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4 years ago

10

What determines an object’s velocity? A. speed and direction B. speed at a particular instant C. average direction over the dura

tion D. direction at a particular instant

2 answers:

belka [17]4 years ago

5 0

speed and direction is the answer

SIZIF [17.4K]4 years ago

3 0

A.) Speed and distance

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If the third charge (–| q 3 |) is placed at point P, but not held fixed, it will experience a force and accelerate away from the

Mazyrski [523]

Complete Question

The complete question is shown on the first uploaded image

Answer:

A

The potential of this system is $U=6.75*10^{-7}J$

B

The electric potential at point p is $V_p= -900V$

C

The work required is $W= 9*10^{-7}J$

D

The speed of the charge is $v=600m/s$

Explanation:

A sketch to explain the question is shown on the second uploaded image

Generally the potential energy for a system of two charges is mathematically represented as

$U = \frac{kq_1 q_2}{d}$

where k is the electrostatic constant with a value of $k = 9*10^9 N m^2 /C^2$

q is the charge with a value of $q = 1*10^{-9}C$

d is the distance given as $d =5m$

Now we are given that $q_1 = q$ and $q_2 = 3q$ and

Now substituting values

$U = \frac{9*10^9 *1*10^{-9} * 3*10^{-9}}{5}$

$U=6.75*10^{-7}J$

The electric potential at point P is mathematically obtained with the formula

$V_p = V_{-q} + V_{-3q}$

I.e the potential at $q_1$ plus the potential at $q_2$

Now potential at $q_1$ is mathematically represented as

$V_{-q} = \frac{-kq}{s}$

and the potential at $q_2$ is mathematically represented as

$V_{-3q} = \frac{-3kq}{s}$

Now substituting into formula for potential at P

$V_p = \frac{-kq}{s} + \frac{-3kq}{s} = -\frac{4kq}{s}$

$= \frac{4*9*10^9 *1*10^{-9}}{4*10^{-2}}$

$V_p= -900V$

The Workdone to bring the third negative charge is mathematically evaluated as

$W =\Delta U = \frac{kq_1q_3}{s} + \frac{kq_2q_3}{s}$

$= \frac{kq*q}{s} + \frac{kq*3q}{s}$

$= \frac{4kq^2}{s}$

$= \frac{4* 9*10^9 * (1*10^{-9})^2}{4*10^{-2}}$

$W= 9*10^{-7}J$

From the Question are told that the charge $q_3$ would a force and an acceleration which implies that all its potential energy would be converted to kinetic energy.This can be mathematically represented as

$\Delta U = W = \frac{1}{2} m_{q_3} v^2$

$9*10^{-7} = \frac{1}{2} m_{q_3} v^2$

Where $m_{q_3} = 5.0*10^{-12}kg$

Now making v the subject we have

$v = \sqrt{\frac{9*10^{-12}}{5*10^{-12}*0.5} }$

$v=600m/s$

8 0

4 years ago

An airplane flies at a speed of 250 mi/hr, 40 degrees south of east. A wind blows at a speed of 35 mi/hr 30 degrees west of sout

alina1380 [7]

Compute the components of the given vectors. Let $P$ denote the plane's velocity vector, and $W$ the wind. Then

$P=\langle P_x,P_y\rangle=\langle250\cos(-40^\circ),250\sin(-40^\circ)\rangle$

$\implies P=\langle191.5,-160.7\rangle$

$W=\langle35\cos(-120^\circ),35\sin(-120^\circ)\rangle=\langle-17.5,-30.3\rangle$

The resultant velocity (rounded) is

$P+W=\langle174,-191\rangle$

with magnitude $\sqrt{174^2+(-191)^2}=258$ and direction $\tan^{-1}\frac{-191}{174}=-47.7^\circ$ , or about 258 mi/hr at 47.7 degrees south of east.

3 0

3 years ago

Tema [17]

Answer:

B.

1-3hours

Explanation:

This is because a diabetic patients have increased blood sugar or glucose concentration. After eating a meal, the blood glucose concentration will be increased as it is been accumulated . Therefore it is best diabetic patient exercise at that hour to reduce it's blood glucose concentration.

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3 years ago

Michelle recently started selling her invention: A bed that looks like it floats in mid-air. The bed is actually suspended by ma

Tamiku [17]

Answer:

Explanation:

designer

illusionist

engineer

entrepreneur

salesperson

human

inventor

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3 years ago

Two gliders collide on a frictionless air track that is aligned along the x axis. Glider A has an initial velocity of +4.0 m/s a

DedPeter [7]

Answer:

As collision is elastic,thus we can use conservation of momentum equation

mA=0.2 kg

(vB)1=0 m/s.......................as it is on rest before collision

(vA)1=4 m/s

(vA)2=-1 m/s

(vB)2=2 m/s

using equation

(mA*vA+mB*vB)1= (mA*vA+mB*vB)2

Where 1 and 2 represents before and after collision

(0.2*4)+(mB*0)=(0.2*-1)+(mB*2)

0.8=-0.2+(2mB)

mass of object B=mB=0.3 Kg

6 0

3 years ago