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3 years ago

A student reaches into a bag of objects.which property of the objects can be observed by using the sense of touch

Physics

2 answers:

matrenka [14]3 years ago

8 0

Id shoot for texture

enot [183]3 years ago

6 0

They can tell the shape of the object by using the sense of touch

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State whether plastic is biodegradable or non-biodegradable ? Give reasons for your answer.

Gnom [1K]

Answer:

non biodegradable

Explanation:

It is non biodegradable because plastic cannot dispose off easily ..

3 0

3 years ago

you push a 51 kg box with a force of 485 N. the friction force on the box is 232 N. calculate the acceleration of the crate.

vichka [17]

Answer:

a = 4.96 m/s²

Explanation:

Given,

The mass of the box, m = 51 Kg

The magnitude of the applied force, Fₐ = 485 N

The friction force on the box, Fₓ = 232 N

The net force acting on the box is,

F = Fₐ - Fₓ

Substituting the given values in the above equation

F = 485 - 232

= 253 N

The acceleration of the crate is given by

a = F/m

= 253 / 51

= 4.96 m/s²

Hence, the acceleration of the crate is, a = 4.96 m/s²

3 0

3 years ago

Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.

8_murik_8 [283]

The electric potential is a scalar unit, so we don't have to struggle with the vectors. The formula that gives electric potential is

$V = \frac{1}{4\pi\epsilon_0}\frac{q}{r}$

1) At point a, the electric potential is the sum of the potentials due to q1 and q2. So,

$V_a = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}$

The distance from the center of the square to one of the corners is $\sqrt2 L/2 = 0.035m$

$V_a = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.035} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.035} = 0$

The answer is zero, because the point charges are at equal distances and their magnitudes are also equal but their directions are opposite.

2) $V_b = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}$

$r_1 = 0.05\sqrt2m\\r_2 = 0.05m$

$V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05\sqrt2} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.05}\\V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05} (\frac{1}{\sqrt2}-1)\\V_b = \frac{1}{4\pi\epsilon_0} (4\times 10^{-5})(-0.29)\\V_b = (-\frac{2.9\times10^{-6}}{\pi\epsilon_0})[tex]3) The work done on q3 by q1 and q2 is equal to the difference between energies. This is the work-energy theorem. So,[tex]W = U_b - U_a$

$U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_3}{r} = Vq_3$

$W = q_3(V_b - V_a) = q_3(V_b - 0)\\W = (-2\times10^{-6})(-\frac{2.9\times10^{-6}}{\pi\epsilon_0})\\W = \frac{5.8\times10^{-12}}{\pi\epsilon_0}$

4 0

3 years ago

What are the dates of the atlantic hurricane season

Bad White [126]

Answer:

Starts on Saturday, June 1

and ends on

Saturday, November 30

Explanation:

6 0

3 years ago

Read 2 more answers

The X-ray source Cygnus X-1 has a mass of at least 11 solar masses and a diameter of only about one-quarter the diameter of the

SOVA2 [1]

Answer:

Answer is It was deduced from the rate at which it glimmers.

Refer below.

Explanation:

The X-ray source Cygnus X-1 has a mass of at least 11 solar masses and a diameter of only about one-quarter the diameter of the Earth. With such a small diameter it must be a compact object, and with such a large mass it can't be a white dwarf or a neutron star, so a black hole is the only possibility remaining. The diameter of Cygnus X-1 found:

It was deduced from the rate at which it glimmers.

7 0

3 years ago