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stealth61 [152]

3 years ago

9

According to the Big Bang theory timeline, all matter and energy was compressed into a very small volume at an extremely high te

mperature.
a. True
b. False

1 answer:

IceJOKER [234]3 years ago

3 0

The answer i know is True

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The potential difference across a resistor increases by a factor of 4. How

Anton [14]

Answer:

Correct option is C it decreases by a factor of 2

5 0

3 years ago

Read 2 more answers

Suppose you have two small pith balls that are 5.5 cm apart and have equal charges of -29 nc?

zysi [14]

The question is missing, however, I guess the problem is asking for the value of the force acting between the two balls.

The Coulomb force between the two balls is:
$F= k_e \frac{ q_1 q_2}{r^2}$
where $k_e=8.99\cdot10^9~N m^2 C^{-2}$ is the Coulomb's constant, $q_1=q_2=29~nC=29\cdot 10^{-9}~C$ is the intensity of the two charges, and $r=5.5~cm=0.055~m$ is the distance between them.

Substituting these numbers into the equation, we get
$F=2.5~10^{-3}~N$

The force is repulsive, because the charges have same sign and so they repel each other.

6 0

3 years ago

Bobo, the clown, can swim at 2.0 m/s. he must make a landing directly across to the north side of the styx river, which is 100.

victus00 [196]

We are given the following:

Bobo's swimming speed = 2.0 m/s
Width of the river = 100 m
Flowrate of the river = 6.0 m/s due east

First, we need to illustrate the problem. Draw the river with a width of 100 meters. Then, the flow of the river, east at 6 meters per second. Lastly, draw Bobo at one side of the river facing north and an arrow representing swimming speed at 2 meters per second.

Now, we can use the Pythagorean theorem to solve this rate problem.

c^2 = a^2 + b^2

c = speed of Bobo needed
a = speed of Bobo facing north
b = flow rate of the river going east

c^2 = 2^2 + 6^2

c = 6.32 m / s should be his speed to overcome the current and make a landing at the desired location.

6 0

3 years ago

A basketball weighing 0.63 kg is dropped from a height of 6.0 meters onto a court. Use the conservation of energy equation to de

frutty [35]

Answer:

The velocity of the ball at a height of 2.0 meters above the court is approximately 8.85 m/s

Explanation:

The given parameters of the ball are;

The mass of the ball, m = 0.63 kg

The height from which the ball is dropped, h₁ = 6.0 meters

The height at which the velocity of the ball is sought, h₂ = 2 meters

The initial potential energy of the ball, P.E. = m·g·h₁ = 0.63 × 9.8 × 6.0 = 37.044

The initial potential energy of the ball, P.E.₁ = 37.044 J

The potential energy of the ball, when the ball is at 2 meters above the court, P.E.₂ = m·g·h₂ = 0.63 × 9.8 × 2.0 = 12.348

The potential energy of the ball, when the ball is at 2 meters above the court, P.E.₂ = 12.348 J

From M.E> = P.E. + K.E.

Where;

M.E = The total mechanical energy of the ball = Constant

P.E. = The potential energy of the ball

K.E. = The kinetic energy of the ball

By the conservation of energy principle, we have;

The potential energy lost by the ball = The kinetic energy gained by the ball

The potential energy lost by the ball = P.E.₁ - P.E.₂ = 37.044 - 12.348 = 24.696

The potential energy lost by the ball = 24.696 J

The kinetic energy gained by the ball = 1/2·m·v² = 1/2×0.63×v²

Where;

v = The velocity of the ball

∴ The potential energy lost by the ball at 2.0 meters above the court = 24.696 J = The kinetic energy gained by the ball at 2.0 meters above the court = 1/2×0.63×v²

24.696 J = 1/2×0.63 kg ×v²

v² = 24.696 J / (1/2×0.63 kg) = 78.4 m²/s²

∴ v = √(78.4 m²/s²) = 8.85437744847 m/s

The velocity of the ball at a height of 2.0 meters above the court, v ≈ 8.85 m/s.

7 0

3 years ago

A 25 kg box of textbooks rests on a loading ramp that makes an angle α with the horizontal. The coefficient of kinetic friction

Lunna [17]

Up until the moment the box starts to slip, the static friction is maximized with magnitude <em>f</em>, so that by Newton's second law,

• the net force acting on the box parallel to the ramp is

∑ <em>F</em> = <em>mg</em> sin(<em>α</em>) - <em>f</em> = 0

where <em>mg</em> sin(<em>α</em>) is the magnitude of the parallel component of the box's weight; and

• the net force acting perpendicular to the ramp is

∑ <em>F</em> = <em>n</em> - <em>mg</em> cos(<em>α</em>) = 0

where <em>n</em> is the magnitude of the normal force and <em>mg</em> cos(<em>α</em>) is the magnitude of the perpendicular component of weight.

From the second equation we have

<em>n</em> = <em>mg</em> cos(<em>α</em>)

and <em>f</em> = <em>µn</em> = <em>µmg</em> cos(<em>α</em>), where <em>µ</em> is the coefficient of static friction. Substituting these into the first equation gives us

<em>mg</em> sin(<em>α</em>) = <em>µmg</em> cos(<em>α</em>) ==> <em>µ</em> = tan(<em>α</em>) ==> <em>α</em> = arctan(0.35) ≈ 19.3°

8 0

2 years ago