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3 years ago

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The famous black planet, haunch, has a radius of 106 m, a gravitational acceleration at the surface of 4 m/s2 , and the tangenti

al speed of any point on its equator is 103 m/s. a famous haunch ant has a mass of 40 kg and is standing on a spring scale at the equator. what is the reading of the spring scale?

1 answer:

levacccp [35]3 years ago

6 0

Gravity on the surface = 4 m/s^2
Now, the acceleration due to centripetal motion, a = v^2/R

Where,
v= 10^3 m/s, R = 10^6 m
Then,
a = (10^3)^2/(10^6) = 1 m^2/s

The net gravitational acceleration = 4-1 = 3 m/s^2

The reading on the spring scale = ma = 40*3 = 120 N

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f an electron rotates around the nucleus at that radius, what would be its speed according to the planetary model? (enter your a

Dima020 [189]

According to the planetary model, the speed of an electron revolving around the nucleus is given as follows;

v = e/√4 $\pi$ ∈₀mr.

The orbital speed of an electron, orbiting around a nucleus in a circular orbit of radius 50 consistent with the Bohr model, regularly called a planetary version, the electrons encircle the nucleus of the atom in precise allowable paths referred to as orbits. whilst the electron is in one of these orbits, its energy is fixed.

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1 year ago

Unlike electric field lines, magnetic field lines are continuous and always form closed loops.

Verizon [17]

Answer:

True

Explanation:

Magnetic field refers to a region around a magnetic material or magnet, where a magnetic effect i felt. The magnetic lines of force are continuous lines that always travels from north to south pole and forms a closed loop. It is for these reason that magnetic poles exists as dipoles and not mono-poles. Electric field lines however travel from positive to negative and cannot return to the positive hence they do not form closed loops, They can only travel in just one direction.

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3 years ago

in 1994, Vladimir Kurlivich , from Belarus, set the world record as the world's strongest weight lifter. he did this by lifting

tensa zangetsu [6.8K]

The free-body diagram of your question is; 2 downward forces (253 kg mass of barbell & 133 kg body mass of Kurlovich) acting together on a point supported by 2 upward forces as normal forces exerted by Kurlovich's feet.

Solving the normal forces exerted by 2 feet :

Summation of Forces Vertical = 0

2 Dowwnard Forces = 2 Upward Forces (2F)

253 + 133 = 2F

2F = 386 Kgs

F = 386 / 2

F = 193 Kgs (Normal Force Exerted by Each Foot)

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3 years ago

A meter stick is at rest on frictionless surface. A hockey puck is going towards the 30cm mark on the stick and is traveling per

faust18 [17]

Answer:

A) $0.306k$

B) $0.1325k$

C) $v_{s_{f}} =7.65(-i)+17.35(-j)$

D) $w=41.64 rads^{-1}$

Explanation:

Given:

hockey puck is moving towards 30cm mark perpendicular to the stick
$m_{s} = 0.05kg$
$m_{h} =0.17kg$
$v_{h_{i} } = 9 ms^{-1}$
after collision the puck is deflected 30°
$v_{h_{f} } =4.5 ms^{-1}$

To find the initial angular momentum about origin which is the 50th mark of the metre scale (It's COM) : angular momentum $L$ = $mv$ x $r$ where, v - is the velocity of the puck perpendicular to the radial vector

r - is the radius vector

∴

A) $L_{i} = m_{h}r*v_{h_{i} }\\L_{i}= 0.17*\frac{50-30}{100} (-i)*9(-j)\\L_{i} = 0.306 k$

B) after collision , it moves 30° from original path;

and it's speed = $\frac{9}{2} =4.5ms^{-1}$ ;

∴the perpendicular velocity $v_{per}$ = 4.5 $cos30$ = $2.25\sqrt{3}$ $ms^{-1}$

⇒ $L=m_{h}r*v=0.17*0.2(-i)*2.25\sqrt{3}(-j) \\L= 0.1325 k$

C) <em><u>since the net external force on the system is zero , the total momentum of the system can be conserved .</u></em>

thus ,

$m_{s}v_{s_{i}}+m_{h}v_{h_{i}}=m_{s}v_{s_{f}}+m_hv_{h_{f}}\\0+0.17*9(-j)=0.05*v_{s_{f}}+0.17*(2.25(i)+2.25\sqrt{3}(-j))\\$

solving this we get,

⇒ $v_{s_{f}} =7.65(-i)+17.35(-j)$

D) <u><em>since there is no external torque about the system ,the angular momentum can be conserved.</em></u>

$L_{h_{i}}= L_{h_{f}} + Iw$

where ,

$w$ is the angular velocity of the stick.

$I$ is the moment of inertia of the stick about COM :

$I =\frac{m_{s}l^{2}}{12} \\m_{s}=0.05kg\\l=1m\\I = 0.004167 kgm^{2}$

∴

⇒ $0.306k=0.1325k+0.004167w\\w=41.64 rads^{-1}$

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4 years ago

Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass g

Mrac [35]

Answer:62.82 kg

Explanation:

Given

Net External force(F)=49 N

Astronaut acceleration (a) measured to be $0.78 m/s^2$

According Newton's second law of motion we get

F=ma

$49=m\times 0.78$

$m=\frac{49}{0.78}$

m=62.82 kg

7 0

3 years ago