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3 years ago

In which situation can you be at rest and moving at the same time

1 answer:

5 0

Answer:Being at rest is a relative term. Relative to earth we are at rest when we are sitting down. However, you are moving because the earth is moving its orbit around the sun.

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How do P-Waves and S-Waves form?

marissa [1.9K]

P waves are produced by all earthquakes. They are compression waves that form when rocks break due to pressure in the Earth. S waves are secondary waves that are also created during an earthquake. They travel at a slower speed than the p-waves.

S waves are the waves that come after the earthquake and P waves

6 0

3 years ago

Jose’s lab instructor gives him a solution of sodium phosphate that is buffered to a pH of 4. Because of an error that he made w

Kazeer [188]

I am thinking that maybe the problem is not with the calibration. It might be that the buffered solution is already expired since at this point the solution is already not stable and will give a different pH reading than what is expected.

7 0

3 years ago

7.22 Ignoring reflection at the air–water boundary, if the amplitude of a 1 GHz incident wave in air is 20 V/m at the water surf

Serga [27]

Answer:

z = 0.8 (approx)

Explanation:

given,

Amplitude of 1 GHz incident wave in air = 20 V/m

Water has,

μr = 1

at 1 GHz, r = 80 and σ = 1 S/m.

depth of water when amplitude is down to 1 μV/m

Intrinsic impedance of air = 120 π Ω

Intrinsic impedance of water = $\dfrac{120\pi}{\epsilon_r}$

Using equation to solve the problem

$E(z) = E_0 e^{-\alpha\ z}$

E(z) is the amplitude under water at z depth

E_o is the amplitude of wave on the surface of water

z is the depth under water

$\alpha = \dfrac{\sigma}{2}\sqrt{\dfrac{(120\pi)^2}{\Epsilon_r}}$

$\alpha = \dfrac{1}{2}\sqrt{\dfrac{(120\pi)^2}{80}}$

$\alpha =21.07\ Np/m$

now ,

$1 \times 10^{-6} = 20 e^{-21.07\times z}$

$e^{21.07\times z}= 20\times 10^{6}$

taking ln both side

21.07 x z = 16.81

z = 0.797

z = 0.8 (approx)

5 0

3 years ago

If the sprinter from the previous problem accelerates at that rate for 20 m, and then maintains that velocity for the remainder

kakasveta [241]

Question:

A 63.0 kg sprinter starts a race with an acceleration of 4.20m/s square. What is the net external force on him? If the sprinter from the previous problem accelerates at that rate for 20m, and then maintains that velocity for the remainder for the 100-m dash, what will be his time for the race?

Answer:

Time for the race will be t = 9.26 s

Explanation:

Given data:

As the sprinter starts the race so initial velocity = v₁ = 0

Distance = s₁ = 20 m

Acceleration = a = 4.20 ms⁻²

Distance = s₂ = 100 m

We first need to find the final velocity (v₂) of sprinter at the end of the first 20 meters.

Using 3rd equation of motion

(v₂)² - (v₁)² = 2as₁ = 2(4.2)(20)

v₂ = 12.96 ms⁻¹

Time for 20 m distance = t₁ = (v₂ - v ₁)/a

t₁ = 12.96/4.2 = 3.09 s

He ran the rest of the race at this velocity (12.96 m/s). Since has had already covered 20 meters, he has to cover 80 meters more to complete the 100 meter dash. So the time required to cover the 80 meters will be

Time for 100 m distance = t₂ = s₂/v₂

t₂ = 80/12.96 = 6.17 s

Total time = T = t₁ + t₂ = 3.09 + 6.17 = 9.26 s

T = 9.26 s

5 0

3 years ago

A youngster having a mass of 50.0 kg steps off a 1.00 m high platform. If she keeps her legs fairly rigid and comes to rest in 1

attashe74 [19]

The average act on her during the deceleration is 4.47 meters per second.

Explanation:

Given:

youngster mass m = 50.0 kg

She steps off a 1.00 m high platform that is s = 1 meter

She comes to rest in the 10-meter second

To Find:

The average force and momentum

Formulas:

p = m * v

F * Δ t = Δ p

vf^2= vi^2+2as

Solution:

a = 9.8 m/s

vi = 0

vf^2= 0+2(9.8)(1)

vf^2 = 19.6

vf = 4.47 m/s .

Therefore the average force is 4.47 m/s.

5 0

3 years ago