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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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In an attempt to reduce the extraordinarily long travel times for voyaging to distant stars, some people have suggested travelin

alexandr402 [8]

Answer:

a) $v=0.999124c$

b) $E=7.566*10^{22}$

c) $E_a=760 times\ larger$

Explanation:

From the question we are told that

Distance to Betelgeuse $d_b=430ly$

Mass of Rocket $M_r=20000$

Total Time in years traveled $T_d=36years$

Total energy used by the United States in the year 2000 $E_{2000}=1.0*10^20$

Generally the equation of speed of rocket v mathematically given by

$v=\frac{2d}{\triangle t}$

$v=860ly/ \triangle t$

where

$\triangle t=\frac{\triangle t'}{(\sqrt{1-860/ \triangle t)^2}}$

$\triangle t=\frac{36}{(\sqrt{1-860/ \triangle t)^2}}$

$\triangle t=\sqrt{(860)^2+(36)^2}$

$\triangle t=860.7532$

Therefore

$v=\frac{860ly}{ 860.7532}$

$v=0.999124c$

Generally the equation of the energy E required to attain prior speed mathematically given by

$E=\frac{1}{\sqrt{1-(v/c)^2} }-1(20000kg)(3*10^8m/s)^2$

$E=7.566*10^{22}$

c)Generally the equation of the energy $E_a$ required to accelerate the rocket mathematically given by

$E_a=\frac{E}{E_{2000}}$

$E_a=\frac{7.566*10^{22}}{1.0*10^{20}}$

$E_a=760 times\ larger$

8 0

3 years ago

A hayride wagon is going down a spooky country road at 15 m/s when a Scarecrow appears in the roadway. The man at the wheel of t

nexus9112 [7]

Answer:

d = 68.18 m

Explanation:

Given that,

Initial velocity, u = 15 m/s

Finally it comes to stop, v = 0

Acceleration, a = -1.65 m/s²

Time, t = 2.5 s

We need to find the distance covered by the hayride before coming to a stop. Let d is the distance covered. Using third equation of motion to find it :

$v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{-(15)^2}{2\times -1.65}\\\\d=68.18\ m$

So, the hayride will cover a distance of 68.18 m.

6 0

3 years ago

A ball is thrown vertically downward at 10 m/s. what is it’s speed 1s and 2s later

soldi70 [24.7K]

Answer:

20 m/s

30 m/s

Explanation:

Given:

v₀ = -10 m/s

a = -9.8 m/s²

When t = 1 s:

v = v₀ + at

v = (-10 m/s) + (-9.8 m/s²) (1 s)

v = -19.8 m/s

When t = 2 s:

v = v₀ + at

v = (-10 m/s) + (-9.8 m/s²) (2 s)

v = -29.6 m/s

Rounded to one significant figures, the speed of the ball at 1 s and 2 s is 20 m/s and 30 m/s, respectively.

3 0

3 years ago

You comb your hair and the comb becomes negatively charged. Strictly speaking, how will the mass of your hair change? A. it wil

Step2247 [10]

Answer:

C. it will not change.

Explanation:

While combing, the rubbing of the comb with the hair, transfer of electron takes place from the hair to the comb and the comb becomes negatively charged. But, this transfer of electron does not make any considerable change in the mass of the hair. This is because the mass of an electron is highly negligible. Now, neglecting the mass of an electron, the transfer of the electrons from the hair to the comb makes charging of the comb, but no loss of mass in the hair. So, the mass of hair will no change.

7 0

3 years ago

An electron experiences a magnetic force with a magnitude of 4.90×10−15 n when moving at an angle of 60.0 ∘ with respect to a ma

Leya [2.2K]

Missing questions: "find the speed of the electron".

Solution:
the magnetic force experienced by a charged particle in a magnetic field is given by
$F=qvB \sin \theta$
where
q is the particle charge
v its velocity
B the magnitude of the magnetic field
$\theta$ the angle between the directions of v and B.

Re-arranging the formula, we find:
$v= \frac{F}{qB \sin \theta}$
and by substituting the data of the problem (the charge of the electron is $q=1.6 \cdot 10^{-19} C$ ), we find the velocity of the electron:
$v= \frac{F}{qB \sin \theta}= \frac{4.90 \cdot 10^{-15}N}{(1.6 \cdot 10^{-19}C)(3.70 \cdot 10^{-3} T)(\sin 60^{\circ})}=9.56 \cdot 10^6 m/s$

4 0

3 years ago