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Talja [164]
3 years ago
7

A marble with a mass of 1.5 grams rolls at a constant rate at a constant rate of 0.50 cm/s across a table. What is the magnitude

of the marbles's momentum?
Physics
2 answers:
Leno4ka [110]3 years ago
3 0
Ok so momentum is the product of mass and velocity. Its SI unit is Kg m/s

The marble's velocity is 0.5 cm/s and it's mass is 1.5.

Multiply this -> 0.5 * 1.5 = 0.75 g cm/s

Usually everything in Physics is converted into SI units therefore the SI measurement of the marble is 0.00000075 Kg m/s


Llana [10]3 years ago
3 0

Answer:

Momentum, p=7.5\times 10^{-6}\ kg-m/s

Explanation:

It is given that,

Mass of the marble, m = 1.5 gm = 0.0015 kg

Speed of the marble, v = 0.5 cm/s = 0.005 m/s

We need to find the momentum of the marble. It can be given by the product of mass and velocity i.e.

p= m\times v

p= 0.0015\ kg\times 0.005\ m/s

p = 0.0000075 kg-m/s

p=7.5\times 10^{-6}\ kg-m/s

So, the marbles's momentum is 7.5\times 10^{-6}\ kg-m/s. Hence, this is the required solution.

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Both the X- and Y-components of a vector are doubled. Which of the following describes what happens to the resulting Vector?
Aliun [14]

Answer: C. Magnitude increases by 2

Explanation:

7 0
4 years ago
A thin, uniformly charged insulating rod has a linear charge density λ = 3 nC/m and lies along the x axis from x = 1m to x = 3m.
Contact [7]

Answer:

A) V_A = 11.93~V

B) The vector definition of E-field is

\vec{E} = -1.13\^x + 2.41\^y

where magnitude is E = 2.66 N/m.

Explanation:

The potential of a uniformly charged rod can be found by the method of integration. We will first choose an infinitesimal part on the rod. We will compute the potential of this part at point A. Then we will integrate this potential over the entire rod.

We will use the following formula for electric potential:

V = \frac{1}{4\pi \epsilon_0}\frac{Q}{r}

Let us choose the infinitesimal part a distance 'x' from the origin. Then the distance between this point and point A is

r = \sqrt{x^2+4^2}

The infinitesimal length is 'dx', and the potential of this length is dV. Let's apply the formula:

dV = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{\sqrt{x^2 + 4^2}}

Here, the charge Q is equal to the charge density multiplied by the length. Q = λdx

Now we have to integrate this infinitesimal potential over the rod:

V = \int\limits^3_1 {dV} \, dx = \frac{1}{4\pi \epsilon_0}\int\limits^3_1 {\frac{\lambda}{\sqrt{x^2 + 16}} \, dx

By using an integral table, this can be calculated:

V = \frac{3\times 10^{-9}}{4\pi\epsilon_0}\ln(|\sqrt{x^2+16}+x|)\left \{ {{x=3} \atop {x=1}} \right. \\V = 11.93~V

B) The electric field can be found by a similar approach, but a different formula:

\vec{E} = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}\^r

Let's apply this formula to the infinitesimal part we have chosen.

dE_x = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{x^2 + 4^2}\cos(\theta)\\dE_y = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{x^2 + 4^2}\sin(\theta)

By the geometry sine and cosine terms can be found:

\sin(\theta) = \frac{4}{\sqrt{x^2+16}}\\\cos(\theta) = \frac{x}{\sqrt{x^2 + 16}}

The x- and y-components of the E-field can be found separately by integrating the infinitesimal parts over the entire rod.

E_x = \int\limits^3_1 {dE_x} \, dx = \frac{\lambda}{4\pi\epsilon_0}\int\limits^3_1 {\frac{x}{(x^2+16)^{3/2}}} \, dx  = 1.13(-\^x)\\E_y = \int\limits^3_1 {dE_y} \, dx = \frac{4\lambda}{4\pi\epsilon_0}\int\limits^3_1 {\frac{1}{(x^2+16)^{3/2}}} \, dx  = 2.41(\^y)

So, the final E-field is

\vec{E} = -1.13\^x + 2.41\^y

The magnitude of the E-field is

E = 2.66 N/m

6 0
3 years ago
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Answer:

<h2> B. Components</h2>

Explanation:

Splitting vectors into components makes it easier to perform operations on them, typically categorizing vectors into horizontal and vertical components which are the two components makes it very easy to perform operations, since they are similar terms and operations cab be done easily

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