All categories

4 years ago

The apparent displacement of an object as it is viewed from two different positions is known as

1 answer:

3 0

This phenomenon is known as parallax. The simplest example is this: if you are sitting in the passenger seat, the gas tank needle may appear to read empty. However, if you are the driver, you will see that you are not on empty, but should get some gas.

You might be interested in

A team of scientists is working to determine if a meteor created a landform. One team includes geologists

Anuta_ua [19.1K]

Answer:

your answer should be astronomy

Explanation:

6 0

2 years ago

What word is used to describe an acid or base that can destroy body tissue, clothing, and many other?

lara31 [8.8K]

Corrosive. It’s something that tends to cause corrosion, and it means to destroy or damage things slowly by chemical action.

7 0

3 years ago

If the 20-mm-diameter rod is made of A-36 steel and the stiffness of the spring is k = 55 MN/m , determine the displacement of e

Alinara [238K]

Answer:

σ = 1.09 mm

Explanation:

Step 1: Identify the given parameters

rod diameter = 20 mm

stiffness constant (k) = 55 MN/m = 55X10⁶N/m

applied force (f) = 60 KN = 60 X 10³N

young modulus (E) = 200 Gpa = 200 X 10⁹pa

Step 2: calculate length of the rod, L

$K = \frac{A*E}{L}$

$L = \frac{A*E}{K}$

$A=\frac{\pi d^{2}}{4}$

d = 20-mm = 0.02 m

$A=\frac{\pi (0.02)^{2}}{4}$

A = 0.0003 m²

$L = \frac{A*E}{K}$

$L = \frac{(0.0003142)*(200X10^9)}{55X10^6}$

L = 1.14 m

Step 3: calculate the displacement of the rod, σ

$\sigma = \frac{F*L}{A*E}$

$\sigma = \frac{(60X10^3)*(1.14)}{(0.0003142)*(200X10^9)}$

σ = 0.00109 m

σ = 1.09 mm

Therefore, the displacement at the end of A is 1.09 mm

3 0

3 years ago

Consider a concave mirror that has a focal length f. In terms of f, determine the object distances that will produce a magnifica

Aleks04 [339]

We have that the magnification of each focal length is given respectively as

A) has $u=3\frac{f}{2}$

B) has $u=4\frac{f}{3}$

C) has $u=5\frac{f}{4}$

From the question we are told that:

Focal Length F

Generally, the equation for Magnification is mathematically given by

$M=\frac{-v}{u}$

Therefore

$v=2u$

For A

$M=-2$

Therefore

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

$\frac{1}{f}=\frac{1}{u}+\frac{1}{2u}$

Therefore

$u=3\frac{f}{2}$

For B

$M=-3$

Therefore

$v=3u$

Where

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

$\frac{1}{f}=\frac{1}{u}+\frac{1}{3u}$

Therefore

$u=4\frac{f}{3}$

For C

$M=-4$

Therefore

$v=4u$

Therefore

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

$\frac{1}{f}=\frac{1}{u}+\frac{1}{4u}$

Therefore

$u=5\frac{f}{4}$

Conclusion

From the calculations above we can rightly say that the magnifications of the values above are

A has $u=3\frac{f}{2}$

B has $u=4\frac{f}{3}$

C has $u=5\frac{f}{4}$

For more information on this visit

brainly.com/question/14468351

3 0

3 years ago

EMF. Currents in dc transmission lines can be 100 A or more. Some people have expressed concern that the electromagnetic fields

mrs_skeptik [129]

Answer:

The magnetic field B of the transmission line is 4.5·10^(-6)T. The ratio B/Be is 0.0009% (there is no risk to human health)

Explanation:

we will consider a dc transmission line as an infinite longitudinal line in the Z-axis with a dc current of 180 A. To solve the problem we will use ampere's law. We can consider the geometry of the problem (Rotational symmetry in respect of Z-axis) than the field B is annular with center in the transmission line.

$\vec{B}(r,\varphi,z)=B(r)\vec{\varphi}$

$\displaystyle\oint_{C} \vec{B}(r)\,\vec{dl}=\mu I_c$

We will use a circular amperian curve C. Therefore:

$\displaystyle\oint_{C} \vec{B}(r)\,\vec{dl}=\int_{0}^{2\pi} B(r)\vec{\varphi}\,\vec{\varphi}rd\varphi=B(r)r\int_{0}^{2\pi} \,d\varphi=B(r)r2\pi=\mu I_c$

$B(r)=\displaystyle\frac{\mu 180A}{r2\pi}$

For a height of 8.0m (r=8m):

$B(8m)=\displaystyle\frac{\mu 180A}{8m2\pi}=4.5\cdot10^{-6}T$

Compared to the earth magnetic field:

$\displaystyle\frac{B}{B_e}\%= \frac{B}{B_e}100=0.0009\%$

6 0

3 years ago