Un objeto de 5 Kg se mueve a 20 m⁄s, que trabajo habrá que realizar para que su velocidad se duplique:

When a carousel is in motion, the movement of the carousel horse can BEST be described as

lara [203]

Kinetic energy because the horse is in motion

7 0

3 years ago

Two equally charged, 1.00 g spheres are placed with 2.00 cm between their centers. when released, each begins to accelerate at 2

Leya [2.2K]

1) Force = m*a = 1.00 g * (1kg / 1000 g) * 225 m/s^2 = 0.225 N

2) Charge

Force = K (charge)^2 /(distance)^2 => charge = √ [Force * distance^2 / k]

k = 9.00 * 10^9 N*m^2 / C^2

charge = √ [0.225 N * (0.02 m)^2 / 9.00* 10^9 N*m^2 / C^2 ]

charge = 0.0000001 C = 0.0001 mili C

3 0

3 years ago

Three collinear forces,F1=45N west,F2=63N east and an unknown force F3 are applied to an object.The resultant force of the three

Greeley [361]

Take east to be the positive direction. Then the resultant force from adding F₁ and F₂ is

F₁ + F₂ = (-45 N) + 63 N = 18 N

which is positive, so it's directed east.

To this we add a third force F₃ such that the resultant is 12 N pointing west, making it negative, so that

18 N + F₃ = -12 N

F₃ = -30 N

So F₃ has a magnitude of 30 N and points west.

6 0

3 years ago

Put the pairs of atoms in order, with the pair that has the biggest electronegativity difference between the two atoms at the to

mojhsa [17]

Electronegativity is the measure of the tendency of an atom to attract a bonding pair of electrons. In the periodic table, electronegativity increase across the period because the charges on the nucleus increase. The correct arrangement for the atoms given above is as follows
Flourine and Francium
Chlorine and Cesium
Nitrogen and Sodium
Phosphorus and Lithium
Nitrogen and Sulphur.

5 0

3 years ago

A long, hollow, cylindrical conductor (inner radius 3.4 mm, outer radius 7.3 mm) carries a current of 36 A distributed uniformly

Elden [556K]

Answer:

a. $B= 9.45 \times10^{-3} T$

b. $B= 0.820 T$

c. $B= 0.0584 T$

Explanation:

First, look at the picture to understand the problem before to solve it.

a. d1 = 1.1 mm

Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:

To solve the equations we have to convert all units to those of the international system. (mm→m)

$B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\$

μ0 is the constant of proportionality

μ0=4πX10^-7 N*s2/c^2

b. d2=3.6 mm

Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:

J: current density

c: outer radius

b: inner radius

The cilinder's current is negative, as it goes on opposite direction than the wire's current.

$J= \frac {-I_{c}}{\pi(c^{2}-b^{2} ) }}$

$J=\frac{-36}{\pi(5.33\times10^{-5}-1.16\times10^{-5}) } =-274.80\times10^{3} A/m^{2}$

$B=\frac{u_{0}(I_{w}-JA_{s})}{2\pi d_{2} } \\A_{s}=\pi (d_{2}^{2}-b^2)=4.40\times10^{-6} m^2\\$

$B=\frac{6.68\times10^{-5}}{8.14\times10^{-5}} =0.820 T$

c. d3=7.4 mm

Here, the point is located out of the cilinder. Therefore, we have to consider both, the conductor's current and the wire's current as follows:

$B=\frac{u_{0}(I_w-I_c)}{2\pi d_3 } =\frac{2.011\times10^-5}{3.441\times10^{-4}} =0.0584 T$

As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.

3 0

3 years ago