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3 years ago

How can a magnet attract or repel another magnet even if they are not touching?

Physics

1 answer:

lakkis [162]3 years ago

4 0

Answer:

Magnetic forces are non contact forces; they pull or push on objects without touching them. Magnets are only attracted to a few 'magnetic' metals and not all matter. Magnets are attracted to and repel other magnets.

Explanation:

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1. While driving to Palm Desert you notice that some of the windmills near the freeway are spinning 20 times per minute. The len

igomit [66]

Answer:

a) $\alpha = -0.233 rad/s^{2}$

b) the rotational acceleration will remain the same, $\alpha = -0.233 rad/s^{2}$

c) When r = 36 m, $a_{c} = 157.25 m/s^{2}$

When r = 18 m, $a_{c} = 78.63 m/s^{2}$

d) When r = 36 m, $a_{c} = 39.31 m/s^{2}$

When r = 36 m, $a_{c} = 19.66 m/s^{2}$

Explanation:

The windmills are spinning 20 times per minute, the number of spins in 1 second = 20/60 = 1/3 spins/sec

frequency, f = 1/3 spins/sec

there are 1/3 spins in 1 second,

there will be 50 spins in 50/ (1/3) seconds = 150 seconds

i.e time taken to make 50 spins = 150 seconds

Time interval between 20 and 50 spins, Δt = 150 - 60 = 90

Δt = 90

Angular frequency at 20 spins:

w₁ = 2 π f

w₁ = 2 π * 1/3 = 2π/3 = 2.09 rad/s

Since the blade stops at 50 spins, angular frequency at 50 spins, w₂ = 0

The rotational acceleration,

$\alpha = \frac{\triangle w}{\triangle t} \\\alpha = \frac{w_{2} - w_{1} }{\triangle t} \\\alpha = \frac{0 - 2.09 }{90} \\\alpha = -0.233 rad/s^{2}$

b) The magnitude of the rotational acceleration does not depend on the radius, r. It depends on the angular frequency, therefore the magnitude of the rotational acceleration doe not change.

c) The centripetal acceleration is given by the formula, $a_{c} = w^{2} r$

While the windmill spins at 20 times per minute, the angular speed gotten, w = 2.09 rad/s

When r = 36 m

$a_{c} = 2.09^{2} * 36\\a_{c} = 157.25 m/s^{2}$

At the halfway point, r = 18 m

$a_{c} = 2.09^{2} * 18\\a_{c} = 78.63 m/s^{2}$

d) If the angular velocity of the blades were cut in halves:

w = 2.09/2

w = 1.045 rad/s

When r = 36 m

$a_{c} = 1.045^{2} * 36\\a_{c} = 39.31 m/s^{2}$

When r = 18 m

$a_{c} = 1.045^{2} * 18\\a_{c} = 19.66 m/s^{2}$

8 0

3 years ago

A current of 6.0 A runs through a circuit for 2.5 minutes.

Mekhanik [1.2K]

Current is measured as charge per unit time. To get change, simply multiply the current with time:

$6A*2.5mins = 6A*(2.5*60)seconds = 900Coulombs$

5 0

3 years ago

Read 2 more answers

An astronaut travels to a star system 3.9 lyly away at a speed of 0.90 cc . Assume that the time needed to accelerate and decele

kakasveta [241]

Total time elapsed is =8.2y

The starting event is the astronaut leaving Earth. The finishing event is the astronaut arriving at the star system. The time between these events on Earth is:

Δt=3.9ly/0.9c

Δt=4.3y

For the astronaut, two events occur at the same position and can be measured with just one clock. Hence,

Δτ

$= \sqrt{1 - \frac{v {}^{2} }{c {}^{2} } } \times Δt$

Δτ

$= \sqrt{1 - ( \frac{0.9c}{c} ) {}^{2} } \times (4.3y)$

Δτ=1.8ly

The total elapsed time is:

T elapsed=Δt+3.9

T elapsed=4.3+3.9

T elapsed=8.2y

learn more about time from here: brainly.com/question/28208983

#SPJ4

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1 year ago

Kate walks up to an automated teller machine and pushes a button. The pressure exerted by her finger is transformed into electri

lozanna [386]

The answer is Transduction .

8 0

3 years ago

The pressure exerted by a gas is 2.0 atm while it has a volume of 350 mL. What would be the volume of this sample of gas at stan

34kurt

Answer:

700 mL or 0.0007 m³

Explanation:

P₁ = Initial pressure = 2 atm

V₁ = Initial volume = 350 mL

P₂ = Final pressure = 1 atm

V₂ = Final volume

Here the temperature remains constant. So, Boyle's law can be applied here.

P₁V₁ = P₂V₂

$\frac{P_1V_1}{P_2}=V_2\\\Rightarrow V_2=\frac{2\times 350}{1}\\\Rightarrow V_2=700\ mL$

So, volume of this sample of gas at standard atmospheric pressure would be 700 mL or 0.0007 m³

7 0

3 years ago