All categories

3 years ago

Please help! I have another part to this too.

Physics

1 answer:

JulsSmile [24]3 years ago

5 0

first one is true, there's no net force acting on it thats greater than another or making it unbalanced, if there was the object would be in some kind of motion

All scientist use meters, that way scientist can share information across country without needing to convert the data.

3. is air resistance

4. The large rock

You might be interested in

The Grinch is trying to push a 5000-kg sled up the side of Mount Crumpit (whose side make an angle of 60 degrees above the horiz

navik [9.2K]

Answer:

Explanation:

height lost h = 300 sin60

= 259.8 m .

gravitational potential energy = mgh

= 5000 x 9.8 x 259.8

= 1273 x 10⁴ J

Kinetic energy

= 1/2 mv²

= .5 x 5000 x 60²

= 900 x 10⁴ J

there is loss of kinetic energy = 373 x 10⁴ J.

This loss in energy is due to kinetic friction that came into action when the sled slipped downwards.

The lost energy is converted into heat energy or thermal energy.

8 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

The tire of your bicycle needs air so you attach a bicycle pump to it and begin to push down on the pump’s handle. If you exert

dusya [7]

Answer:

12.5 J

Explanation:

Force, F = 25 N

Distance, d = 0.5 m

The direction of force and the displacement is same.

Work is defined as the product of force in the direction of displacement and the displacement.

Work = Force x displacement x CosФ

Where, Ф be the angle between force and the displacement

Here, Ф = 0°

So, W = 25 x 0.5 x Cos0°

W = 12.5 J

6 0

3 years ago

Which statement is true about the theory of plate tectonics and the theory of continental drift?:

Rufina [12.5K]

The answer is A I had this question. Before

5 0

3 years ago

Read 2 more answers

The bohr model of the atom addressed the problem of

cupoosta [38]

Rutherford's nuclear model stated that a cloud of negative electrons surround the nucleus however scientists realized that electrons in a cloud around the nucleus of an atom would be attracted to the nucleus, causing the atom to collapse. Thus Bohr's model proposed that electrons were contained in shells and they orbit the nucleus at fixed distances.

7 0

3 years ago