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andreyandreev [35.5K]

3 years ago

6

Which describes the electromagnetic force only? Check all that apply. is attractive is repulsive has an infinite range has a ver

y small range produces light produces electricity

1 answer:

Oxana [17]3 years ago

7 0

Answer:Electromagnetic force, like all forces, is measured in Newtons. Electrostatic forces are described by Coulomb’s law, and both electric and magnetic forces are covered by the Lorentz force law. However, Maxwell’s four equations provide the most detailed description of electromagnetism.

Explanation:

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In a Little League baseball game, the 145 g ball reaches the batter with a speed of 15.0 m/s. The batter hits the ball, and it l

Oduvanchick [21]

Answer: 5.075Ns

Explanation:

Given the following :

Mass of ball = 145g

Initial Speed of ball = 15m/s

Final speed of ball when hit by the batter = - 20m/s ( Opposite direction)

The impulse of a body is represented using the relation:

Force(f) * time(t) = mass (m) * (final Velocity(V) - initial velocity(u))

Therefore, using:

m(v - u) = impulse

Mass of ball = 145 / 1000 = 0.145kg

Impulse = 0.145(- 20 - 15)

Impulse = 0.145(-35)

Impulse = 5.075Ns

3 0

3 years ago

A friend rides, in turn, the rims of three fast merry-go-rounds while holding a sound source that emits isotropically at a certa

Aliun [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The Ranking of the curve according to their speed would be equal Rank because $v_1 =v_2 =v_3$

b

The first frequency would have a higher rank compared to the other two which will have the same ranking when ranked with respect to their angular velocities because

$w_1 >w_2 = w_3$

c

The ranking of the second third frequency would be the same but their ranking would be greater than that of the first frequency because

$r_2 =r_3 >r_1$

Explanation:

Mathematically Frequency can be represented as

$F = \frac{v}{\lambda}$

Where $\lambda$ is the wavelength and v is the velocity

Now looking at the diagram we see that

For the first frequency we have

Let the wavelength be $\lambda_1 = \lambda$ , and the frequency $F_1 = F$

For the second frequency

Let the wavelength be $\lambda_2 = 2 \lambda$ , and the frequency $F_2 = \frac{F}{2}$

For the third frequency

Let the wavelength be $\lambda_3 = 2\lambda$ , and the frequency $F_3 = \frac{F}{2}$

To obtain v for each of the frequency we make v the subject in the equation above for each frequency

So,

For the first frequency we have

$v_1 = \lambda_1 F_1 = \lambda F$

For the second frequency

$v_2 = \lambda_2 F_2 = 2 \lambda*\frac{F} {2} = \lambda F$

For the third frequency

$v_3 = \lambda_3 F_3 = 2 \lambda*\frac{F} {2} = \lambda F$

Hence

The Ranking of the curve according to their speed would be equal Rank because $v_1 =v_2 =v_3$

Mathematically angular speed can be represented as

$w = 2 \pi f$

For the first frequency we have

$w_1 = 2\pi F_1 = 2 \pi F$

For the second frequency

$w_2 = 2 \pi F_2 = 2 \pi \frac{F}{2} = \pi F$

For the third frequency

$w_3 = 2 \pi F_3 = 2 \pi \frac{F}{2} = \pi F$

Hence

The first frequency would have a higher rank compared to the other two which will have the same ranking when ranked with respect to their angular velocities because

$w_1 >w_2 = w_3$

Mathematically the relationship between the angular velocity and the linear velocity can be represented as

$v = wr$

=> $r = \frac{v}{w}$

Since the linear velocity is constant we have that

$r \ \alpha \ \frac{1}{w}$

This means that r varies inversely to the angular velocity ,What this means for ranking due to the radius is that the ranking of the second third frequency would be the same but their ranking would be greater than that of the first frequency because

$r_2 =r_3 >r_1$

5 0

3 years ago

A smooth circular hoop with a radius of 0.400 m is placed flat on the floor. A 0.325-kg particle slides around the inside edge o

aleksandrvk [35]

Answer:

a) W = - 6.825 J, b) θ = 1.72 revolution

Explanation:

a) In this exercise the work of the friction force is negative and is equal to the variation of the kinetic energy of the particle

W = ΔK

W = K_f - K₀

W = ½ m v_f² - ½ m v₀²

W = ½ 0.325 (5.5² - 8.5²)

W = - 6.825 J

b) find us the coefficient of friction

Let's use Newton's second law

fr = μ N

y-axis (vertical) N-W = 0

fr = μ W

work is defined by

W = F d

the distance traveled in a revolution is

d₀ = 2π r

W = μ mg d₀ = -6.825

μ = $\frac{ -6.825}{d_o \ mg}$

The total work as the object stops the final velocity is zero v_f = 0

W = 0 - ½ m v₀²

W = - ½ 0.325 8.5²

W = - 11.74 J

μ mg d = -11.74

we subtitle the friction coefficient value

( $\frac{-6.8525 }{d_o mg}$ ) m g d = -11.74

6.825 $\frac{d}{d_o}$ = 11.74

d = 11.74/6.825 d₀

d = 1.7201 2π 0.400

d = 4.32 m

this is the total distance traveled, the distance and the angle are related

θ = d / r

θ = 4.32 / 0.40

θ = 10.808 rad

we reduce to revolutions

θ = 10.808 rad (1rev / 2π rad)

θ = 1.72 revolution

3 0

3 years ago

The best leaper in the animal kingdom is the puma, which can jump to a height of 3.7m when leaving the ground at an angle of 45

zheka24 [161]

Answer:

v = 12 m/s

Long, boring, and convoluted explanation:

First, let's lay out our information.

- <em>max height = 3.7 m</em>

- <em>0 = 45°</em>

<em>- gravitational acceleration constant = 9.8 </em> $\frac{m}{s^2}$ <em />

<em />

Since the puma leaves the ground at a <em> 45 °</em> angle, its motion will follow a curved path as seen in many projectile motion problems, where the object is being influenced solely by the force of gravity. And because the puma leaves the ground at an angle, its initial velocity is broken down into its horizontal and vertical components. We were also told, though indirectly, that the max height is <em>3.7m</em> because the puma can reach up to that height. Gravity is always given to be <em>9.8 </em> $\frac{m}{s^2}$ <em />

<em />

Because we are dealing with maximum height and gravity, we have to use the vertical component of the velocity, <em>vsin ( θ )</em> , and not the horizontal component, <em>vcos ( θ )</em> .

Given its max height, the acceleration due to gravity, and the angle, we can now solve for the speed at which the puma leaves the ground using the following equation: <em>vsin ( θ ) = </em> $\sqrt{2hg}$

Where <em> vsin ( θ )</em> is the vertical component of the initial velocity and <em>h</em> and <em>g</em> are max height and gravitational acceleration constant respectively.

Plugin, rearrange and solve

v sin ( θ ) = $\sqrt{2hg}$

v sin ( 45 ∘ ) = √ 2 × 3.7 × 9.8

v ( 0.71 ) = $\sqrt{72.52}$

v ( 0.71 ) = 8.52

v = 8.52 /0.71

v = 12 m s

<em />

<em />

4 0

3 years ago

A diffraction grating has 300 lines per mm. If light of wavelength 630 nm is sent through this grating, what is the highest orde

True [87]

Answer:

The order of maximum is $n = 5$

Explanation:

From the question we are told that

The diffraction grating is k = 300 lines per mm = 300000 lines per m

The wavelength is $\lambda = 630 \ nm = 630 *10^{-9} \ m$

Generally the condition for constructive interference is mathematically represented as

$dsin \theta = n * \lambda$

Here n is the order maximum

d is the distance the grating which is mathematically represented as

$d = \frac{1}{k}$

=> $d = \frac{1}{300000}$

=> $d = 3.3*10^{-6}\ m$

So

$n = \frac{dsin \theta}{ \lambda}$

at maximum $sin\theta = 1$

$n = \frac{d}{\lambda}$

=> $n = \frac{3.3*10^{-6}}{630 *10^{-9}}$

=> $n = 5$

8 0

3 years ago