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2 years ago

Determine the poles of the magnet. Look at the three compass readings that are on top of the magnet. Label the

Physics

2 answers:

melisa1 [442]2 years ago

6 0

Answer:

the red pointer on the magnet ( grey region) : points towards north

red pointer outside the magnet ( white region) is pointing towards south

Explanation:

please see the attached image

Ilya [14]2 years ago

5 0

Answer:

ALL WILL ATTRACT EACH OTHER ..

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A skateboarder rolls down the sidewalk with an initial velocity of 0 m/s. If her acceleration is

VikaD [51]

Answer:

80m/s

Explanation:

to find it you have to work it out by using the formula distance divided by speed to find time.

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2 years ago

The primary job of a(n) ____ <br> is to increase the power of a modified radio wave.

amm1812

The primary job of a(n) ____

is to increase the power of a modified radio wave.

Answer:

<h2>amplifier</h2>

Explanation:

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2 years ago

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Light with a wavelength of 725 nm passes through the slits of a double-slit experiment. The slits of the experiment are a distan

crimeas [40]

Answer:

3.10°

Explanation:

I took the test

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2 years ago

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Two converging lenses are separated by 27.0 cm. The focal length of each lens is 8.90 cm. An object is placed 33.0 cm to the lef

Ymorist [56]

Answer:

1/i + 1/o = 1/f thin lens equation

i = 33 * 8.9 / (33 - 8.9) = 12.2 cm to right of first lens

27 - 12.2 = 14.8 cm to left of second lens

i = 14.8 * 8.9 / (14.8 - 8.9) = 22,3 cm to right of second lens

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2 years ago

A 13561 N car traveling at 51.1 km/h rounds

Minchanka [31]

Answer:

a) The centripetal acceleration of the car is 0.68 m/s²

b) The force that maintains circular motion is 940.03 N.

c) The minimum coefficient of static friction between the tires and the road is 0.069.

Explanation:

a) The centripetal acceleration of the car can be found using the following equation:

$a_{c} = \frac{v^{2}}{r}$

Where:

v: is the velocity of the car = 51.1 km/h

r: is the radius = 2.95x10² m

$a_{c} = \frac{(51.1 \frac{km}{h}*\frac{1000 m}{1 km}*\frac{1 h}{3600 s})^{2}}{2.95 \cdot 10^{2} m} = 0.68 m/s^{2}$

Hence, the centripetal acceleration of the car is 0.68 m/s².

b) The force that maintains circular motion is the centripetal force:

$F_{c} = ma_{c}$

Where:

m: is the mass of the car

The mass is given by:

$P = m*g$

Where P is the weight of the car = 13561 N

$m = \frac{P}{g} = \frac{13561 N}{9.81 m/s^{2}} = 1382.4 kg$

Now, the centripetal force is:

$F_{c} = ma_{c} = 1382.4 kg*0.68 m/s^{2} = 940.03 N$

Then, the force that maintains circular motion is 940.03 N.

c) Since the centripetal force is equal to the coefficient of static friction, this can be calculated as follows:

$F_{c} = F_{\mu}$

$F_{c} = \mu N = \mu P$

$\mu = \frac{F_{c}}{P} = \frac{940.03 N}{13561 N} = 0.069$

Therefore, the minimum coefficient of static friction between the tires and the road is 0.069.

I hope it helps you!

3 0

3 years ago