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bazaltina [42]
3 years ago
6

What magnitude charge creates a 1.0 n/c electric field at a point 1.0 m away?

Physics
1 answer:
Stolb23 [73]3 years ago
8 0

Answer:

1.1\cdot 10^{-10}C

Explanation:

The electric field produced by a single point charge is given by:

E=k\frac{q}{r^2}

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have

E = 1.0 N/C (magnitude of the electric field)

r = 1.0 m (distance from the charge)

Solving the equation for q, we find the charge:

q=\frac{Er^2}{k}=\frac{(1.0 N/c)(1.0 m)^2}{9\cdot 10^9 Nm^2c^{-2}}=1.1\cdot 10^{-10}C

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Go with the last one because you need potential energy for kinetic energy
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What is the relationship between temperature and height in the troposphere ,stratosphere,mesophere,thermosphere,exosphere
Soloha48 [4]

In general the higher we go the cooler it becomes . temperature fall continues within the lowermost layer of our atmosphere, known as the troposphere. Above the troposphere, the stratosphere exists and in this region the temperature increases with an increase in altitude. The mesosphere is the part of the atmosphere that lies above the stratosphere. In this region, the temperature falls with an increase in altitude. And finally as we continue rising up, we reach the thermosphere, where the temperature increases with increased altitude.

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4 years ago
If the magnetic force is 3.5 × 10–2 N, how fast is the charge moving?
Sveta_85 [38]

Answer:

The velocity of charge is 1.1×10⁴

Option(D)

Explanation:

F = q(v×B)

3.5×10-²= (8.4×10-⁴)*(v)*(6.7×10-³)*sin35

v = (3.5)/(8.4×6.7×0.57) × 10⁵

v = 3.5/32.08 × 10⁵

v = 0.109 × 10⁵

v ≈ 1.1 × 10⁴ m/s

6 0
3 years ago
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A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g
natali 33 [55]

(a) 5.65 times the Earth's radius

The escape velocity for a projectile on Earth is

v_e=\sqrt{\frac{2GM}{R}}

where

G is the gravitational constant

M is the Earth's mass

R is the Earth's radius

If the projectile has an initial speed of 0.421 escape speed,

v=0.421 v_e

So its initial kinetic energy will be

K=\frac{1}{2}m(0.421 v)^2=0.089 m(\sqrt{\frac{2GM}{R}})^2=0.177 \frac{GMm}{R}

where m is the mass of the projectile

At the point of maximum altitude, all this energy is converted into gravitational potential energy:

K=U\\0.177 \frac{GMm}{R}=\frac{GMm}{r}

where r is the distance from the Earth's centre reached by the projectile. We can write r as a multiple of R, the Earth's radius:0.177 \frac{GMm}{R}=\frac{GMm}{nR}

And solving the equation we find

n=\frac{1}{0.177}=5.65

So, the projectile reaches a radial distance of 5.65 times the Earth's radius.

b) 2.36 times the Earth's radius

The kinetic energy needed to escape is:

K=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}

This time, the projectile has 0.421 times this energy:

K=0.421 \frac{GMm}{R}

Again, at the point of maximum altitude, all this energy will be converted into potential energy:

0.421 \frac{GMm}{R}=\frac{GMm}{nR}

and by solving for n we find

n=\frac{1}{0.421}=2.36

So, the projectile reaches a radial distance of 2.36 times the Earth's radius.

c) E=U=\frac{GMm}{R}

The least initial mechanical energy needed for the projectile to escape Earth is equal to the gravitational potential energy of the projectile at the Earth's surface:

E=U=\frac{GMm}{R}

Indeed, the kinetic energy of the projectile must be equal to this value. In fact, if we use the formula of the escape velocity inside the formula of the kinetic energy, we find

K_e=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}

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Answer:

Since there is only one path for the charges to flow through, the current is the same through each resistor. The equivalent resistance of a set of resistors in a series connection is equal to the algebraic sum of the individual resistances.

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