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3 years ago

A person began running due east and covered 15 km in 2.0 hr. What is the average velocity of the person?

Physics

2 answers:

g100num [7]3 years ago

6 0

Velocity = distance / time

v = (15*1000)m / (2*60*60)s

v=2.08 m / s

kotegsom [21]3 years ago

6 0

ave vel = total distance/total time in a specified direction

15km/2hr due east=7.5k/h d east

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Which of the following describes the electron sharing between hydrogen and fluorine A.Hydrogen and fluorine share one electron w

atroni [7]

Hydrogen has one electron in its outermost shell, while fluorine has seven electron in its outermost shell, hence both hydrogen and fluorine needs a single electron to complete its outermost shell.
That's why there is a single bond between hydrogen and fluorine.
Hence both hydrogen and fluorine share one electron with each other, so option "A" is correct.

5 0

3 years ago

Water flows with constant speed through a garden hose that goes up to 27.5 cm high. if the water pressure is 132kpa at the botto

sergejj [24]

Answer:

The pressure at the top of the step is 129.303 kilopascals.

Explanation:

From Hydrostatics we find that the pressure difference between extremes of the water column is defined by the following formula, which is a particular case of the Bernoulli's Principle ( $v_{bottom}\approx v_{top}$ ):

$p_{bottom}-p_{top} = \rho\cdot g\cdot \Delta h$ (1)

$p_{bottom}$ , $p_{top}$ - Total pressures at the bottom and at the top, measured in pascals.

$\rho$ - Density of the water, measured in kilograms per cubic meter.

$\Delta h$ - Height difference of the step, measured in meters.

If we know that $p_{bottom} = 132000\,Pa$ , $\rho = 1000\,\frac{kg}{m^{3}}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $\Delta h = 0.275\,m$ , then the pressure at the top of the step is:

$p_{top} = p_{bottom}-\rho\cdot g\cdot \Delta h$

$p_{top} = 132000\,Pa-\left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.275\,m)$

$p_{top} = 129303.075\,Pa$

$p_{top} = 129.303\,kPa$

The pressure at the top of the step is 129.303 kilopascals.

6 0

2 years ago

How am I supposed to solve this?

RSB [31]

Answer:

4.02 km/hr

Explanation:

5 km/hr = 1.39 m/s

The swimmer's speed relative to the ground must have the same direction as line AC.

The vertical component of the velocity is:

uᵧ = us cos 45

uᵧ = √2/2 us

The horizontal component of the velocity is:

uₓ = 1.39 − us sin 45

uₓ = 1.39 − √2/2 us

Writing a proportion:

uₓ / uᵧ = 121 / 159

(1.39 − √2/2 us) / (√2/2 us) = 121 / 159

Cross multiply and solve:

159 (1.39 − √2/2 us) = 121 (√2/2 us)

220.8 − 79.5√2 us = 60.5√2 us

220.8 = 140√2 us

us = 1.115

The swimmer's speed is 1.115 m/s, or 4.02 km/hr.

7 0

2 years ago

Read 2 more answers

What lasts longer... a perm, or a chemical relaxer for your hair ?

Dmitrij [34]

Answer:

Perm lasts longer

Explanation:

Cuz a chemical relaxer only lasts a few weeks and a perm lasts a few months hope this helps

7 0

2 years ago

A man does 4,475 J of work in the process of pushing his 2.50 103 kg truck from rest to a speed of v, over a distance of 26.0 m.

Tcecarenko [31]

Answer:

a) 1.89 m/s b) 172.1 N

Explanation:

Applying the work-energy theorem, if we can neglect the friction between truck and road, the total change in kinetic energy must be equal to the work done by the external forces.
This work, is just 4,475 J.
So we can write the following equation:

$\Delta K = \frac{1}{2} * m*v^{2} = 4,475 J$

where m= mass of the truck = 2.5*10³ kg.
So, we can find the speed v, as follows:

$v =\sqrt{\frac{2*W}{m}} =\sqrt{\frac{2*4,475J}{2.5e3kg} } = 1.89 m/s$

The work done by the man, is just the horizontal force applied, times the displacement produced by the force horizontally:

$W = F*d$

We can solve for F, as follows:

$F = \frac{W}{d} = \frac{4,475 J}{26.0m} = 172.1 N$

4 0

2 years ago