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olga_2 [115]
3 years ago
6

A hockey puck has a coefficient of kinetic friction of μk = .35. If the puck feels a normal force (FN) of 5 N, what is the frict

ional force that acts on the puck?
Physics
1 answer:
alina1380 [7]3 years ago
3 0

Answer:

The frictional force is  F_f =  1.75 \  N

Explanation:

From the question we are told that

     The coefficient of kinetic force is  μk = 0.35

     The normal force felt by the puck is  F_N  =  5 \  N

Generally the frictional force that acts on the puck is mathematically represented as

          F_f =  \mu_k  *  F_N

=>       F_f =  0.35  *  5

=>       F_f =  1.75 \  N

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A car travels a distance of 320 km in 4 hours. What is your average speed in meters per second?
Andreas93 [3]

Answer:

22.2 m/s

Explanation:

First, we need to convert km to m by multiplying by 1000. This means that the car traveled 320 000 meters.

Next, we convert hours to minutes by multiplying by 3600 (the number of seconds in an hour). This means that overall, the car traveled 320 000 m in 14 400 seconds.

The average speed can be found by using the equation \frac{distance}{time}. After substitution, this gives the fraction \frac{320 000}{14 400}, which reduces to 22 \frac{2}{9} m/s, or about 22.2 m/s.

4 0
3 years ago
a tank circuit contains a capacitor and an inductor that produce 30 of reactance at the resonant frequency. the inductor has a q
Romashka-Z-Leto [24]

The total circuit current at the resonant frequency is 0.61 amps

What is a LC Circuit?

  • A capacitor and an inductor, denoted by the letters "C" and "L," respectively, make up an LC circuit, also referred to as a tank circuit, a tuned circuit, or a resonant circuit.
  • These circuits are used to create signals at particular frequencies or to receive signals from more complicated signals at particular frequencies.

Q =15 = (wL)/R

wL = 30 ohms = Xl

R = 2 ohms

Zs = R + jXl = 2 +j30 ohms where Zs is the series LR impedance

| Zs | = 30.07 <86.2° ohms

Xc = 1/(wC) = 30 ohms

The impedance of the LC circuit is found from:

Zp = (Zs)(-jXc)/( Zs -jXc)

Zp = (2+j30)(-j30)/(2 + j30-j30) = (900 -j60)2 = 450 -j30 = 451 < -3.81°

I capacitor = 277/-j30 = j9.23 amps

I Zs = 277/(2 +j30) = (554 - j8,310)/904 = 0.61 - j9.19 amps

I net = I cap + I Zs = 0.61 + j0.04 amps = 0.61 < 3.75° amps

Hence, the total circuit current at the resonant frequency is 0.61 amps

To learn more about LC Circuit from the given link

brainly.com/question/29383434

#SPJ4

5 0
1 year ago
Describe two examples of how humans use electromagnetic waves.
Alex73 [517]
Electromagnetic waves are used in everyday life. You are looking at your computer screen right now. The light that is coming off of the screen is visible light, a form of electromagnetic radiation. Electromagnetic waves are also used to send information. For example, AM or FM radios are radio waves that transfer sound information to your local radio.
5 0
3 years ago
Read 2 more answers
An airplane flies 20km in a direction 60 degrees north of east, then 30 km straight east, then 10km straight north. How far and
Fed [463]

Answer:

48.4 km, 34.3° north of east

Explanation:

Let's say east is the +x direction and north is the +y direction.

Adding up the x components of the vectors:

x = 20 cos 60 + 30 + 0

x = 40 km

Adding up the y components of the vectors:

y = 20 sin 60 + 0 + 10

y = 27.3 km

The magnitude of the displacement is:

d = √(x² + y²)

d = 48.4 km

The direction is:

θ = atan(y/x)

θ = 34.3° north of east

6 0
3 years ago
The field-line representation of the e-field in a certain region in space is shown below. The dashed lines represent equipotenti
lorasvet [3.4K]

You need to go on google an dlook this up

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3 years ago
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