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Lostsunrise [7]
3 years ago
9

Can the distance coverd by a body ever be negative? What about its distance and displacement​

Physics
1 answer:
Tju [1.3M]3 years ago
4 0

answer in photo enjoy

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What is the purpose of the report​
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The purpose of report : Reports communicate information which has been compiled as a result of research and analysis of data and of issues .

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Which two spheres are interacting when there is a rainbow ? Please help make you brainiest.
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Explanation:

7 0
3 years ago
Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t
Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

M=-\frac{s'}{s}

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

M=-\frac{15 cm}{12 cm}=-1.25

D. Real and inverted

The magnification equation can be also rewritten as

M=\frac{y'}{y}

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

y'=My

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

M=\frac{5}{9}=-\frac{s'}{s}

which can be rewritten as

s'=-M s = -\frac{5}{9}s

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

s'=-Ms

and then we can substitute it into the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}

and we can solve for s:

\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0
4 years ago
If rho(x,y) is the density of a wire (mass per unit length), then
lidiya [134]

Answer:

See description

Explanation:

With the given information we have:

x(t) = 1 + cos(t)\\ y(t)=sin(t)\\ \rho(x,y) = 3x

the interval is [0,\pi ]

now the mass m has the given expression:

m = \int \rho(x,y) dS

we will use the formula for a line integral and let:

dS=\sqrt{x'(t)^2 + y'(t)^2}=\sqrt{cos(t)^2 + sin(t)^2}dt=dt

therefore we have:

m=\int \rho(x,y)dS=\int\limits^\pi_0 {3*x}dS=\int\limits^\pi _0{3*(1+cos(t))dS\\=\int\limits^\pi _0{3*(1+cos(t))dt

we solve the integral:

m=3*\int\limits^\pi _0{(1+cos(t))dt= 3*(t+sin(t))\limits^\pi _0=3*\pi=9.42

7 0
3 years ago
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