All categories

3 years ago

A bicyclist travels 30 km in 2 hours, what is their average speed?

Physics

1 answer:

Mamont248 [21]3 years ago

3 0

Explanation:

average speed = total distance ÷ total time taken

=30÷2 = 15 km/h

You might be interested in

Please help me ,create a slogan that promotes the role of internet in education

aleksklad [387]

Answer:

¨Facts you didn´t know¨ or ¨unknown facts¨

Explanation:

7 0

3 years ago

PLEASE HELPPPP !!!! Scientific investigations must be well ____ to be sure the data collected will help answer the scientists qu

konstantin123 [22]

Answer-

Analyzed or conducted

Hope this helped

3 0

2 years ago

A sprinter set a high school record in track and field, running 200.0 m in 20.6 s . what is the average speed of the sprinter in

Paraphin [41]

Answer : The average speed of the sprinter is, 34.95 Km/hr

Solution :

Average velocity : It is defined as the distance traveled by the time taken.

Formula used for average velocity :

$v_{av}=\frac{d}{t}$

where,

$v_{av}$ = average velocity

d = distance traveled = 200 m

t = time taken = 20.6 s

Now put all the given values in the above formula, we get the average velocity of the sprinter.

$v_{av}=\frac{200m}{20.6s}\times \frac{3600}{1000}=34.95Km/hr$

conversion :

(1 Km = 1000m)

(1 hr = 3600 s)

Therefore, the average speed of the sprinter is, 34.95 Km/hr

8 0

3 years ago

Read 2 more answers

Describe the principle of conversation of energy:

exis [7]

Answer: Energy can neither be created nor destroyed, rather it is converted from one form to another

Explanation:

The principle of conversation of energy explains how energy is conserved in nature by being converted from one form to another such that no energy is created nor destroyed.

Practical examples include:

- electrical pressing iron that converts electrical energy to heat energy

- solar panels that converts solar energy to electrical energy

- Car batteries that converts chemical energy to light energy etc

7 0

3 years ago

A beam of microwaves with λ = 0.9 mm is incident upon a 9 cm slit. At a distance of 1.5 m from the slit, what is the approximate

liq [111]

Answer:

3 cm

Explanation:

According to the question,

$D=1.5 m$ .

$d=9 cm$ .

$\lambda =0.9 mm$ .

Now the approximate slit's image width is equal to width of central maxima.

And width of central maxima is twice the width from center to first maxima

So,

$y=2\frac{\lambda (D)}{d}$ .

Substitute all the variable in above equation.

$y=\frac{(2)0.9\times 10^{-2} m(1.5 m) }{0.09 m}$ .

$y=3 cm$ .

5 0

3 years ago