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4 years ago

10

Velocity can be either positive or negative, depending on the direction of the displacement. The time interval, Ît, is always po

sitive. Why?

1 answer:

m_a_m_a [10]4 years ago

7 0

Answer:The time interval is always positive quantity because time always moves forward. Time can not be reversed.

Explanation:

Velocity is defined as rate of change of position nothing but the displacement. Displacement can be positive or negative.

In case of time interval, time is always a positive quantity. it is impossible to reverse the time. To obtain any displacement we need to measure the time taken. which has to be positive. It always moves forward.

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You charge an initially uncharged 65.7-mf capacitor through a 39.1-Ï resistor by means of a 9.00-v battery having negligible int

uysha [10]

In a RC-circuit, with the capacitor initially uncharged, when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
$Q(t) = Q_0 (1-e^{-t/\tau})$
where t is the time, $Q_0 = CV$ is the maximum charge on the capacitor at voltage V, and $\tau = RC$ is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.

1) Using $R=39.1 \Omega$ and $C= 65.7 mF=65.7\cdot 10^{-3}F$ , the time constant of the circuit is:
$\tau = RC=(39.1 \Omega)(65.7 \cdot 10^{-3}F)=2.57 s$

2) To find the charge on the capacitor at time $t=1.95 \tau$ , we must find before the maximum charge on the capacitor, which is
$Q_0 = CV=(65.7 \cdot 10^{-3}F)(9 V)=0.59 C$
And then, the charge at time $t=1.95 \tau$ is equal to
$Q(1.95 \tau) = Q_0 (1-e^{-t/\tau})=(0.59 C)(1-e^{-1.95})=0.51 C$

3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be $Q_0 = 0.59 C$ . We can see this also from the previous formule, by using $t=\infty$ :
$Q(t) = Q_0 (1-e^{-\infty})=Q_0(1-0) = 0.59 C$

4 0

3 years ago

Two point charges each carrying a charge of + 4.5 E - 6 C are located 4.5 meters away from each other. How strong is the electro

BARSIC [14]

Answer:

0.009 N, repulsive

Explanation:

The electrostatic force between two electric charges is given by:

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is the separation between the two charges

In this problem, we have

$q_1 =q_2 = +4.5\cdot 10^{-6}C$ are the two charges

r = 4.5 m is their separation

Substituting into the equation, we find

$F=(9\cdot 10^9 Nm^2 C^{-2})\frac{(+4.5\cdot 10^{-6} C)(4.5\cdot 10^{-6} C)}{(4.5 m)^2}=0.009 N$

Moreover, the force is repulsive. In fact, the following rules apply:

- When two charges have same sign, they repel each other

- When two charges have opposite signs, they attract each other

7 0

3 years ago

A meteoroid is in a circular orbit 600 km above the surface of a distant planet. The planet has the same mass as Earth but has a

AVprozaik [17]

<h2>Answer:</h2><h2>The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/ $s^{2}$ </h2>

Explanation:

A meteoroid is in a circular orbit 600 km above the surface of a distant planet.

Mass of the planet = mass of earth = 5.972 x $10^{24}$ Kg

Radius of the earth = 90% of earth radius = 90% 6370 = 5733 km

The acceleration of the meteoroid due to the gravitational force exerted by the planet = ?

By formula, g = $\frac{GM}{r^{2} }$

where g is the acceleration due to the gravity

G is the universal gravitational constant = 6.67 x $10^{-11}$ $m^{3} kg^{-1} s^{-2}$

M is the mass of the planet

r is the radius of the planet

Substituting the values, we get

g = $\frac{(6.67 * 10^{-11}) (5.972 * 10^{24}) }{5733^{2} }$

g = 12.12 m/ $s^{2}$

The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/ $s^{2}$

6 0

4 years ago

A uniform solid disk of mass 5.00 kg and diameter 47.0 cm starts from rest and rolls without slipping down a 40.0 ∘ incline that

saveliy_v [14]

Answer:

The speed it reaches the bottom is

$v=6.51m/s$

Explanation:

Given: $m=5.0kg$ , $r=47cm\frac{1m}{100cm}=0.47m$

Using the conservation of energy theorem

$U_i=K_E+K_{ER}$

$m*g*h=\frac{1}{2}*m*v^2+\frac{1}{2}*I*w^2$

$v=r*w$ , $I=\frac{1}{2}*m*r^2$

$m*g*h=\frac{1}{2}*m*(r*w)^2 +\frac{1}{2}*[\frac{1}{2} *m*r^2]*w^2$

$m*g*h=\frac{3}{4}*m*r^2*w^2$

$g*h=\frac{3}{4}*r^2*w^2$

Solve to w'

$w^2=\frac{4*g*h}{3*r^2}$

$h=x*sin(30)=6.5m*sin(30)=3.25m$

$w=\sqrt{\frac{4*9.8m/s^2*3.25m}{3*(0.235m)^2}}$

$w=27.74rad/s$

$v=27.74rad/s*0.235m=6.51m/s$

7 0

4 years ago

3. Hahn determined that barium (atomic number 56) was one of the elements created when a uranium atom (atomic number 92) split .

creativ13 [48]

According to a periodic table, Krypton was created during the fission of Uranium.

<h3>What is the atomic number?</h3>

<em>Atomic</em> number is a characteristic associated with an element and indicates its number of protons, when a fision occurs, the total number protons is conserved.

Thus, the fission of uranium is led by two elements with <em>atomic</em> numbers 56 and 36. According to a periodic table, those <em>atomic</em> numbers are associated to elements Barium (Ba) and Krypton (Kr), respectively.

According to a periodic table, Krypton was created during the fission of Uranium. $\blacksquare$

To learn more on fission, we kindly invite to check this verified question: brainly.com/question/6572079

8 0

2 years ago