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mixer [17]
3 years ago
15

A runner completes the 200-meter dash with a time of 19.80 seconds. What was the runner's average speed in miles per hour?

Physics
1 answer:
andriy [413]3 years ago
4 0

Answer:

v = 22.54 mph.

Explanation:

Given that,

Distance moved, d = 200 m

Time, t = 19.8 s

We need to find the runner's average speed.

We know that,

1 mile = 1609.34 m

200 m = 0.124 miles

19.8 seconds = 0.0055 h

So,

Speed = distance/time

v=\dfrac{0.124}{0.0055}\\\\v=22.54\ mph

So, the runner's average speed is 22.54 mph.

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Derive the formula 2as=v² -u² where the formula have have usual meanings
attashe74 [19]

Answer:The main formula is v² = u² + 2as

Explanation:

S=½(u +v)t

t = v+u/a

S=½(v-u)(v+u/a)

S=½v²+uv-uv-u²/a

2as=v²-u²

5 0
3 years ago
What are 3 ways you can save energy on earth day or any day?
Ber [7]
By unplugging unused devices, by turning off any unused lights, and by switching your lightbulbs to something more energy efficient.
8 0
4 years ago
What is the wavelength of an earthquake wave if it has a speed of 12 km/s and a frequency of 15 Hz
professor190 [17]
First change km/ s into m/s, then use the formula
Lambda = velocity/ frequency
3 0
3 years ago
Two boats leave the same port at the same time, with boat A traveling north at 15 knots (nautical miles per hour) and boat B tra
Mrrafil [7]

Answer:

The chance in distance is 25 knots

Explanation:

The distance between the two particles is given by:

s^2 = (x_A - x_B)^2+(y_A - y_B)^2  (1)

Since A is traveling north and B is traveling east we can say that their displacement vector are perpendicular and therefore (1) transformed as:

s^2 = x_B^2+y_A^2 (2)

Taking the differential with respect to time:

\displaystyle{2s\frac{ds}{dt}= 2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt}}  (3)

where \displaystyle{\frac{dx_B}{dt}}=v_B and \displaystyle{\frac{dx_A}{dt}}=v_A are the respective given velocities of the boats. To find s and x_B we make use of the given position for A, y_A=30, the Pythagoras theorem and the relation between distance and velocity for a movement with constant velocity.

\displaystyle{y_A = v_A\cdot t\rightarrow t = \frac{y_A}{v_A}=\frac{30}{15}=2 h

with this time, we know can now calculate the distance at which B is:

\displaystyle{x_B = v_B\cdot t= 20 \cdot 2 = 40\ nmi

and applying Pythagoras:

\displaystyle{s = \sqrt{x_B^2+y_A^2}=\sqrt{30^2 + 40^2}=\sqrt{2500}=50}

Now substituting all the values in (3) and solving for  \displaystyle{\frac{ds}{dt} } we get:

\displaystyle{\frac{ds}{dt} = \frac{1}{2s}(2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt})}\\\displaystyle{\frac{ds}{dt} = 25 \ knots}

4 0
3 years ago
44. A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down
Vladimir [108]

Answer:

18.4 m

Explanation:

(a)

The known variables in this problem are:

u = 1.40 m/s is the initial vertical velocity (we take downward direction as positive direction)

t = 1.8 s is the duration of the fall

a = g = 9.8 m/s^2 is the acceleration due to gravity

(b)

The vertical distance covered by the life preserver is given by

d=ut + \frac{1}{2}at^2

If we substitute all the values listed in part (a), we find

d=(1.40 m/s)(1.8 s)+\frac{1}{2}(9.8 m/s^2)(1.8 s)^2=18.4m

8 0
3 years ago
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