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2 years ago

Usa test prep- Look at the picture below to answer. If u answer correctly i will mark you as brainliest!

Physics

2 answers:

Diano4ka-milaya [45]2 years ago

7 0

A. Since the nucleus consists of protons and neutrons, they have a mass of 1 units while electrons that are surrounding the nucleus has almost negligible weight.

Plsssss mark me as brainliest

Vladimir [108]2 years ago

4 0

Answer:

Explanation:

The greatest concentration of atomic mass is in the nucleus because it is made up of protons and neutrons. The electrons surrounding the nucleus don't have as much mass as protons or neutrons.

Hopefully this helps...

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If a particular atom of an element has five protons a neutral atom of this element would have ________ electrons. A. 0 B. 3 C. 5

Bezzdna [24]

C. 5

a neutral atom has no electrical charge. protons are positive and electrons are negative, they need to be the same to make it a neutral atom.

5 0

2 years ago

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aliya0001 [1]

You have to figure it out

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2 years ago

A 2011 Porsche 911 Turbo S goes from 0-27 m/s in 2.5 seconds. What is the car's acceleration?

Natalka [10]

Answer:

-10.8m/s^2

Explanation:

a=change in velocity/change in time

-27 m/s/2.5=10.8m/s^2

or if its not negative

27m/s/2.5=10.8m/s^2

3 0

2 years ago

A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the

algol13

Stress required to cause slip on in the direction [ 1 1 0 ] is 7.154 MPa

<u>Explanation:</u>

Given -

Stress Direction, A = [1 0 0 ]

Slip plane = [ 1 1 1]

Normal to slip plane, B = [ 1 1 1 ]

Critical stress, Sc = 2.92 MPa

Let the direction of slip on = [ 1 1 0 ]

Let Ф be the angle between A and B

cos Ф = A.B/ |A| |B| = [ 1 0 0 ] [1 1 1] / √1 √3

cos Ф = 1/√3

σ = Sc / cosФ cosλ

For slip along [ 1 1 0 ]

cos λ = [ 1 1 0 ] [ 1 0 0 ] / √2 √1

cos λ = 1/√2

Therefore,

σ = 2.92 / 1/√3 1/√2

σ = √6 X 2.92 MPa = 2.45 X 2.92 = 7.154MPa

Therefore, stress required to cause slip on in the direction [ 1 1 0 ] is 7.154MPa

4 0

3 years ago

A large container contains a large amount of water. A hole is drilled on the wall of the container, at a vertical distance h = 0

barxatty [35]

Answer:

Velocity = 3.25[m/s]

Explanation:

This problem can be solved if we use the Bernoulli equation: In the attached image we can see the conditions of the water inside the container.

In point 1, (surface of the water) we have the atmospheric pressure and at point 2 the water is coming out also at atmospheric pressure, therefore this members in the Bernoulli equation could be cancelled.

The velocity in the point 1 is zero because we have this conditional statement "The water surface drops very slowly and its speed is approximately zero"

h2 is located at point 2 and it will be zero.

$(P_{1} +\frac{v_{1}^{2} }{2g} +h_{1} )=(P_{2} +\frac{v_{2}^{2} }{2g} +h_{2} )\\P_{1} =P_{2} \\v_{1}=0\\h_{2} =0\\v_{2}=\sqrt{0.54*9.81*2}\\v_{2}=3.25[m/s]$

4 0

3 years ago