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4 years ago

Without electrons atoms would carry A. no charge at all. B. a positive charge. C. a negative charge. D. an alternating charge.

Physics

1 answer:

kykrilka [37]4 years ago

5 0

Answer is B A positive charge

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At each corner of a square of side there are point charges of magnitude Q, 2Q, 3Q, and 4Q

Bad White [126]

Answer:

$\displaystyle |F_t|=10.9\ \frac{KQ^2}{l^2}$

$\displaystyle \theta =68^o$

Explanation:

Electrostatic Force

It's the force that appears between two electrical charges q1 q2 when they are placed at a certain distance d. The force can be computed by using the Coulomb's law:

$\displaystyle F=\frac{KQ_1Q_2}{d^2}$

We have an arrangement of 4 charges as shown in the image below. We need to calculate the total force exerted on the charge 2Q by the other 3 charges. The free body diagram is also shown in the second image provided. The total force on 2Q is the vectorial sum of F1, F2, and F3. All the forces are repulsive, since all the charges have the same sign. Let's compute each force as follows:

$\displaystyle |F_1|=\frac{KQ(2Q)}{l^2}=\frac{2KQ^2}{l^2}$

$\displaystyle |F_2|=\frac{K(2Q)(4Q)}{l^2}=\frac{8KQ^2}{l^2}$

The distance between 3Q and 2Q is the diagonal of the rectagle of length l:

$\displaystyle |d_3|=\sqrt{l^2+l^2}=\sqrt{2}\ l$

The force F3 is

$\displaystyle |F_3|=\frac{K(3Q)(2Q)}{(\sqrt{2l)}^2}=\frac{3KQ^2}{l^2}$

Each force must be expressed as vectors. F1 is pointed to the right direction, thus its vertical components is zero

$\displaystyle \vec{F_1}=\left \langle |F_1|,0 \right \rangle=\left \langle \frac{2KQ^2}{l^2},0 \right \rangle$

F2 is pointed upwards and its horizontal component is zero

$\displaystyle \vec{F_2}=\left \langle 0,\frac{8KQ^2}{l^2} \right \rangle$

F3 has two components because it forms an angle of 45° respect to the horizontal, thus

$\displaystyle \vec{F_3}=\left \langle \frac{3KQ^2}{l^2}\ cos45^o,\frac{3KQ2}{l^2} sin45^o\right \rangle$

$\displaystyle \vec{F_3}=\left \langle \frac{3\sqrt{2}KQ^2}{2l^2},\frac{3\sqrt{2}KQ^2}{2l^2}\right \rangle$

Now we compute the total force

$\displaystyle \vec{F_t}=\vec{F_1}+\vec{F_2}+\vec{F_3}$

$\displaystyle \vec{F_t}=\left \langle \frac{2KQ^2}{l^2},0 \right \rangle +\left \langle 0,\frac{8KQ^2}{l^2} \right \rangle + \left \langle \frac{3\sqrt{2}KQ^2}{2l^2},\frac{3\sqrt{2}KQ^2}{2l^2}\right \rangle$

$\displaystyle \vec{F_t}=\left \langle \left(2+\frac{3\sqrt{2}}{2}\right)\frac{KQ^2}{l^2},\left(8+\frac{3\sqrt{2}}{2}\right) \frac{KQ^2}{l^2}\right \rangle$

$\displaystyle F_t=\left \langle 4.121,10.121 \right \rangle \frac{KQ^2}{l^2}$

Now we compute the magnitude

$\boxed{\displaystyle |F_t|=10.9\ \frac{KQ^2}{l^2}}$

The direction of the total force is given by

$\displaystyle tan\theta =\frac{10.121}{4.121}=2.4558$

$\boxed{\displaystyle \theta =68^o}$

6 0

3 years ago

During World War I, the Germans had a gun called Big Bertha that was used to shell Paris. The shell had an initial speed of 1.4

Ksju [112]

Answer:

The shell hit at a horizontal distance 149.41 km

Explanation:

The shell had an initial speed of 1.4 km/s

Initial speed = 1400 m/s

Inclination = 65.8° to the horizontal

First let us find the time of travel

Consider the vertical motion,

We have equation of motion, s = ut + 0.5 at²

Initial velocity, u = 1400 sin 65.8 = 1276.97 m/s

Acceleration, a = -9.81 m/s²

Displacement, s = 0 m

Substituting

s = ut + 0.5 at²

0 = 1276.97 x t + 0.5 x (-9.81) xt²

t = 260.34 s

Now let us find the horizontal displacement,

Consider the horizontal motion,

We have equation of motion, s = ut + 0.5 at²

Initial velocity, u = 1400 cos 65.8 = 573.89 m/s

Acceleration, a = 0 m/s²

Time, t = 260.34 s

Substituting

s = ut + 0.5 at²

s = 573.89 x 260.34 + 0.5 x 0 x 260.34²

s = 149407.11 m = 149.41 km

The shell hit at a horizontal distance 149.41 km

5 0

3 years ago

A 5.0-kg object is pulled along a horizontal surface at a constant speed by a 15-n force acting 20° above the horizontal. How mu

puteri [66]

Answer:

84.6 J

Explanation:

The work done by the force is given by

$W=Fd cos \theta$

where

W is the work done

F = 15 N is the force applied

d = 6.0 m is the displacement

$\theta=20^{\circ}$ is the angle between the force's direction and the displacement

Substituting the numbers into the equation, we find

$W=(15 N)(6.0 m)cos 20^{\circ} =84.6 J$

3 0

3 years ago

Two substances were mixed and there was a color change and small flames. What can you infer happened?

Norma-Jean [14]

A chemical change occurred

Explanation:

There are two types of changes, physical changes and chemical changes:

A chemical change occurs when the nature of a substance changes, and new substances are formed. This occurs when the bonds between molecules in the reactants break, and they re-form into new substance. A change in the colors of the substance involved is, for example, a sign that a chemical change occurred
A physical change occurs when there is a change in state of a substance, but no new substances are formed. For instance, a physical change is the melting of ice, or the boiling of water, or the condensation of water..

In this problem, two substances are mixed and there is a color change and small flames. The color change indicates the formation of new substances, so a chemical change has occurred. Moreover, the presence of small flames indicates that a chemical reaction has occurred (burning of oxygen), so a new substance has formed, so this is another indication of a chemical change.

#LearnwithBrainly

7 0

3 years ago

Read 2 more answers

25 A plank AB 3.0 m long weighing 20 kg and

Eva8 [605]

Answer:

Ahhhhhhhhhgghh ahhhhhhh

3 0

3 years ago