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belka [17]
3 years ago
12

You biked to the store in 10 minutes. The store was 3 km away. What was your average speed?

Physics
1 answer:
sineoko [7]3 years ago
4 0

Answer:

speed in km/min:

0.3km/min

speed in km/h:

18km/h

speed in m/s:

5m/s

Explanation:

the average speed is defined as:

s=\frac{d}{t}

where s is the speed, d is the distance traveled, and t is the time

according to the statement:

d=3km = 3,000m

and

t=10min=\frac{1}{6}h=600s (10 minutes are equal to a sixth of an hour which is 600 seconds)

Because the units in which the speed is required are not indicated, here are some options for the answer:

Depending on the units in which you need the speed, are the quantities you should use.

Speed in km / h (kilometers per hour):

s=\frac{3km}{\frac{1}{6}h }\\ \\s=18km/h

speed in m / s (meters per second):

s=\frac{3,000m}{600s}\\ s=5m/s

or the speed in km/min (kilometers per minute)

s=\frac{3km}{10min}\\ s=0.3km/min

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8 0
3 years ago
If you accelerate from rest at 12m/s^2 for 5 seconds, what is your finally velocity?
jeka94

Answer:

v=60m/s

Explanation:

Use the Kinematic Equation:

v=v_o+at

Plug in what is given and solve

v=0+(12)(5)

v=60m/s

3 0
2 years ago
A person travels by car from one city to another with differem constan1 speeds between pairs of cities. She drives for 30.0 min
Montano1993 [528]

Answer:

a.52.9 km/h

b.90 km

Explanation:

We are given that

v_1=89km/h

t_1=30min

v_2=100km/h

t_2=12min

v_3=40km/h

t_3=45 min

Time spend on eating lunch and buying ga=15 min.

a.Total time=30+12+45+15=102 minute=\frac{102}{60}=1.7 hour

1 hour=60 minutes

Distance=speed\times time

d_1=v_1\times t_1=80\times\frac{30}{60}=40km

d_2=100\times \frac{12}{60}=20 km

d_3=40\times \frac{45}{60}=30 km

Total distance=d_1+d_2+d_3=40+20+30=90km

Average speed=\frac{total\;speed}{total\;time}

Using the formula

Average speed=\frac{90}{1.7}=52.9Km/h

b.Total distance between the initial and final city lies along the route=90 km

4 0
3 years ago
Using equations, determine the temperature, pressure and density of the air for a aircraft flying at 19.5 km. Is this aircraft s
Viefleur [7K]

Answer:

a) - 72.5°c

b) pressure = 3625.13 Pa

c) density =  0.063 kg/m^3

d) it is a subsonic aircraft

Explanation:

a) Determine Temperature

Temperature at 19.5 km ( 19500 m )

T = -131 + ( 0.003 * altitude in meters )

  =  -131 + ( 0.003 * 19500 ) = - 72.5°c

b) Determine pressure and density at 19.5 km altitude

Given :

Po (atmospheric pressure at sea level )  = 101kpa

R ( gas constant of air ) = 0.287 KJ/Kgk

T = -72.5°c ≈ 200.5 k

pressure = 3625.13 Pa

hence density = 0.063 kg/m^3

attached below is the remaining part of the solution

C) determine if the aircraft is subsonic or super sonic

Velocity ( v ) = \sqrt{CRT}  =  \sqrt{1.4*287*200.5 } = 283.8 m/s

hence it is a subsonic aircraft

4 0
3 years ago
A kettle is rated at 1 kW, 220 V. Calculate the working resistance of the kettle.
Anna [14]

Explanation:

Power of electric kettle, P = 1 kW

Voltage, V = 220 V

(a) Electric power is given by the formula as follows :

P=\dfrac{V^2}{R}

R is resistance

R=\dfrac{V^2}{P}\\\\R=\dfrac{(220)^2}{10^3}\\\\R=48.4\ \Omega

(b) When connected to a 220 V supply, it takes 3 minutes for the water in the kettle to reach boiling point.

Energy supplied is given by :

E=P\times t

P is power, P=\dfrac{V^2}{R}

E=\dfrac{V^2}{R}t\\\\E=\dfrac{(220)^2}{48.4}\times 180\\\\E=180000\ J\\\\E=180\ kJ

5 0
3 years ago
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