If an object has a volume of 2.5mL and a mass of 10g than what is the density of an object?

How do you graph motion in physics? How do you graph motion in physics? I've seen problems state that an object is in free fall

Dahasolnce [82]

This is Kinematics and the equations in your book.

A speed time graph would plot the speed of something against the teime it was at a speed.

If it were changing it speed constantly, that would be a straight line if acclerating. Total distrance would be the area under the graph.

5 0

3 years ago

An electromagnetic wave moves through electric and magnetic fields. How might the movement of the wave be described?

Elenna [48]

A. The vibrations of the fields are perpendicular to the direction in which the wave moves.

7 0

3 years ago

A 56.0 kg box hangs from a rope. what is the tension in the rope if: the box has vy = 5.30 m/s and is speeding up at 5.10 m/s2 ?

Minchanka [31]

T = tension force in the rope in upward direction

m = mass of the box attached at end of rope = 56 kg

W = weight of the box in downward direction due to gravity

a = acceleration of the box in upward direction = 5.10 m/s²

weight of the box is given as

W = mg

inserting the values

W = (56) (9.8)

W = 548.8 N

force equation for the motion of the box is given as

T - W = ma

inserting the values

T - 548.8 = (56) (5.10)

T = 834.4 N

5 0

2 years ago

A uniformly charged ball of radius a and charge –Q is at the center of a hollowmetal shell with inner radius b and outer radius

vlabodo [156]

Answer:

r < a:

$E = \frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}$

r = a:

$E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}$

a < r < b:

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

r = b:

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

b < r < c:

E = 0

r = c:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}$

r < c:

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

Explanation:

Gauss' Law will be applied to each region to find the E-field.

$\int \vec{E}d\vec{a} = \frac{Q_{encl}}{\epsilon_0}$

An imaginary sphere is drawn with radius r, which is equal to the point where the E-field is asked. The area of this imaginary sphere is multiplied by E, and this is equal to the charge enclosed by this imaginary surface divided by ε0.

r<a:

Since the ball is uniformly charged and not hollow, then the enclosed charge can be found by the following method: If the total ball has a charge -Q and volume V, then the enclosed part of the ball has a charge Q_enc and volume V_enc. Then;

$\frac{Q}{V} = \frac{Q_{encl}}{V_{encl}}\\\frac{Q}{\frac{4}{3}\pi a^3} = \frac{Q_{encl}}{\frac{4}{3}\pi r^3}\\Q_{encl} = \frac{Qr^3}{a^3}$

Applying Gauss' Law:

$E4\pi r^2 = \frac{-Qr^3}{\epsilon_0 a^3}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}\\E = \frac{r}{4\pi a^3}\frac{Q}{\epsilon_0}$

The minus sign determines the direction of the field, which is towards the center.

At r = a:

$E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}$

At a < r < b:

The imaginary surface is drawn between the inner surface of the metal sphere and the smaller ball. In this case the enclosed charge is equal to the total charge of the ball, -Q.

$E4\pi r^2 = \frac{-Q}{\epsilon_0}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

At r = b:

$E = -\frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}$

Again, the minus sign indicates the direction of the field towards the center.

At b < r < c:

The hollow metal sphere has a net charge of +2Q. Since the sphere is a conductor, all of its charges are distributed across its surface. No charge is present within the sphere. The smaller ball has a net charge of -Q, so the inner surface of the metal sphere must possess a net charge of +Q. Since the net charge of the metal sphere is +2Q, then the outer surface of the metal should possess +Q.

Now, the imaginary surface is drawn inside the metal sphere. The total enclosed charge in this region is zero, since the total charge of the inner surface (+Q) and the smaller ball (-Q) is zero. Therefore, the Electric region in this region is zero.

E = 0.

At r < c:

The imaginary surface is drawn outside of the metal sphere. In this case, the enclosed charge is +Q (The metal (+2Q) plus the smaller ball (-Q)).

$E4\pi r^2 = \frac{Q}{\epsilon_0}\\E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

At r = c:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}$

3 0

3 years ago

Why is the battery run flat with reference to the energy transformations that take place

andreev551 [17]

Answer:

Rechargeable batteries can still go flat after repeated use because the materials involved in the reaction lose their ability to charge and regenerate

Explanation:

they go flat because they a rechargeable and sometimes don't always work

thank you hope u have a good day

7 0

2 years ago