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matrenka [14]
3 years ago
11

Brainliest to first to answer. What is the maximum stress which a material can withstand when it is pulled apart?

Physics
2 answers:
Fofino [41]3 years ago
6 0

Answer:

stress tension tensile strength

Explanation:

USPshnik [31]3 years ago
3 0

Answer:

stress tension tensile strength

Explanation:

The maximum stress which a material can withstand when it is pulled apart is its: stress tension tensile strength.

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Which of the following is an example of newton second law of motion?
zaharov [31]

Answer: Its answer C: A wheelbarrow is more difficult to move as More objects are placed inside.

Explanation: The greater the mass of the object the more force is needed to make it move.

Hope this helps!! :)

4 0
3 years ago
When gases, liquids, or solids are in contact with a moving object, the flow of
Gwar [14]

Answer:

Heat

Friction is what causes heat.

Brainliest always helps.

5 0
2 years ago
Explain how to find the angle between two nonzero vectors. Choose the correct answer below. A. The angle between two nonzero vec
kozerog [31]

Answer:

θ = Cos⁻¹[A.B/|A||B|]

A. The angle between two nonzero vectors can be found by first dividing the dot product of the two vectors by the product of the two vectors' magnitudes. Then taking the inverse cosine of the result

Explanation:

We can use the formula of the dot product, in order to find the angle between two non-zero vectors. The formula of dot product between two non-zero vectors is written a follows:

A.B = |A||B| Cosθ

where,

A = 1st Non-Zero Vector

B = 2nd Non-Zero Vector

|A| = Magnitude of Vector A

|B| = Magnitude of Vector B

θ = Angle between vector A and B

Therefore,

Cos θ = A.B/|A||B|

<u>θ = Cos⁻¹[A.B/|A||B|]</u>

Hence, the correct answer will be:

<u>A. The angle between two nonzero vectors can be found by first dividing the dot product of the two vectors by the product of the two vectors' magnitudes. Then taking the inverse cosine of the result</u>

3 0
2 years ago
How much water will flow in 30 secs through 200 mm of capillary tube of 1.50 mm in diameter, if the pressure difference across t
Paladinen [302]

The water outflow in 30 secs through 200 mm of the capillary tube is mathematically given as

Qo=1.6 \times 10^{2} \mathrm{~mL}

<h3>What is the water outflow in 30 secs through 200 mm of the capillary tube?</h3>

\begin{aligned}\Delta P &=6660 \mathrm{~m} / \mathrm{m}^{2} \\\mu &=8.01 \times 10^{-4} \text { Pas } \\t &=30 \mathrm{~s} \\L &=200 \mathrm{~mm}=200 \times 10^{-3} \mathrm{~m} \\D &=1.5 \mathrm{~mm}=1.5 \times 10^{-3} \mathrm{~m} \Rightarrow \gamma=\frac{1.5 \times 10^{-3}}{2} \mathrm{~m}\end{aligned}

Generally, the equation for Rate of flow of Liquid is  mathematically given as

\\$$Q=\frac{\pi r^{4} \times \Delta P}{8 \mu L}

$$

Where dP is pressure difference r is the radius

\mu is the viscosity of water

L is the length of the pipe

Q=\frac{\pi \times\left(\frac{1.5 \times 10^{-3}}{2}\right)^{4} \times 6660}{8 \times 8.01 \times 10^{-4} \times 200 \times 10^{-3}}

Q=5.2 \mathrm{~mL} / \mathrm{s}

In $30s the quantity that flows out of the tube

&Qo=5.2 \times 30 \\&Qo=1.6 \times 10^{2} \mathrm{~mL}

In conclusion, the quantity that flows out of the tube

Qo=1.6 \times 10^{2} \mathrm{~mL}

Read more about the flows rate

brainly.com/question/27880305

#SPJ1

5 0
1 year ago
A parallel-plate capacitor has 2.10 cm × 2.10 cm electrodes with surface charge densities ±1.00×10-6 C/m2. A proton traveling pa
Darya [45]

Answer:

x=0.53x10^{-3} m

Explanation:

Using Gauss law the field is uniform so

E=ζ/ε

Charge densities ⇒ζ=1.x10x^{-6} \frac{C}{m^{2}}

ε=8.85x10^{-12} \frac{C^{2}}{n*m^{2}}

E=\frac{1x10^{-6}\frac{C}{m^{2}}}{8.85x^{-12}\frac{C^{2} }{N*m^{2}}} \\E=0.11299 x10^{-6} \frac{N}{C}

Force of charge is

F_{q}=q*E\\F_{q}=1.6x10^{-19}C*0.11299x10^{6}\frac{N}{C} \\F_{q}=1.807x10^{-14} N

F_{q}=m*a\\a=\frac{F_{q}}{m}=\frac{1.807x10^{-13}N}{1.67x10^{-27}}\\ a=1.082x0^{14} \frac{m}{s^{2}} \\t=\frac{x}{v}\\ x=2.1cm\frac{1m}{100cm}=0.021m \\v=6.7x10^{6}\frac{m}{s} \\ t=\frac{0.021m}{6.7x10^{6}\frac{m}{s}} \\t=3.13x10^{-9}s

So finally knowing the acceleration and the time the distance can be find using equation of uniform motion

x_{f}=x_{o}+\frac{1}{2}*a*t^{2}\\ x_{o}=0\\x_{f}=\frac{1}{2} a*t^{2}=\frac{1}{2}*1.082x10^{14}\frac{m}{s^{2} } *(3.134x^{-9}s)^{2}  \\x_{f}=0.53x^{-3}m

5 0
2 years ago
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