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BlackZzzverrR [31]
3 years ago
6

Two massive particles with identical charge are launched into the uniform field between two plates from the same launch point wi

th the same velocity. They both impact the positively charged plate, but the second one does so four times as far as the first. What sign is the charge? What physical difference would give them different impact points (quantify as a percent)? How does this compare to the gravitational projectile motion case?
Physics
1 answer:
Svetach [21]3 years ago
6 0
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<h2>The interaction between charged object is a non-contact force that acts over some distance of separation.Charge,charge and distance.Every <em>e</em><em>l</em><em>e</em><em>c</em><em>t</em><em>r</em><em>i</em><em>c</em><em>a</em><em>l</em><em> </em><em>i</em><em>n</em><em>t</em><em>e</em><em>r</em><em>a</em><em>c</em><em>t</em><em>i</em><em>o</em><em>n</em><em> </em><em>i</em><em>n</em><em>v</em><em>o</em><em>l</em><em>v</em><em>e</em><em>s</em><em> </em><em>a</em><em> </em><em>f</em><em>o</em><em>r</em><em>c</em><em>e</em><em> </em><em>t</em><em>h</em><em>a</em><em>t</em><em> </em><em>h</em><em>i</em><em>g</em><em>h</em><em>l</em><em>i</em><em>g</em><em>h</em><em>t</em><em>s</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>i</em><em>m</em><em>p</em><em>o</em><em>r</em><em>t</em><em>a</em><em>n</em><em>c</em><em>e</em><em> </em><em>o</em><em>f</em><em> </em><em>t</em><em>h</em><em>e</em><em>s</em><em>e</em><em> </em><em>t</em><em>h</em><em>r</em><em>e</em><em>e</em><em> </em><em>v</em><em>a</em><em>r</em><em>i</em><em>a</em><em>b</em><em>l</em><em>e</em><em>s</em></h2>

er

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Heat transfer from a hot surface to a cool surface in direct contact is called
navik [9.2K]
This is an example of conduction
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3 years ago
In the Biomedical and Physical Sciences building at MSU there are 135 steps from the ground floor to the sixth floor. Each step
satela [25.4K]

Answer:

W = 16.5 Kj

P = 49.9 Watt

E = 16471

Explanation:

m = 73.5kg

t = 5min 30sec = (5×60) + 30 = 330sec

each step = 16.6cm = 0.166m

h = 135×0.166 = 22.41 m

g = 10 m/s²

(i) W = F × s = W × h = mgh

W = 73.5×10×22.41 = 16471.35

W = 16.5 Kj

(ii) Power = workdone/time

P = 16471.35/330

P = 49.9 Watt

(iii) The energy burnt in this process = 16471

4 0
3 years ago
You push a disk-shaped platform tangentially on its edge 2.0 m from the axle. The platform starts at rest and has a rotational a
Natali [406]

Answer: 40.84 m

Explanation:

Given

Radius of the disk, r = 2m

Velocity of the disk, v = 7 rad/s

Acceleration of the disk, α = 0.3 rad/s²

Here, we use the formula for kinematics of rotational motion to solve

2α(θ - θ•) = ω² - ω•²

Where,

ω• = 0

ω = v/r = 7/2

ω = 3.5 rad/s

2 * 0.3(θ - θ•) = 3.5² - 0

0.6(θ - θ•) = 12.25

(θ - θ•) = 12.25 / 0.6

(θ - θ•) = 20.42 rad

Since we have both the angle and it's radius, we can calculate the arc length

s = rθ = 2 * 20.42

s = 40.84 m

Thus, the needed distance is 40.84 m

7 0
2 years ago
Four point charges of magnitudes +3q, -q, +2q, and -4q are arranged in the corners of a square of side length L. The charge -q c
mafiozo [28]

Answer:

d) 0 V

Explanation:

It can be showed that the potential due to a point charge q, to a distance d from the charge, can be expressed as follows:

V = \frac{k*q}{r}

where k = \frac{1}{4*\pi*\epsilon0} = 9e9 N*m2/C2

As the potential is an scalar, and is linear with the charge, we can apply the superposition principle, which means that we can find the potential due to one of the charges, as if the other were not present.

By symmetry, all four charges are at the same distance from the center, so we can write the total potential, as follows:

V = \frac{k}{d} ( q1 + q2 + q3 + q4) (1)

where d, is the semi-diagonal of the square, that we can find applying Pythagorean theorem, as follows:

d = \sqrt{\frac{L^{2}}{4} + \frac{L^{2}}{4} } = L*\frac{\sqrt{2}}{2}

Replacing by the values in (1) we have:

V = \frac{9e9N*m2/C2}{\frac{L}{2}*\sqrt{2} }* ( +3q -q + 2q + -4q)  = 0 V

which is equal to the option d).

6 0
3 years ago
American eels (Anguilla rostrata) are freshwater fish with long, slender bodies that we can treat as a uniform cylinder 1.0 m lo
madam [21]

Answer:

option D

Explanation:

given,

uniform length of cylinder = 1 m

diameter of the cylinder = 10 cm = 0.1 m

Eels have been recorded to spin = 14 rev/s

camera records at = 120 frames per second

time = \dfrac{1}{120}\ s/frame

angle at which eel rotate = ?

ω = 14 rev/s

ω = 14 x 2 π rad/s

ω = 28 π rad/s

angle at which eel rotate

 θ = ω t

θ = 28\pi\times \dfrac{1}{120}

θ = 0.733 rad

θ =0.733 \times \dfrac{180^0}{2\pi}

θ =42^0

Hence, the correct answer is option D

4 0
3 years ago
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