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frozen [14]
3 years ago
7

The electric field strength E 0 is measured at a perpendicular distance R from an infinitely large, thin sheet that contains a u

niform positive charge density σ . In terms of the stated properties and the permittivity of free space ε 0 , write an expression for the electric field strength E at a perpendicular distance 2 R from the sheet.
Physics
1 answer:
Brilliant_brown [7]3 years ago
7 0

Answer:

Electric Field Strength E₀ = E₀ = Constant

Explanation:

The Electric Filed Strength E₀ to an Infinite uniformly charge large sheet is constant how far is it i.e. it is independent of the distance away from the uniformly charge sheet.

Formula: E₀ = σ / 2 ε₀  

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What is the force due to gravity of a 38 kg student?
alukav5142 [94]

Answer:

F_g = 372.78 N

Explanation:

Formula for force of gravity is;

F_g = mg

Where;

m is mass

g is acceleration due to gravity

We are given;

Mass = 38 kg

Acceleration due to gravity has a constant value of 9.81 m/s²

Thus;

F_g = 38 × 9.81

F_g = 372.78 N

6 0
3 years ago
Consider the two moving boxcars in Example 5. Car 1 has a mass of m1 = 65000 kg and a velocity of v01 = +0.80 m/s. Car 2 has a m
Amiraneli [1.4K]

Answer:

1.034m/s

Explanation:

We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,

m_1 = 65000kg\\v_1 = 0.8m/s\\m_2 = 92000kg\\v_2 = 1.2m/s

<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,

V_{cm} = \frac{m_1v_1+m_2v_2}{m_1+m_2}

Substituting,

V_{cm} = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

V_{cm} = 1.034m/s

Part B)

For the Part B we need to apply conserving momentum equation, this formula is given by,

m_1v_1+m_2v_2 = (m_1+m_2)v_f

Where here v_f is the velocity after the collision.

v_f = \frac{m_1v_1+m_2v_2}{m_1+m_2}

v_f = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

v_f = 1.034m/s

8 0
3 years ago
What is the angular displacement of a minute hand of a clock after 3 minutes?​
Inessa [10]

Answer:

π/10 rads

Explanation:

It takes an hour (60 minutes) for the minute's hand to turn a full circle or achieve an angular rotation of

2πl rad.

Now, number of periods of 3 minutes in an hour is;

Number of periods = 60/3 = 20 periods

Thus, 3 minutes rotation accounts for 1/20 of 2π the rotation of the minute's hand in an hour.

Thus;

Angular displacement = (1/20) * 2π = π/10 rads

6 0
2 years ago
Newton’s law of universal gravitation a. is equivalent to Kepler’s first law of planetary motion. b. can be used to derive Keple
Finger [1]
Kepler derived his three laws of planetary motion entirely from
observations of the planets and their motions in the sky.

Newton published his law of universal gravitation almost a hundred
years later.  Using some calculus and some analytic geometry, which
any serious sophomore in an engineering college should be able to do,
it can be shown that IF Newton's law of gravitation is correct, then it MUST
lead to Kepler's laws.  Gravity, as Newton described it, must make the planets
in their orbits behave exactly as they do.

This demonstration is a tremendous boost for the work of both Kepler
and Newton.
5 0
3 years ago
Read 2 more answers
Two objects gravitationally attract with a force of 100 N. If the mass of one object is doubled and the mass of the other object
barxatty [35]

Hello!

Recall the equation for gravitational force:
F_g = \frac{Gm_1m_2}{r^2}

Fg = Force of gravity (N)
G = Gravitational constant

m1, m2 = masses of objects (kg)
r = distance between the objects' center of masses (m)

There is a DIRECT relationship between mass and gravitational force.

We are given:
F_g = 100N

If we were to double one mass and triple another, according to the equation:
F'_g = \frac{G(2m_1)(3m_2)}{r^2} = 6(\frac{G(m_1)(m_2)}{r^2}) = 6F_g

Thus:
6 * F_g = 6 * 100 = \boxed{600N}

5 0
3 years ago
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