Between the ball and the player’s head, there are forces. Which of Newton’s laws does this represent? Support your choice.

g Two long parallel wires are a center-to-center distance of 2.50 cm apart and carry equal anti-parallel currents of 2.70 A. Fin

schepotkina [342]

Answer:

864 mT

Explanation:

The magnetic field due to a long straight wire B = μ₀i/2πR where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, i = current in wire, and R = distance from center of wire to point of magnetic field.

The magnitude of magnetic field due to the first wire carrying current i = 2.70 A at distance R which is mid-point between the wires is B = μ₀i/2πR.

Since the other wire also carries the same current at distance R, the magnitude of the magnetic field is B = μ₀i/2πR.

The resultant magnetic field at B is B' = B + B = 2B = 2(μ₀i/2πR) = μ₀i/πR

Now R = 2.50 cm/2 = 1.25 cm = 1.25 × 10⁻² m and i = 2.70 A.

Substituting these into B' = μ₀i/πR, we have

B' = 4π × 10⁻⁷ H/m × 2.70 A/π(1.25 × 10⁻² m)

B = 10.8/1.25 × 10⁻⁵ T

B = 8.64 × 10⁻⁵ T

B = 864 × 10⁻³ T

B = 864 mT

4 0

3 years ago

These are equal and opposite forces that do not cause a change in position or motion.

inysia [295]

Answer:

Balanced forces.

Explanation:

Forces that are opposite in direction, and, equal in size do not cause a change in motion or position are known as balanced forces. When it is applied to an object at rest, the object will not show any movement.

A good example of a balanced force is when we are trying to push a wall, the wall will push back with an opposite but equal force, so, this time neither the wall nor you will move.

5 0

4 years ago

An automobile travels for 90 minutes along a highway. The graph shows the progress of the automobile, with its start point desig

bagirrra123 [75]

i think its 20 km/h but man, how would i know

3 0

3 years ago

The frequency of a light wave in air is 4.6 × 10¹⁴ Hz. What is the wavelength of this wave after it enters a pool of water?

Mashutka [201]

Answer:

490 nm

Explanation:

Wave length: This is the distance of a complete cycle covered by a wave.

The expression for frequency, wave length and speed is given as,

c = λf.................. Equation 1

Where c = speed of light in water, λ = wave length of light, f = frequency of light.

Also,

n = v/c............. Equation 2

Where n = refractive index of water, v = speed of sound in air.

Making c the subject of the equation,

c = v/n.............. Equation 3

Substitute equation 3 into equation 2

v/n = λf

make λ the subject of the equation

λ = v/nf............... Equation 4

Given: f = 4.6 × 10¹⁴ Hz

Constant: n = 1.33, v = 3.0×10⁸ m/s.

Substitute into equation 4

λ = 3.0×10⁸ /(1.33×4.6 × 10¹⁴ )

λ = 0.49×10⁻⁶ m

λ = 490 nm.

Hence the right option is 490 nm

3 0

3 years ago

A spherical conductor has a radius of 14.0 cm and a charge of 42.0 µC. Calculate the electric field and the electric potential a

Yakvenalex [24]

Answer:

0 MN/C, 2.697 MV

4.1953 MN/C, 1.2586 MV

19.2642857143 MN/C, 2.697 MV

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Electric field at r = 8 cm

E = 0 (inside)

Electric potential is given by

$V=\dfrac{kq}{R}\\\Rightarrow V=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.14}\\\Rightarrow V=2697000\ V=2.697\ MV$

Electric potential is 2.697 MV

r = 30 cm

$E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.3^2}\\\Rightarrow E=4195333.33\ MN/C=4.1953\ MN/C$

Electric field is 4.1953 MN/C

$V=\dfrac{kq}{r}\\\Rightarrow V=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.3}\\\Rightarrow V=1258600\ V=1.2586\ MV$

Electric potential is 1.2586 MV

r = R = 14 cm

$E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.14^2}\\\Rightarrow E=19264285.7143\ N/C=19.2642857143\ MN/C$

The electric field is 19.2642857143 MN/C

$V=\dfrac{kq}{r}\\\Rightarrow V=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.14}\\\Rightarrow V=2697000\ V=2.697\ MV$

The potential is 2.697 MV

7 0

3 years ago