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3 years ago

if a piece of sea floor has moved 50 km in 5 million years what is the yearly rate of sea-floor motion?

Physics

2 answers:

BartSMP [9]3 years ago

6 0

<span>0.0001 km / year or 10^-5 km/year just take 50 km and divide it by 5 million</span>

Sonbull [250]3 years ago

5 0

50 km = 50,000 meters

   (50,000 meters) / (5 million years)

   =    (5 x 10⁴) / (5 x 10⁶)    (meters/year)

   = 1 x 10⁻²    meter/year

   = 1 centimeter per year

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Why does ice reflect more energy compared to water?

Stells [14]

Answer:

while ice is made by water again it melts and becomes water. water is colourless and odourless and has no taste but ice is only cold and hard. water is used for drinking and other things. but is for freshness and it never flows

Explanation:

so ice reflect more energy compared to water

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2 years ago

What is the correct formula for power? A. Power = work / time B. Power = work * time C. Power = force * distance D. Power = work

steposvetlana [31]

Answer:

A. Power = Work / Time

Explanation:

Power is the amount of work done over time, or rather the rate of work, which is given by the unit of watts (W). Since work is defined by Force * Displacement, we can also say Power = Force * Displacement / Time.

3 0

3 years ago

Read 2 more answers

The pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25°C, the pres

12345 [234]

Answer:0.0704 kg

Explanation:

Given

initial Absolute pressure $(P_1)$ =210+101.325=311.325

$T_1=25^{\circ}\approx 298 K$

$V=0.025 m^3$

$T_2=50^{\circ}\approx 323 K$

as the volume remains constant therefore

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$\frac{311.325}{298}=\frac{P_2}{323}$

$P_2=337.44 KPa$

therefore Gauge pressure is 337.44-101.325=236.117 KPa

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$m_1=0.91 kg$

Final mass $m_2=\frac{P_2V}{RT_2}=\frac{311.325\times 0.025}{0.0287\times 323}$

$m_2=0.839$

Therefore $m_1-m_2$ =0.91-0.839=0.0704 kg of air needs to be removed to get initial pressure back

4 0

3 years ago

A hydrogen atom in a galaxy moving with a speed of 6.65×106 m/???? away from the Earth emits light with a wavelength of 5.13×10−

Mumz [18]

Answer:

The observed wavelength on Earth from that hydrogen atom is $5.24\times 10^{-7}\ m$ .

Explanation:

Given that,

The actual wavelength of the hydrogen atom, $\lambda_a=5.13\times 10^{-7}\ m$

A hydrogen atom in a galaxy moving with a speed of, $v=6.65\times 10^6\ m/s$

We need to find the observed wavelength on Earth from that hydrogen atom. The speed of galaxy is given by :

$v=c\times \dfrac{\lambda_o-\lambda_a}{\lambda_a}$

$\lambda_o$ is the observed wavelength

$\lambda_o=\dfrac{v\lambda_a}{c}+\lambda_a\\\\\lambda_o=\dfrac{6.65\times 10^6\times 5.13\times 10^{-7}}{3\times 10^8}+5.13\times 10^{-7}\\\\\lambda_o=5.24\times 10^{-7}\ m$

So, the observed wavelength on Earth from that hydrogen atom is $5.24\times 10^{-7}\ m$ . Hence, this is the required solution.

8 0

2 years ago

You drop a 20 Ton object and a

Alborosie

Explanation:

it dosent depend on the weights of the items. I'll reach the ground at same time taking as no air friction or restrictions.

i.e

v = u + gt

whte v is final velocity of the object

u is initial velocity of the object

g is acceleration due to gravity and

t is time. thanks

please if found helpful rate brainliest

7 0

1 year ago