All categories

3 years ago

What is the speed of a walking person in m/s if the person travels 2000 m in 35 minutes?

Physics

1 answer:

iren [92.7K]3 years ago

4 0

Distance = 2000m

Time = 35 minutes = 35×60 sec = 2100 sec

We know, Speed = Distance/Time

Therefore, Speed

= 2000m/2100s = 20/21 m/s

Answer: 20/21 m/s

Hope it helps! Please do comment

You might be interested in

A stone is thrown upward at an angle. what happens to the horizontal component of its velocity as it rises? as it falls?

Ne4ueva [31]

<span>Rising or falling, it does not change.</span>

5 0

3 years ago

How long would it take for Sofia to walk 300 meters if she is walking at a velocity of 2.5 m/s?

Vaselesa [24]

Answer:

Time=120seconds

Explanation:

S=300m

V=2.5m/s

t=?

V=S/t

t=S/V

t=300/2.5

t=120 second

7 0

4 years ago

Read 2 more answers

How many calendar days pass between the new moon and full moon?

german

The complete cycle of phases lasts 29.531 days.

From New Moon to Full Moon is half of that . . . 14.765 days,
which is very close to 2 weeks.

4 0

3 years ago

Read 2 more answers

A sealed tank containing seawater to a height of 12.8 m also contains air above the water at a gauge pressure of 2.90 atm . Wate

exis [7]

Answer:

The velocity of water at the bottom, $v_{b} = 28.63 m/s$

Given:

Height of water in the tank, h = 12.8 m

Gauge pressure of water, $P_{gauge} = 2.90 atm$

Solution:

Now,

Atmospheric pressue, $P_{atm} = 1 atm = 1.01\tiems 10^{5} Pa$

At the top, the absolute pressure, $P_{t} = P_{gauge} + P_{atm} = 2.90 + 1 = 3.90 atm = 3.94\times 10^{5} Pa$

Now, the pressure at the bottom will be equal to the atmopheric pressure, $P_{b} = 1 atm = 1.01\times 10^{5} Pa$

The velocity at the top, $v_{top} = 0 m/s$ , l;et the bottom velocity, be $v_{b}$ .

Now, by Bernoulli's eqn:

$P_{t} + \frac{1}{2}\rho v_{t}^{2} + \rho g h_{t} = P_{b} + \frac{1}{2}\rho v_{b}^{2} + \rho g h_{b}$

where

$h_{t} - h_{b} = 12.8 m$

Density of sea water, $\rho = 1030 kg/m^{3}$

$\sqrt{\frac{2(P_{t} - P_{b} + \rho g(h_{t} - h_{b}))}{\rho}} = v_{b}$

$\sqrt{\frac{2(3.94\times 10^{5} - 1.01\times 10^{5} + 1030\times 9.8\times 12.8}{1030}} = v_{b}$

$v_{b} = 28.63 m/s$

5 0

4 years ago

Find the sum of the vectors: 40m/s2 Northeast, 10 m/s2 Northeast

podryga [215]

Answer:

Explanation:

Since both vectors are pointing on the same direction (Northeast), the sum of them will point in that same direction, and its magnitud will be the sum of the magnitudes of each vector (40m/s2+10m/s2). This problem is just a problem in one dimension. The sum of the vectors is then 50m/s2 Northeast.

8 0

4 years ago