I need help on these 3 questions. <br> I need the original formula used and the givens

A car travels 10.0 m/s. What is its velocity in km/h?

Vitek1552 [10]

Since 1m/s=3.6 km/h, we can conclude that 10.0m/s = 36 km/h

5 0

3 years ago

What is the change in potential energy of a squirrel that jumps from a tree branch that is 4.5 meters off of the ground and land

quester [9]

Answer:

16.3 joules...................

3 0

3 years ago

In the flow past a compression corner, the upstream Mach number and pressure are 3.5 and 1 atm, respectively. Downstream of the

yulyashka [42]

Answer:

$\theta=23.7^{\circ}$

Explanation:

The ratio of pressure 2 to 1 us 5.48/1= 5.48 rounded off as 5.5.

Referring to table A.2 of modern compressible flow then $M_{\beta_1}=2.2$

Also

$M_{\beta_1}=M_1 sin \beta$ and making $sin\beta$ the subject of the formula then

$sin\beta=\frac {2.2}{3.5}\\\beta=38.94^{\circ}$

Making reference to $\theta-\beta-M$ diagram then

$\theta=23.7^{\circ}$

4 0

3 years ago

What are the physics terms behind watching TV?

GalinKa [24]

Everyone knows that one of their favorite past times is sitting in front of the television and watching movies, shows, or playing video games. However with this almost motionless, lazy activity comes a great deal of static physics and mechanics.

When you are sitting down enjoying whatever show it is you may be watching, you actually have several forces acting on you concurrently. For example, by sitting on the couch with no extra weight on you, your weight is equivalent to the normal force, or the force of the couch on you. In addition to the force of the couch of you, if you are leaning on an arm or laying down, a similar force acts on you, except at an angle or incline. The general rule for laying on the couch watching television is that whatever force you exert on an object, that object exerts the same force in the opposite direction, or 180 degrees around.

3 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

I need help on these 3 questions. I need the original formula used and the givens