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Hatshy [7]
3 years ago
5

How long will a plane have to fly continuously with 1100 miles per hour in order to cover the same distance as that from Earth t

o the next star, Proxima Centauri?
Physics
1 answer:
vaieri [72.5K]3 years ago
4 0

Explanation :

It is given that,

Velocity of plane, v = 1100 mi/h

Proxima Centauri is a red dwarf star. Its distance from Earth to  Proxima Centauri is 4.243 light year.

Or we can say that the distance between Earth and  Proxima Centauri is,

d=2.49\times 10^{13}\ miles

We know that, time t=\dfrac{d}{s}

t=\dfrac{2.49\times 10^{13}\ miles}{1100\  miles/h}

t=0.22\times 10^{11}\ h

or

t = 2.5 years ( approx )

Hence, this is the required solution.

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A runner with a mass of 60kg accelerates at 2.2m/s2. What is the runner’s net force?
abruzzese [7]
As we know that, f=ma where
f= net force
m=mass of body
a=acceleration
Substitute m and a in the formula and you will get the answer
3 0
3 years ago
Biologists think that some spiders "tune" strands of their web to give enhanced response at frequencies corresponding to those a
Nat2105 [25]

Answer:

Explanation:

Given

diameter d=20 \mu m

density \rho =1300 kg/m^3

frequency \nu =150 Hz

Length of silk strand L=14 cm

Velocity in the string is as follows

\nu =\sqrt{\frac{T}{\mu }}

The expression for Fundamental Frequency

f=\frac{\nu }{2l}

f=\frac{1}{2l}\times \sqrt{\frac{T}{\mu }}

f=\frac{1}{2l}\times \sqrt{\frac{T}{\frac{m}{l}}}

f=\frac{1}{2l}\times \sqrt{\frac{Tl}{\rho V}}

Squaring

f^2=\frac{1}{4l^2}\times \frac{Tl}{\rho V}

T=4\rho \cdot A\cdot l^2\cdot f^2

T=4\times 1300\times \frac{\pi }{4}(20\times 10^{-6})^2\times (0.14)^2\times 150^2

T=7.2\times 10^{-4} N

8 0
4 years ago
A 3.42 kg mass hanging vertically from a spring on the Earth (where g = 9.8 m/s2) undergoes simple oscillatory motion. If the sp
sineoko [7]

Answer:

Time period of oscillation on moon will be equal to 3.347 sec

Explanation:

We have given mass which is attached to the spring m = 3.42 kg

Spring constant K = 12 N/m

We have to find the period of oscillation

Period of oscillation is equal to T=2\pi \sqrt{\frac{m}{K}}, here m is mass and K is spring constant

So period of oscillation T=2\times 3.14\times \sqrt{\frac{3.42}{12}}

T=2\times 3.14\times 0.533=3.347sec

So time period of oscillation will be equal to 3.347 sec

As it is a spring mass system and from the relation we can see that time period is independent of g

So time period will be same on earth and moon

3 0
3 years ago
Two children of mass 18 kg and 29 kg sit balanced on a seesaw with the pivot point located at the center of the seesaw. If the c
DedPeter [7]

Answer:

3.085 [m].

Explanation:

1) The rule:

m₁*g*l₁=m₂*g*l₂, where m₁ and l₁ - the mass and distance for the small child, m₂ and l₂ - for the big child;

2) according to the condigion l₁+l₂=5, then

3) it is possible to make up the system:

\left \{ {{l_1+l_2=5} \atop {m_1*l_1=m_2*l_2}} \right. \ = > \ \left \{ {{l_1=5-l_2} \atop {18*(5-l_2)=29*l_2}} \right. \ = > \ \left \{ {{l_1=\frac{145}{47}} \atop {l_2=\frac{90}{47}}} \right.

4) finally, l₁=145/47≈3.085 [m].

7 0
2 years ago
1. Since sleep is so important, we might wonder why people so often fail to get a sufficient amount
ziro4ka [17]

Answer: 9/10

Explanation:

because it's really important and makes you energetic

8 0
3 years ago
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