:

Statement A

At all times, angular velocity is $\omega = 0\,{\rm{rad/s}$

Angular acceleration is the rate of change in angular velocity with respect to time.

Angular velocity and angular acceleration are related by

${\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t$

Which when re-arranged becomes

$\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}$

There’s no change in angular velocity anytime when the angular velocity is $\omega = 0\,{\rm{rad/s}}$

The equation can be modified as follows:

$\begin{array}{c}\\\alpha = \frac{{0\,{\rm{rad/s}} - 0\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}$

Therefore, the angular acceleration becomes zero hence statement A is valid.

Statement B

Angular acceleration is the rate of change in angular velocity with respect to time.

Angular velocity and angular acceleration are related by

${\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t$

Which when re-arranged becomes

$\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}$

There’s no change in angular velocity anytime when the angular velocity is $\omega = 10\,{\rm{rad/s}}$ .The final and initial velocities remain the same.

The equation can be modified as follows:

$\begin{array}{c}\\\alpha = \frac{{10\,{\rm{rad/s}} - 10\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}$

Therefore, the angular acceleration becomes zero and statement B is valid

Statement C

Angular velocity is defined as the change in the angular position with respect to time.

Angular velocity and angular displacement are related by

$\theta = \omega t$

Which can also be modified as:

${\theta _{\rm{f}}} - {\theta _{\rm{i}}}$

Note that the final position is ${\theta _{\rm{f}}}$ and initial position is ${\theta _{\rm{i}}}$

Modifying the equation to find the angular velocity we obtain

$\omega = \frac{{{\theta _{\rm{f}}} - {\theta _{\rm{i}}}}}{t}$

When the angular displacement has the same value at all times, the equation becomes

$\begin{array}{c}\\\omega = \frac{{{\theta _{\rm{i}}} - {\theta _{\rm{i}}}}}{t}\\\\ = 0\\\end{array}$

The angular velocity becomes zero.

Angular acceleration and angular velocity are related by

${\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t$

The expression above can be rearranged as follows:

$\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}$

At all times, the angular velocity is $\omega = 0\,{\rm{rad/s}}$ hence initial and final velocities remain the same

We obtain

$\begin{array}{c}\\\alpha = \frac{{0\,{\rm{rad/s}} - 0\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}$

Therefore, the angular acceleration becomes zero and statement C is valid.

Therefore, statements A,B and C are consistent .

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Two hockey pucks on an ice rink are held together with a compressed spring between them. The pucks are released and the spring p

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Where the m1 and m2 is the mass of two object. U1 and U2 are the initial velocity of hockey pucks. V1 and V2 are the final velocity of Hockey Pucks.

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6 0

3 years ago

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5.00-kg particle starts from the origin at time zero. Its velocity as a function of time is given by v =6t^2 i + 2t j where v is

otez555 [7]

The concept of derivatives and integrals allows to find the results for the questions are the motion of a particle where the speed depends on time are:

a)the position is: r = 2 t³ i + t² j

b) the position of the body on the y-axis is a parabola and on the x-axis it is a cubic function

c) The acceleration is: a = 12 t i + 2 j

d) the force is: F = 60 t i + 10 j

e) the torque is: τ = 40 t³ k^

f) tha angular momentum is: L = 4t³ - 6 t² k^

g) The kinetic energy is: K = 2 m t² (9t² +1)

h) The power is: P = 2m (36 t³ + 2t)

Kinematics studies the movement of bodies, looking for relationships between position, speed and acceleration.

a) They indicate the function of speed.

v = 6 t² i + 2t j

Ask the function of the position. The velocity is defined by the variation of the position with respect to time

v = $\frac{dr}{dt}$

dr = v dt

we substitute and integrate.

∫ dr = ∫ (6 t² i + 2t j) dt

r - 0 = $6 \frac{t^3 }{3} \ \hat i + 2 \frac{t^2}{2 \ \hat j }$

r = 2 t³ i + t² j

b) The position of the body on the y axis is a parabola and on the x axis it is a cubic function.

c) Acceleration is defined as the change in velocity with time.

$a = \frac{dv}{dt}$

a = $\frac{d}{dt} \ (6t^2 i + 2t j)$

a = 12 t i + 2 j

d) Newton's second law states that force is proportional to mass times the body's acceleration.

F = ma

F = m (12 t i + 2 j)

F = 5 12 t i + 2 j

F = 60 t i + 10 j

e) Torque is the vector product of the force and the distance to the origin.

τ = F x r

The easiest way to write these expressions is to solve for the determinant.

$\tau = \left[\begin{array}{ccc}i&j&k\\F_x&F_y&F_z\\x&y&z\end{array}\right]$

$\tau = \left[\begin{array}{ccc}i&j&k\\60t &10&0\\2t^3 &t^2&0\end{array}\right]$

τ = (60t t² - 2t³ 10) k

τ = 40 t³ k ^

f) Angular momentum

L = r x p

L =rx (mv)

L = m (rxv)

The easiest way to write these expressions is to solve for the determinant.

$\left[\begin{array}{ccc}i&j&k\\2t^3 &t^2&0\\6t^2&2t&0\end{array}\right]$

L = (4t³ - 6 t²) k

g) The kinetic energy is

K = ½ m v²

K = ½ m (6 t² i + 2t j) ²

K = m 18 t⁴ + 2t²

K = 2 m t² (9t² +1)

h) Power is work per unit time

P = $\frac{dW}{dt}$ dW / dt

The relationship between work and kinetic energy

W = ΔK

P = $2m \ \frac{d}{dt} ( 9 t^4 + t^2)$

p = 2m (36 t³ + 2t)

In conclusion with the concept of derivatives and integrals we can find the results for the questions are the motion of a particle where the speed depends on time are:

a) The position is: r = 2 t³ i + t² j

b) The position of the body on the y-axis is a parabola and on the x-axis it is a cubic function

c) The acceleration is: a = 12 t i + 2 j

d) The force is: F = 60 t i + 10 j

e) The torque is: τ = 40 t³ k^

f) The angular momentum is: L = 4t³ - 6 t² k^

g) The kinetic energy is: K = 2 m t² (9t² +1)

h) The power is: P = 2m (36 t³ + 2t)

Learn more here: brainly.com/question/11298125

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2 years ago

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