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3 years ago

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Find the minimum acceleration attained on the interval 0<=t<=4 by the particale whos velocity is given by v(t)=t^2-4t^2-3t

+3

1 answer:

anyanavicka [17]3 years ago

8 0

We know that vt = t^2 - 4t^2 - 3t + 3

a = v' = 3t^2 - 8t - 3

a'= 6t - 8 = 0

6t = 8

t = 8/6

hope this helps

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Alberto is walking east at 10 m/s from his house to his grandma's house. It takes him 30s to walk 50 m/s more to his grandma's h

Pachacha [2.7K]

Answer:

Acceleration = 1.33m/s² East

Explanation:

Given the following data;

Initial velocity = 10m/s

Final velocity = 50m/s

Time = 30secs

To find the acceleration;

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.

Mathematically, acceleration is given by the equation;

$Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}$

$a = \frac{v - u}{t}$

Where,

a is acceleration measured in $ms^{-2}$

v and u is final and initial velocity respectively, measured in $ms^{-1}$

t is time measured in seconds.

Substituting into the equation, we have;

$a = \frac{50 - 10}{30}$

$a = \frac{40}{30}$

Acceleration = 1.33m/s² East

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3 years ago

A sphere of radius 5.15 cm and uniform surface charge density +12.1 µC/m2 exerts an electrostatic force of magnitude 35.9 ✕ 10-3

rosijanka [135]

The radius of the sphere is r=5.15 cm=0.0515 m, and its surface is given by
$A=4 \pi r^2 = 4 \pi (0.0515 m)^2 = 0.033 m^2$

So the total charge on the surface of the sphere is, using the charge density
$\rho=+1.21 \mu C/m^2 = +1.21 \cdot 10^{-6} C/m^2$ :
$Q= \rho A = (+1.21 \cdot 10^{-6} C/m^2)(0.033 m^2)=4.03 \cdot 10^{-8}C$

The electrostatic force between the sphere and the point charge is:
$F=k_e \frac{Qq}{r^2}$
where
ke is the Coulomb's constant
Q is the charge on the sphere
$q=+1.75 \muC = +1.75 \cdot 10^{-6}C$ is the point charge
r is their separation

Re-arranging the equation, we can find the separation between the sphere and the point charge:
$r=\sqrt{ \frac{k_e Q q}{F} }= \sqrt{ \frac{(8.99 \cdot 10^9 Nm^2 C^{-2})(4.03 \cdot 10^{-8} C)(1.75 \cdot 10^{-6}C)}{35.9 \cdot 10^{-3}N} }=0.133 m=13.3 cm$

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3 years ago

A lowly high diver pushes off horizontally with a speed of 2.00 m/s from the platform edge 10.0 m above the surface of the water

VladimirAG [237]

Answer:

(a) 1.6m

(b) 6.86m

(c) 2.86s

Explanation:

This is a horizontal shooting problem, so we will use the following equations.

For horizontal distance x we use:

$x(t)=x_{0}+v_{0}t$

Where $x(t)$ is the horizontal distance at a time $t$ , $x_{0}$ is the initial horizontal position which in this case will be zero: $x_{0}=0$ . And $v_{0}$ is the initial speed: $v_{0}=2m/s$

So to solve (a):

(a) At what horizontal distance from the edge is the diver 0.800 s after pushing off?

we have that the time is: $t=0.8s$ so the horizontal distance is:

$x(0.8)=(2m/s)(0.8s)1.6m$

<u>The answer for (a) is 1.6m</u>

Now, to solve (b) we need the equation for vertical distance:

$y(t)=y_{0}-\frac{1}{2}gt^2$

Where $y(t)$ is the vertical distance at a time $t$ , $y_{0}$ is the initial vertical distance: $y_{0}=10m$ And $g$ is the gravitational acceleration: $g=9.81m/s^2$ }

(b) At what vertical distance above the surface of the water is the diver just then?

The time is the same as in the last question: $t=0.8s$

$y(0.8)=10m-\frac{1}{2}(9.81m/s^2)(0.8)^2$

$y(0.8)=10m-\frac{1}{2}(9.81m/s^2)(0.64s^2)$

$y(0.8)=10m-\frac{1}{2}(6.2784m)$

$y(0.8)=10m-(3.1392)$

$y(0.8)=6.86m$

<u>the answer to (b) is 6.86m</u>

(c) At what horizontal distance from the edge does the diver strike the water?

To solve (c) we first need to know how long it took to reach the height of the water, that is, for what time y = 0

Using $y(t)=y_{0}-\frac{1}{2}gt^2$

if $y(t)=0$

$0=10-\frac{1}{2}gt^2\\-10=-\frac{1}{2}gt^2\\20=gt^2\\\frac{20}{g}=t^2\\ \sqrt{\frac{20}{g}}=t$

and since $g=9.81m/s^2$

$t=\sqrt{\frac{20}{9.81} } =\sqrt{2.039}=1.43s$

At $t=1.43s$ the car hits the water, so the horizontal distance at that time is:

$x(t)=x_{0}+v_{0}t$

$x(1.43)=(2m/s)(1.43s)=2.86m$

<u>the answer to (c) is 2.86s</u>

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3 years ago

Two points charge of 4\mu C and 2\mu C are placed at theopposite corners of a rectangle. What is the potential difference Va- Vb

bulgar [2K]

Answer:

$Va-Vb=168KV$

Explanation:

From the question we are told that

Two points charge of 4\mu C and 2\mu C

Generally we find the Va and Vb individually to find there difference

Given a rectangle with two equal sides each,Assume lengths for bot sides

Length L=0.3

Breath B=0.4

Diagonal D= $\sqrt{0.3^2+0.4^2} =0.5$

at opposite sides

Mathematically Va can represented as

$Va =k(\frac{4*10^_-_6}{0.3} +\frac{-2*10^_-_6}{0.5} )$

$Va =9*10^9(\frac{4*10^_-_6}{0.3} +\frac{-2*10^_-_6}{0.5} )$

$Va =9*10^9(0.00001333333-0.000004} )$

$Va =84000V$

$Va =84KV$

Mathematically Vb is represented as

$Va =k(\frac{-4*10^_-_6}{0.3} +\frac{2*10^_-_6}{0.5} )$

$Va =9*10^9(\frac{-4*10^_-_6}{0.3} +\frac{+2*10^_-_6}{0.5} )$

$Va =9*10^9(-0.00001333333+0.000004} )$

$Va =-84000V$

$Va =-84KV$

Therefore

$Va-Vb=84-(-84)\\Va-Vb=84+84\\Va-Vb=168KV$

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Answer:

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