What is a likely direct effect of a community program for local residents?

You get a new pair of sunglasses as a birthday gift from a friend. When wrapping the present, your friend has removed the price

Gwar [14]

Answer:

Explanation:By using the sunglasses while looking at If your sunglasses are polarized, instead of only seeing the surface of a lake or river, you will suddenly be able to see through the glare and into the water below.

Another one is by comparing two sunglasses, hold your glasses and another one simultaneously and look through both pairs at the same time. Then, rotate one pair of sunglasses about 60 degrees. If both pairs of glasses are polarized, the overlapping area will darken as they filter out excess light. If your pair isn't polarized, however, you won't notice any difference.

Also it can be determined by simple computer test If you rotate your glasses sideways while looking at a computer monitor through polarized glasses, portions of your screen will become blank or go dark. The same also applicable to LCD display screens such as the ones on a gas pump.

6 0

3 years ago

Two people, one with mass m1 and the other with mass m2, stand on a stationary sled with mass M on a frozen lake. Assume that th

lozanna [386]

Answer:

Part a)

Velocity of sled

$v = \frac{m_1 s}{m_1 + m_2 + M}$

velocity of first man who jump off

$v_1 = -\frac{(m_2 + M) s}{m_1 + m_2 + M}$

Part b)

Velocity of sled

$v_f = (\frac{m_1 s}{m_1 + m_2 + M}) + (\frac{m_2}{m_2 + M})s$

Also the speed of second person is given as

$v_2 = (\frac{m_1 s}{m_1 + m_2 + M}) - \frac{Ms}{m_2 + M}$

Part c)

change in kinetic energy of sled + two people is given as

$KE = \frac{1}{2}Mv_f^2 + \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$

Explanation:

As we know that here we we consider both people + sled as a system then there is no external force on it

So here we can use momentum conservation

since both people + sled is at rest initially so initial total momentum is zero

now when first people will jump with relative velocity "s" then let say the sled + other people will move off with speed v

so by momentum conservation we have

$0 = m_1(v - s) + (m_2 + M)v$

$v = \frac{m_1 s}{m_1 + m_2 + M}$

so velocity of the sled + other person is

$v = \frac{m_1 s}{m_1 + m_2 + M}$

velocity of first man who jump off

$v_1 = \frac{m_1 s}{m_1 + m_2 + M} - s$

$v_1 = -\frac{(m_2 + M) s}{m_1 + m_2 + M}$

Part b)

now when other man also jump off with same relative velocity

so let say the sled is now moving with speed vf

so by momentum conservation we have

$(m_2 + M)(\frac{m_1 s}{m_1 + m_2 + M}) = m_2(v_f - s) + Mv_f$

$(m_2 + M)(\frac{m_1 s}{m_1 + m_2 + M}) + m_2s = (m_2 + M)v_f$

Now we have

$v_f = (\frac{m_1 s}{m_1 + m_2 + M}) + (\frac{m_2}{m_2 + M})s$

Also the speed of second person is given as

$v_2 = (\frac{m_1 s}{m_1 + m_2 + M}) + (\frac{m_2}{m_2 + M})s - s$

$v_2 = (\frac{m_1 s}{m_1 + m_2 + M}) - \frac{Ms}{m_2 + M}$

Part c)

change in kinetic energy of sled + two people is given as

$KE = \frac{1}{2}Mv_f^2 + \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$

here we know all values of speed as we found it in part a) and part b)

4 0

3 years ago

A shopper walks westward 5.4 meters and then eastward 7.8 meters

RoseWind [281]

Answer:

13.2 meters

Explanation:

(5.4) + (7.8)

6 0

3 years ago

Which statement is best supported by the diagram?

Iteru [2.4K]

The ball's maximum potential energy at point-R.

(Potential energy depends on height.)

8 0

3 years ago

Read 2 more answers

On Earth a ball is thrown straight downward with an initial speed of 1 meter

Montano1993 [528]

-- Gravity adds 9.8 m/s to the downward speed of any object, every second ... as long as there are no other forces messing with it.

-- In 0.6 sec, gravity added (0.6x9.8)= 5.88 m/s to the downward speed of this ball.

-- This ball didn't start from zero. You threw it down with an initial speed of 1 m/s. So after 0.6 sec, with the help of gravity, its speed is

(1) + (5.88) = 6.88 m/s .

Pick choice-C .

Notice that we don't care how high it was off the ground when you threw it, just as long as it was high enough to keep falling for 0.6 sec without hitting the ground.

7 0

4 years ago