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statuscvo [17]
3 years ago
9

A particle R moves with a velocity of 6 m/s due East and another particle S moves at 8 m/s due South. What is the magnitude of t

he velocity of R w.r.t S?
Physics
1 answer:
cluponka [151]3 years ago
6 0

Answer:

10

Explanation:

We can look at this problem as a triangle, r is the hypotenuse so if we take the square root of 6^2+8^2 we get 10

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What is the mass of a large ship that has a momentum of 1.60x10^9 kg*m/s and is moving at a velocity of 10m/s?
Elden [556K]

Answer:

160000000 kg.

Explanation:

p=mv

p=1.6x10^9

v=10m/s

rearrange and substitute:

(1.6x10^9)=m(10)

m=(1.6x10^9)/10

m= 1.6x10^8 kg.

7 0
3 years ago
What are used by organisms for long-term energy storage
kifflom [539]

Answer:

Fatty acids.

Explanation:

-Long term energy storage is stored in the form of triglycerides .

-They are efficient storing molecules. They are more efficient thatn glycogen(cabohydrates).

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4 0
3 years ago
A meter stick A hurtles through space at a speed v = 0.25c relative to you, with its length aligned with the direction of motion
yaroslaw [1]

Answer:

L_0\approx1.0328\ m

Explanation:

Given:

  • relativistic length of stick A, L=1\ m
  • relativistic velocity of stick A with respect to observer, v=0.25c=7.5\times 10^{7}\ m.s^{-1}

<em>Since the object is moving with a velocity comparable to the velocity of light  with respect to the observer therefore the length will appear shorter according to the theory of relativity.</em>

<u> Mathematical expression of the theory of relativity for length contraction:</u>

L=\frac{L_0}{\gamma}

where:

L = relativistic length

L_0= original length at rest

\gamma = Lorentz factor =\frac{1}{\sqrt{1-\frac{v^2}{c^2} } }

\Rightarrow 1=\frac{L_0}{\frac{1}{\sqrt{1-\frac{(0.25c)^2}{c^2} } }}

L_0=\frac{1}{\sqrt{1-\frac{(0.25c)^2}{c^2} } }

L_0\approx1.0328\ m

4 0
3 years ago
A satellite is put in a circular orbit about Earth with a radius equal to 35% of the radius of the Moon's orbit. What is its per
Debora [2.8K]

Answer:

0.21 lunar month

Explanation:

the radius of moon = r₁

time period of the moon = T₁ = 1 lunar month

The radius of the satellite = 0.35 r₁

Time period of satellite

The relation between time period and radius

              T\ \alpha\ \sqrt{r^3}

now,

              \dfrac{T_2}{T_1}=\dfrac{\sqrt{r_2^3}}{\sqrt{r_1^3}}

              \dfrac{T_2}{T_1}=\dfrac{\sqrt{0.35^3r_1^3}}{\sqrt{r_1^3}}

              \dfrac{T_2}{1}=\sqrt{0.35^3}

                              T₂ = 0.21 lunar month

hence, the time period of revolution of satellite is equal to 0.21 lunar month

6 0
3 years ago
Determine the correlation between coronal mass ejections from the Sun to the accumulation of the rare and valuable isotope He3 t
Reil [10]

Most ejections originate from active regions on the Sun's surface, such as groupings of sunspots associated with frequent flares. These regions have closed magnetic field lines, in which the magnetic field strength is large enough to contain the plasma.

6 0
3 years ago
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