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3 years ago

What is strain?

Physics

1 answer:

Sholpan [36]3 years ago

3 0

Answer:

C) The ratio of the change in an object's length to its original length when stretched or compressed.

Explanation:

The formula for strain is:

Strain = Change in Length/Origin Length

Hence, it can be described as the percentile of change in the dimension with respect to the original dimension. So, whenever a tensile or a compressive force is applied on a body, its length changes. The ratio of this change to original length is called strain. So, the correct option is:

C) The ratio of the change in an object's length to its original length when stretched or compressed.

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How fast should a rocket ship move for its mass to be dilated to 167% of its rest mass?

andrey2020 [161]

Answer:

The two forces acting on rockets at the moment of launch are the thrust upwards and the weight downwards. Weight is the force due to gravity and is calculated (at the Earth’s surface) by multiplying the mass (kilograms) by 9.8.The resultant force on each rocket is calculated using the equation resultant force = thrust – weight.

Hopefully, this answer helps you! :)

6 0

3 years ago

If a substance has a density of 2.7g/cm3 and a mass of 86.4g, what is its volume?

Gennadij [26K]

Answer:

32cm³

Explanation:

Given parameters:

Density of substance = 2.7g/cm³

Mass of substance = 86.4g

Unknown:

Volume of substance = ?

Solution:

Density is the mass per unit volume of a substance.

Density = $\frac{mass}{volume}$

Since the unknown is volume we solve for it;

mass = density x volume

86.4 = 2.7 x volume

volume = $\frac{86.4}{2.7}$ = 32cm³

7 0

3 years ago

Total potential and kinetic energy of an object.

Ne4ueva [31]

Answer:

The Total Mechanical Energy

As already mentioned, the mechanical energy of an object can be the result of its motion (i.e., kinetic energy) and/or the result of its stored energy of position (i.e., potential energy). The total amount of mechanical energy is merely the sum of the potential energy and the kinetic energy.

Explanation:

8 0

3 years ago

Question 24 1 pts Find the voltage in an extension cord having a 0.0600 22 resistance and through which a 5.00A current is flowi

Neporo4naja [7]

The voltage in the extension cord is 30 V.

The problem above can be solved using ohm's law

⇒ Formula:

V = IR.................. Equation 1

⇒ Where:

V = Voltage in the extension cord

I = Current flowing through the extension cord
R = Resistance of the extension cord.

From the question, I think there was a slight error in the value of the current given it suppose to be 500 A, and not 5.00 A

⇒ Given:

I = 500 A
R = 0.06 ohms

⇒ Substitute these values into equation 1

V = 500(0.06)
V = 30 V

Hence the voltage in the extension cord is 30 V

Learn more about voltage here: brainly.com/question/4429782

6 0

2 years ago

Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i

s2008m [1.1K]

Answer:

The final speed for the mass 2m is $v_{2y}=-1,51\ v_{i}$ and the final speed for the mass 9m is $v_{1f} =0,85\ v_{i}$ .

The angle at which the particle 9m is scattered is $\theta = -66,68^{o}$ with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

Conservation of linear momentum:

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

$\vec{p}=m\vec{v}$

$p_{xi} =p_{xf}$

In the x axis before the collision we have

$p_{xi}=9m\ v_{i} - 2m\ v_{i}$

and after the collision we have that

$p_{xf} =9m\ v_{1x}$

In the y axis before the collision $p_{yi} =0$

after the collision we have that

$p_{yf} =9m\ v_{1y} - 2m\ v_{2y}$

$p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}$

then

$p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}$

Conservation of kinetic energy:

$\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}$

$\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}$

Putting in one side of the equation each speed we get

$\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}$

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the speed of the 9m particle:

$v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}$

the magnitude of the final speed of the particle 9m is

$v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }$

$v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}$

The tangent that the speed of the particle 9m makes with the -y axis is

$tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}$

As a vector the speed of the particle 9m is:

$\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}$

As a vector the speed of the particle 2m is:

$\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}$

8 0

3 years ago