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3 years ago

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https://docs.googl e.com/presentation/d/e/2PACX-1vQ2d6GrYHS0WmSfvZlgMZ5dOH9UuHobdUDdWLernNTXKuH-DdWzkDsf6CBBqBbE1GGXNQ_2HmHncUHw

/pub?start=false&loop=false&delayms=3000 Here a screen shot

2 answers:

Rom4ik [11]3 years ago

7 0

Nice presentation:) also can I have brainliest I need one more to get my next stage/level

jeka943 years ago

4 0

Answer:

cool

Explanation:

You might be interested in

A certain rain cloud at an altitude of 1.80 km contains 3.20 107 kg of water vapor. How long would it take for a 2.90-kW pump to

kvasek [131]

Answer:

2255 days

Explanation:

height, h = 1.8 km = 1800 m

amount of water, m = 3.2 x 10^7 kg

Power, P = 2.9 kW = 2900 W

Let t be the time taken

Energy required to lift the water,

E = m g h

E = 3.2 x 10^7 x 9.8 x 1800 = 5.65 x 10^11 J

Power, P = Energy / time

t = E / P = (5.65 x 10^11) / 2900

t = 1.95 x 10^8 second

t = 2255 days

thus, the time taken is 2255 days.

7 0

4 years ago

You're driving your new sports car at 85 mph over the top of a hill that has a radius of curvature of 525 m.

Bumek [7]

Explanation:

It is given that,

Speed of the sports car, v = 85 mph = 37.99 m/s

The radius of curvature, r = 525 m

Let $W_N$ is the normal weight and $W_A$ is the apparent weight of the person. Its apparent weight is given by :

$W_A=mg-\dfrac{mv^2}{r}$

So, $\dfrac{W_A}{W_N}=\dfrac{mg-\dfrac{mv^2}{r}}{mg}$

$\dfrac{W_A}{W_N}=\dfrac{g-\dfrac{v^2}{r}}{g}$

$\dfrac{W_A}{W_N}=\dfrac{9.8-\dfrac{(37.99)^2}{525}}{9.8}$

$\dfrac{W_A}{W_N}=0.719$

or

$\dfrac{W_A}{W_N}=71.9\%$

Hence, this is the required solution.

5 0

4 years ago

A tank whose bottom is a mirror is filled with water to a depth of 20.0 cm. A small fish floats motionless 7.0 cm under the surf

liberstina [14]

Answer:

The apparent depth of (a) the fish is 5.3 cm and (b) the image of the fish is 24.8 cm.

Explanation:

According to the following equation:

$\frac{n_{w} }{s} +\frac{n_{a} }{s'} = \frac{n_{a}- n_{w}}{R_{c} } \\$

where <em>nw</em> and <em>na</em> is the refractive indices of water (1.33) and air (1.00) respectively; <em>s</em> is the depth of the fish below the surface of the water; s' is the apparent depth of the fish from normal incidence and Rc is the radius of curvature of the mirror at the bottom of the tank.

Note that the bottom of the tank is assumed to be a flat mirror, therefore the radius of curvature is very large (R⇒∞).

Therefore, the above equation can be expressed as:

$\frac{n_{w}}{s} +\frac{n_{a}}{s'}=0$

Now we can solve for the apparent depth of the fish.

(a) $s'=-(\frac{n_{a}}{n_{w}})x s$ (Make s' subject of the formula from the above equation)

$s'=(\frac{1.00}{1.33} )x7cm$

∴ $s'=5.3 cm.$

(b) The motionless fish floats 13 cm above the mirror, therefore the image of the fish will be situated at 13 + 20 =33 cm away from the real fish.

Therefore, s = 33 cm

$s'=-(\frac{n_{a}}{n_{w}})x s$

$s'=(\frac{1.00}{1.33} )x 33 cm$

$s'=24.8 cm.$

NB: Here, it is assumed that the water is pure, as impurities may alter the refractive index of water.

8 0

4 years ago

8. What variables shift both the long-run and short-run aggregate-supply curves?

vfiekz [6]

Variables shift both the long-run and short-run aggregate-supply curves are productivity, labor Wage Costs and taxes and other costs.

<h3>What is aggregate-supply?</h3>

The entire quantity of goods and services that enterprises in a national economy intend to sell during a certain time period is known as aggregate supply

Variables shift both the long-run and short-run aggregate-supply curves are ;

1. Productivity

2. Labor Wage Costs.

3. Taxes and other costs -

4. Material Prices.

5. Size of labor force.

6. Stock of Capital.

Hence, variables shift both the long-run and short-run aggregate-supply curves are productivity, labor Wage Costs and taxes and other costs.

To learn more about the aggregate-supply, refer:

brainly.com/question/16419230

#SPJ1

4 0

2 years ago

Write the ordered pair that represents vec yz .Then find the magnitude of vec yz Y(- 4, 12), (1, 19)

sladkih [1.3K]

Answer with Explanation:

We are given that

Y=(-4,12)

Z=(1,19)

We have to find the ordered pair which represents the vector YZ and magnitude of vector YZ.

Vector YZ=Z-Y=<1,19>-<-4,12>

Vector YZ=<5,7>

Magnitude of vector r

$r=xi+yj$

$\mid r\mid=\sqrt{x^2+y^2}$

Using the formula

$\vec{YZ}=\sqrt{5^2+7^2}$

$\vec{YZ}=\sqrt{25+49}=\sqrt{74} units$

7 0

3 years ago