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3 years ago

Aluminium is metal or metalloid

Physics

2 answers:

Pie3 years ago

7 0

Answer:

Aluminum-(Al) is a metal

WARRIOR [948]3 years ago

4 0

Answer:

Aluminium is just on the metal side of the border between metals and metalloids, so it is not considered to be a metalloid

Explanation:

it is a metal

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Rom4ik [11]

The car will gain new momentum if it's velocity is doubled or tripled.

6 0

3 years ago

A scientist observes rock masses that have moved past each other in opposite horizontal directions. Which feature

Reil [10]

Answer:

C. strike-slip fault

Explanation:

The scientist must have observed a strike- slip fault.

A fault is an evidence of brittle deformation of the crust in the presence of applied stress on earth materials. Here, the earth material is the rock subjected to tension.

Where a fault occurs, there must have been movement between two blocks of rocks. The direction of movement helps us to delineate the fault type.

When two blocks moves past each other horizontally, it is a strike-slip fault like rubbing your palms together.
When a block moves in the direction of the dip, it forms a dip-slip fault which results in a fault-block mountain characterized by graben and horst systems.

Option A, Plateau is a table landform usually a mountain with flat peak.

Option B is a bowl shaped stratigraphic pattern in which the youngest sequence is at the core of the strata or a fold.

So, the most fitting option is C, a strike-slip fault.

8 0

3 years ago

Add these two velocity vectors to find the magnitude of their resultant vector.

hammer [34]

The magnitude of their resultant vector is 4.6 meters/seconds

Since we are to add the velocity vectors in order to find the magnitude of their resultant vector.

Hence:

Resultant vector magnitude=5.8 meters/seconds + (1.2 meters/seconds)

Resultant vector magnitude=5.8 meters/seconds-1.2 meters/seconds

Resultant vector magnitude 4.6 meters/seconds

Inconclusion The magnitude of their resultant vector is 4.6 meters/seconds

Learn more here:

brainly.com/question/11134601

6 0

3 years ago

what occurs when a swimmer pushes through the water to swim answers are (A) the water exerts a reaction force on the swimmer (B)

miv72 [106K]

When a swimmer pushes through water to swim they are propelled forward because of the water resistance against the hand and feet. It's A. The water doesn't automatically push the swimmer forward. It releases a reaction after the swimmer pushes through the water.

6 0

3 years ago

Read 2 more answers

You are traveling in a car toward a hill at a speed of 36.4 mph. The car's horn emits sound waves of frequency 231 Hz, which mov

Marina CMI [18]

Answer:

a. The frequency with which the waves strike the hill is 242.61 Hz

b. The frequency of the reflected sound wave is 254.23 Hz

c. The beat frequency produced by the direct and reflected sound is

11.62 Hz

Explanation:

Part A

The car is the source of our sound, and the frequency of the sound wave it emits is given as 231 Hz. The speed of sound given can be used to determine the other frequencies, as expressed below;

$f_{1} = f[\frac{v_{s} }{v_{s} -v} ]$ ..............................1

where $f_{1}$ is the frequency of the wave as it strikes the hill;

f is the frequency of the produced by the horn of the car = 231 Hz;

$v_{s}$ is the speed of sound = 340 m/s;

v is the speed of the car = 36.4 mph

Converting the speed of the car from mph to m/s we have ;

hint (1 mile = 1609 m, 1 hr = 3600 secs)

v = $36.4 mph *\frac{1609 m}{1 mile} *\frac{1 hr}{3600 secs}$

v = 16.27 m/s

Substituting into equation 1 we have

$f_{1}$ = $231 Hz (\frac{340 m/s}{340 m/s - 16.27 m/s})$

$f_{1}$ = 242.61 Hz.

Therefore, the frequency which the wave strikes the hill is 242.61 Hz.

Part B

At this point, the hill is the stationary point while the driver is the observer moving towards the hill that is stationary. The frequency of the sound waves reflecting the driver can be obtained using equation 2;

$f_{2} = f_{1} [\frac{v_{s}+v }{v_{s} } ]$

where $f_{2}$ is the frequency of the reflected sound;

$f_{1}$ is the frequency which the wave strikes the hill = 242.61 Hz;

$v_{s}$ is the speed of sound = 340 m/s;

v is the speed of the car = 16.27 m/s.

Substituting our values into equation 1 we have;

$f_{2} = 242.61 Hz [\frac{340 m/s+16.27 m/s }{340 m/s } ]$

$f_{2}$ = 254.23 Hz.

Therefore, the frequency of the reflected sound is 254.23 Hz.

Part C

The beat frequency is the change in frequency between the frequency of the direct sound and the reflected sound. This can be obtained as follows;

Δf = $f_{2}$ - $f_{1}$

The parameters as specified in Part A and B;

Δf = 254.23 Hz - 242.61 Hz

Δf = 11.62 Hz

Therefore the beat frequency produced by the direct and reflected sound is 11.62 Hz

3 0

3 years ago