What is the acceleration of a skier that goes from 2.50 m/s to 14.5 m/s while traveling 505 m down a slope?

Of waterfalls with a height of more than 50 m , Niagara Falls in Canada has the highest flow rate of any waterfall in the world.

Vinil7 [7]

Answer:

Power output: W=1426.9MW

Explanation:

The power output of the falls is given mainly by its change in potential energy:

$Q=-P_{tot}=-(P_{2}-P_{1})$

The potential energy for any point can be calculated as:

$P=m*g*h$

If we consider the base of the falls to be the reference height, at point 2 h=0, so P2=0, and height at point 1 equals 52m:

$Q=P_{1}=m*g*h$

If we replace m with the mass rate M we obtain the rate of change in potential energy over time, so the power generated:

$W=M*g*h=2.8*10^{3}m^{3}/s*1*10^{3}kg/m^{3}*9.8m/s^{2}*52m =1426.9MW$

5 0

1 year ago

In a 5000 m race, the athletes run 12 1/2 laps; each lap is 400 m.Kara runs the race at a constant pace and finishes in 17.9 min

Ksju [112]

Answer:

No. of laps of Hannah are 7 (approx).

Solution:

According to the question:

The total distance to be covered, D = 5000 m

The distance for each lap, x = 400 m

Time taken by Kara, $t_{K} = 17.9 min = 17.9\times 60 = 1074 s$

Time taken by Hannah, $t_{H} = 15.3 min = 15.3\times 60 = 918 s$

Now, the speed of Kara and Hannah can be calculated respectively as:

$v_{K} = \frac{D}{t_{K}} = \frac{5000}{1074} = 4.65 m/s$

$v_{H} = \frac{D}{t_{H}} = \frac{5000}{918} = 5.45 m/s$

Time taken in each lap is given by:

$(v_{H} - v_{K})t = x$

$(5.45 - 4.65)\times t = 400$

$t = \frac{400}{0.8}$

t = 500 s

So, Distance covered by Hannah in 't' sec is given by:

$d_{H} = v_{H}\times t$

$d_{H} = 5.45\times 500 = 2725 m$

No. of laps taken by Hannah when she passes Kara:

$n_{H} = \frac{d_{H}}{x}$

$n_{H} = \frac{2725}{400} = 6.8$ ≈ 7 laps

3 0

1 year ago

Suppose the U.S. national debt is about $15 trillion. If payments were made at the rate of $1,500 per second, how many years wou

Delvig [45]

Answer:

This question has already been answered.

Explanation:

brainly.com/question/13542582

3 0

1 year ago

At what angle should the roadway on a curve with a 50m radius be banked to allow cars to negotiate the curve at 12 m/s even if t

just olya [345]

Answer:

The banking angle of the road is 16.38 degrees.

Explanation:

Given:

Radius of the roadway on curve, R = 50 m

velocity of the moving car, V = 12 m/s

The banking angle can be calculated by using the formula below:

If there's no frictional force, then net force is equal to centripetal force.

MgTanΦ = (MV^2)/ R

TanΦ = V^2 / gR

Where;

Φ is the banking angle

g is acceleration due to gravity

TanΦ = (12 x 12) / (9.8 x 50)

TanΦ = 0.2939

Φ = arctan (0.2939)

Φ = 16.38 degrees

Therefore, the banking angle of the road is 16.38 degrees.

3 0

1 year ago

The gas tank of Dave’s car has a capacity of 12 gallons. The tank was 38 full before Dave filled it to capacity. It cost him $2.

m_a_m_a [10]

Answer:

$ 18.75

Explanation:

given,

capacity of the Dave’s car = 12 gallons

Assuming that the tank is 3/8 full before

Cost of gas per gallon = $2.50 per gallon of gas

Volume Dave have to fill

= $1 - \dfrac{3}{8}$

= $\dfrac{5}{8}$ of full tank

= $\dfrac{5}{8}\times 12$

volume Dave have to fill = 7.5 gallons

total money Dave spent

= 7.5 gallons x $2.50

= $ 18.75

Dave have to spend $ 18.75 to fill tank to it capacity.

5 0

1 year ago

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