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xeze [42]
3 years ago
8

A point charge of 4.0 µC is placed at a distance of 0.10 m from a hard rubber rod with an electric field of 1.0 × 103 . What is

the electric potential energy of the point charge?
× 10–4 J

What is the electric potential energy of the point charge at 1.3 m?


× 10–3 J
Physics
2 answers:
Afina-wow [57]3 years ago
4 0

The electric potential energy of the point charge is 4.

the electric potential energy of the point charge at 1.3 m is 5.2.

Juliette [100K]3 years ago
4 0

Answer:

Part 1)

U = 4 \times 10^{-4} J

Part 2)

U = 5.2 \times 10^{-3} J

Explanation:

Electric field due to rubber rod is given by

E = 1.0 \times 10^3 N/c

now we also know that

V = E. r

V = (1.0 \times 10^3)(0.10)

V = 1.0 \times 10^2

now potential energy is given as

U = qV

U = (4 \mu C)(1.0 \times 10^2)

U = 4 \times 10^{-4} J

Similarly the electric potential at distance r = 1.3 m

so we have

V = (1.0 \times 10^3)(1.3)

V = 1.3 \times 10^3 Volts

now potential energy is given as

U = (4\mu C)(1.3 \times 10^3)

U = 5.2 \times 10^{-3} J

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kondor19780726 [428]
Define unit vectors as follows:
\hat{i} is in the eastern direction.
\hat{j} is in the northern direction.

The position of the first bird is
\vec{a} = -3.6 \, \hat{i} + 1.8 \, \hat{j}

The position of the second bird is
\vec{b} = - 1.8 \, \hat{i} + 3.6 \, \hat{j}

Let θ = the angle between the net displacement vector for the two birds.
By definition,
\vec{a} . \vec{b} = |a| |b| cos\theta \\\\ \theta = cos^{-1} ( \frac{\vec{a}.\vec{b}}{|a||b|} )

\vec{a}.\vec{b} = (-3.6)(-1.8)+(3.6)(1.8) = 12.96
|a| =  \sqrt{3.24+12.96} =4.025
Similarly,
|b| = 4.025

Therefore
\theta = cos^{-1}  \frac{12.96}{4.025^{2}} =36.9^{o}

Answer:  36.9°
5 0
3 years ago
What is meant by the statement,the linear expansivity of copper is 0.000017k
Aliun [14]

Answer:

The change in length per unit length per degree rise in temperature of copper is 0.000017k

Explanation:

Given that :

The linear expansivity of copper is 0.000017k. This simply means that ; for a given copper length, the length of such copper will increase by 0.000017k for every degree rose in temperature of the copper rod.

Therefore, the change in length per unit length per degree rise in temperature (k) is 0.000017

4 0
3 years ago
1. A 100-kg crate is pulled across a warehouse floor using a rope with a force of 250 N at an angle of 45o from the horizontal.
harkovskaia [24]

Answer:

(a) The net force is 80.394 N

    The acceleration of the crate is 0.804 m/s²

(b) the final velocity of the crate is 5.02 m/s

Explanation:

Given;

mass of the crate, m = 100 kg

applied force, F = 250 N

angle of inclination, θ = 45°

coefficient of friction, μ = 0.12

Applied force in y-direction, F_y = Fsin \theta = 250sin45 = 176.78 \ N

Applied force in x-direction, F_x = Fcos \theta = 250cos45 = 176.78 \ N

The normal force is calculated as;

N + Fy -W = 0

N = W - Fy

N = (100 x 9.8) - 176.78

N = 980 - 176.78 = 803.22 N

The frictional force is given by;

Fk = μN

Fk = 0.12 x 803.22

Fk = 96.386 N

(a) The net force is given by;

F_{net} = F_x - F_k\\\\F_{net} = 176.78-96.386\\\\F_{net} = 80.394 \ N

Apply Newton's second law of  motion;

F = ma

a = \frac{F_{net}}{m}\\\\ a = \frac{80.394}{100}\\\\ a = 0.804 \ m/s^2

(b) the velocity of the crate after 5.0 s

F = ma= \frac{m(v-u)}{t} \\\\Ft =m(v-u)\\\\v-u = \frac{Ft}{m}\\\\ v = \frac{Ft}{m} + u\\\\v = \frac{F_{net}*t}{m} + u\\\\v = \frac{80.394*5}{100} + 1\\\\v = 5.02 \ m/s

7 0
3 years ago
The 5-kg block A has an initial speed of 5 m/s as it slides down the smooth ramp, after which it collides with the stationary bl
Akimi4 [234]

Answer:

The coupled velocity of both the blocks is 1.92 m/s.

Explanation:

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Mass of block A, m_1=5\ kg

Initial speed of block A, u_1=5\ m/s

Mass of block B, m_2=8\ kg

Initial speed of block B, u_2=0

It is mentioned that if the two blocks couple together after collision. We need to find the common velocity immediately after collision. We know that due to coupling, it becomes the case of inelastic collision. Using the conservation of linear momentum. Let V is the coupled velocity of both the blocks. So,

m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{5\times 5+0}{(5+8)}\\\\V=1.92\ m/s

So, the coupled velocity of both the blocks is 1.92 m/s. Hence, this is the required solution.

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an electron, a proton and a deuteron move in a magnetic field with same momentum perpendicularly. the ratio of the radii of thei
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If an electron, a proton, and a deuteron move in a magnetic field with the same momentum perpendicularly, the ratio of the radii of their circular paths will be:

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<h3>How is the ratio of the perpendicular parts obtained?</h3>

To obtain the ratio of the perpendicular parts, one begins bdy noting that the mass of the proton = 1m, the mass of deuteron = 2m, and the mass of the alpha particle  = 4m.

The ratio of the radii of the parts can be obtained by finding the root of the masses and dividing this by the charge. When the coefficients are substituted into the formula, we will have:

r = √m/e : √2m/e : √4m/2e

When resolved, the resulting ratios will be:

1: √2 : 1

Learn more about the radii of their circular paths here:

brainly.com/question/16816166

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6 0
2 years ago
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