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aivan3 [116]
2 years ago
8

If a negative charge is initially at rest in an electric field, will it move toward a region of higher potential or lower potenti

al?What about a positive charge? How does the potential energy of the charge change in each instance? Explain.
Physics
1 answer:
monitta2 years ago
6 0

Answer with explanation :

The negative sign means that the potential energy decreases by the movement of the electron.

negative charge at rest in an electric field moves toward the region of an electric field , so that its potential energy will diminish and change into the kinetic energy of motion. The total energy remains constant.

Positive charges will move downhill because of convention. It is to stay in accordance with other potential theories, particularly gravity, where the "charge" is mass, that moves downwards in the gravitational potential field expressed by ϕ(r)=−GM|r|ϕ(r)=−GM|r|. In an electronic system, howbeit, positive charges are fixed in their position within a component (e.g., a wire), therefore instead of the mobile,the negative charges, electrons, move uphill.

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A rod of mass M = 2.95 kg and length L can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m
madam [21]

Answer:

Explanation:

angular momentum of the putty about the point of rotation

= mvR   where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .

= .045 x 4.23 x 2/3 x .95 cos46

= .0837 units

moment of inertia of rod = ml² / 3 , m is mass of rod and l is length

= 2.95 x .95² / 3

I₁ = .8874 units

moment of inertia of rod + putty

I₁ + mr²

m is mass of putty and r is distance where it sticks

I₂  = .8874 + .045 x (2 x .95 / 3)²

I₂ = .905

Applying conservation of angular momentum

angular momentum of putty = final angular momentum of rod+ putty

.0837 = .905 ω

ω is final angular velocity of rod + putty

ω = .092 rad /s .

4 0
2 years ago
Three equal negative point charges are placed at three of the corners of a square of side d. What is the magnitude of the net el
Rina8888 [55]
<span>this  may help you
As far as the field goes, the two charges opposite each other cancel!
So E = kQ / d² = k * Q / (d/√2)² = 2*k*Q / d² ◄
and since k = 8.99e9N·m²/C²,
E = 1.789e10N·m²/C² * Q / d² </span>
8 0
2 years ago
Read 2 more answers
A policeman starts giving chase 60 seconds after a stolen car zooms by at 108 km/hr. At what minimum speed should he drive if he
Iteru [2.4K]

Answer:

30.93 m/s

Explanation:

Given that, the speed of stolen car is,

v_{s} =108km/hr\\v_{s} =108\times \frac{5}{18}m/s\\ v_{s} =30m/s

As policeman start chasing the stolen car after 60 seconds.

Now suppose the speed of policeman car is, v_{p}

The policeman catches the stolen car at a distance of,

S=60km\\S=60000m

Now the distance covered by the policeman in time t is v_{p}\times t

And the distane cover by the thief in stolen car in time(t+60s) is v_{s}\times (t+60sec).

And these distances are equal and they are equal to 60000 m.

Therefore,

v_{p}\times t=v_{s}\times (t+60sec)=60000m

Therfore,

v_{s}\times (t+60sec)=60000m\\30m/s\times (t+60sec)=60000m\\(t+60s)=2000s\\t=1940s

Now use this value to solve for minimum speed of policeman's car.

v_{p}\times 1940=60000\\v_{p}=30.93 m/s

Therefore minimum speed of policeman's car is 30.93 m/s.

6 0
2 years ago
A 15-cm-tall closed container holds a sample of polluted air containing many spherical particles with a diameter of 2.5 μm and a
TEA [102]

Answer:

t = 224 s

Explanation:

given,

length of container = 15 cm = 0.15 m

diameter of spherical particle = 2.5 μm

mass of particle = 1.9 x 10⁻¹⁴ kg

viscosity of air = μ = 1.18 x 10⁻⁵ kg/m.s

time taken by the particle to stop = ?

radius of particle = 2.5/2 = 1.25 μm

volume of particle = \dfrac{4}{3}\pi r^3

                              =\dfrac{4}{3}\pi (1.25 \times 10^{-6})^3

Density =\dfrac{mass}{volume}

Density =\dfrac{1.9 \times 10^{-14}}{\dfrac{4}{3}\pi (1.25 \times 10^{-6})^3}

ρ = 2322 kg/m³

terminal velocity

v_t = \dfrac{2}{9}\ \dfrac{gR^2(\rho - \rho_{air})}{\mu}

v_t = \dfrac{2}{9}\ \dfrac{9.8 \times (1.25 \times 10^{-6})^2(2322 - 1)}{1.18 \times 10^{-5}}

v_t = 6693 x 10⁻⁷ m/s

t = \dfrac{d}{v_t}

t = \dfrac{0.15}{6693 \times 10^{-7}}

t = 224 s

7 0
2 years ago
6. Starting from 1.5 miles away, a car drives toward a speed checkpoint and then passes it. The car travels at a constant rate o
loris [4]
For the answer to the question above, 

<span>To be 0.1 miles away from the check point ,
 the car has to travel 1.4 miles OR 1.6 miles. </span>


53 miles = 60 minutes 

1.4 miles = 1.4 / 53 X 60 = 1.5849056 minutes OR 95.1 seconds 

<span>1.6 miles = 1.6 /53 X 60 = 1.8113207 minutes OR 108.7 seconds 
</span>So the answer is <span>95.1s and 108.7s
I hope my answer helped you</span>
7 0
2 years ago
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