A ball accelerates uniformly at +4.0 m/s2

A dogsled team is shown pulling a man on a sled. Below the picture is a free body diagram with 4 force vectors. The first vector

Scorpion4ik [409]

Answer:

yes

Explanation:

its on edg.

8 0

3 years ago

Read 2 more answers

A standard 1 kilogram weight is a cylinder 41.5 mm in height and 44.0 mm in diameter. what is the density of the material?

Korvikt [17]

Answer;

=15855.40 kg/m^3

Explanation;

Volume (V) of the cylinder = pi x r^2 x h

V = 3.14 x (44/2 x 10^-3)^2 x 41.5 x 10^-3

V = 6.307 x 10^-5 m^3

By density = m/V

mass = 1 kg

density = 1/(6.307 x 10^-5) = 15855.40 kg/m^3

6 0

3 years ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

Distance of the center from each corners $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

Force by each of the charges at the remaining corners:

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

Now, net electric field in the vertical direction:

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

Now, net electric field in the horizontal direction:

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

3 years ago

An electroscope is a fork-shaped device commonly used to detect the presence of charge. The tin leaves of an electroscope will s

velikii [3]

Answer:

All these is caused by the repulsion force.

Explanation:

The electroscope produces a series of electric charges that produce a repulsion force when is putted in contact with a electric charged object.

As the physics law mentions, two different forces are repealed, the electrocospe is charged negatively and the object positively, causing a repulsion force that avoids that both objects touch the other.

7 0

3 years ago

A person observing a flame is interested in the types of electromagnetic waves with slightly shorter and slightly longer wavelen

Nadya [2.5K]

Answer:

Ultra violet rays and infrared rays

Explanation:

The electromagnetic radiation in the order of increasing wavelength is given as

Gamma rays

X rays

Ultra violet rays

Visible radiation

Infrared rays

Microwaves

radiowaves

So, the radiations having wavelength more than the visible radiation are infrared radiations and the radiations which having the wavelength less then the visible radiation is Ultraviolet rays.

So, we observe ultra violet radiations and infrared radiations.

3 0

4 years ago