What does the phrase “constant velocity” indicate?

Two identical metal spheres a and b are connected by a plastic rod. both are initially neutral. 1.0×1012 electrons are added to

Ann [662]

The plastic rod is made of insulator (plastic), so it does not allow charges moving from one sphere to another. This means that all the electrons given to sphere A will remain on sphere A.

The number of electrons initially given to sphere A is $N=1.0 \cdot 10^{12}$ , and since the charge of 1 electron is $e=-1.6 \cdot 10^{-19} C$ , the net charge left on sphere A after the removal of the rod will be
$Q=N e = 1.0 \cdot 10^{12}(-1.6 \cdot 10^{-19} C)=-1.6.0 \cdot 10^{-7} C$

8 0

3 years ago

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A solar system has the five planets shown below. The mass of each planet is proportion

slava [35]

Answer:

Planet C

Explanation:

The figure of the problem is missing: find it in attachment.

The magnitude of the gravitational force between two objects is given by the equation:

$F=\frac{Gm_1 m_2}{r^2}$

where

G is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between the objects

In this problem, we have four planets around planet X, and the mass of each planet is proportional to its size in the figure.

As we can see from the previous equation, the magnitude of the gravitational force is proportional to the mass of the planets: therefore, the planet with largest mass will exert the largest gravitational force on planet X.

From the figure, we see that planet C has the largest size, so the largest mass: therefore, planet C exerts the greatest gravitational force on planet X.

6 0

4 years ago

The position of a particle moving along the x axis depends on the time according to the equation x = ct² - bt³, where x is in me

Kamila [148]

Answer:

(a) $\frac{m}{s^2}$

(b) $\frac{m}{s^3}$

(c) 1 s

(d) 20 m

(e) 1 m

(f) $0\frac{m}{s}$

(g) $-12\frac{m}{s}$

(h) $-36\frac{m}{s}$

(i) $-72\frac{m}{s}$

(j) $-6\frac{m}{s^2}$

(k) $-18\frac{m}{s^2}$

(l) $-30\frac{m}{s^2}$

(m) $-42\frac{m}{s^2}$

Explanation:

Since x is measured in meters and t in seconds, constants a and b must have units that gives meters when multiplied by square and cubic seconds respectivly, so that would mean $\frac{m}{s^2}$ for a and $\frac{m}{s^3}$ for b.

We can get the velocity v equation by deriving the position with respect to t, which gives:

$v=6*t-6*t^2$

And the acceleration a equation by deriving again:

$a=6-12*t$

Now for getting the maximun position between 0 and 4, we must find to points where the positions first derivate is equal to cero and evaluate those points. That is v=0, which gives

$6*t-6*t^2=0\\6t*(1-t)=0\\t=0 or t=1$

For t = 0, x = 0 so the maximun position is archieved at 1 second, which gives x = 1 meter.

For obtaining it's displacement r, we can integrate the velocity from 0 seconds to 4 seconds, which gives the mean value of the position in that interval:

$r=\frac{1}{4}*(2*4^3-3*4^2)m\\r= 20m$

For the remaining questions, we just replace the values of t on the respective equations.

8 0

4 years ago

The moon Phobos orbits Mars

shepuryov [24]

$27.9816 \times 10^{3} s$ is the period of orbit.

Explanation: 

The equation that is useful in describing satellites motion is Newton form after Kepler's Third Law. The period of the satellite (T) and the average distance to the central body (R) are related as the following equation:

$\frac{T^{2}}{R^{3}}=\frac{4 \times \pi^{2}}{G \times M_{c e n t r a l}}$

Where,

T is the period of the orbit

R is the average radius of orbit

G is gravitational constant $6.673 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}$

Here, given data

$M=6.23 \times 10^{23} \mathrm{kg}$

$R=9.38 \times 10^{6} \mathrm{m}$

Substitute the given values, we get T as

$\frac{T^{2}}{\left(9.38 \times 10^{6}\right)^{3}}=\frac{4 \times(3.14)^{2}}{\left(6.673 \times 10^{-11}\right) \times 6.23 \times 10^{23}}$

$T^{2}=\frac{4 \times 9.8596 \times 825.29 \times 10^{18}}{41.57 \times 10^{12}}$

$T^{2}=\frac{32548.12 \times 10^{18-12}}{41.57}=782.97 \times 10^{6}$

Taking square root, we get

$T=27.9816 \times 10^{3} s$

4 0

3 years ago

An object is in circular motion how will the object behave if the centripetal force is removed

Monica [59]

If the force is removed, then there are no other forces acting on the object, the object will continue to move at constant velocity, meaning that it would move in a line tangent to the circular path.

8 0

4 years ago