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3 years ago

It was once recorded that a Jaguar

Physics

1 answer:

Artyom0805 [142]3 years ago

6 0

Answer:

71.85 m/s

Explanation:

Given the following :

Length of skid marks left by jaguar (s) = 290 m

Skidding Acceleration (a) = - 8.90m/s²

Final velocity of jaguar (v) = 0

Speed of Jaguar before it Began to skid =?

Hence, initial speed of jaguar could be obtained using the formula :

v² = u² + 2as

Where

v = final speed of jaguar ; u = initial speed of jaguar(before it Began to skid) ; a = acceleration of jaguar ; s = distance /length of skid marks left by jaguar

0² = u² + (2 × (-8.90) × 290)

0 = u² + (-5,162)

u² = 5162

Take the square root of both sides

u = √5162

u = 71.847 m/s

u = 71.85m/s

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qaws [65]

$2.6×10^6\:\text{m}$

Explanation:

The acceleration due to gravity g is defined as

$g = G\dfrac{M}{R^2}$

and solving for R, we find that

$R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)$

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration $F_c$ experienced by the satellite is equal to the gravitational force $F_G$ or

$F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)$

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$v = \dfrac{2\pi r}{T}$

where r is the radius of the satellite's orbit in meters and T is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

$\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}$

Solving for M, we get

$M = \dfrac{4\pi^2 r^3}{GT^2}$

Putting this expression back into Eqn(1), we get

$R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}$

$\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}$

$\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}$

$\:\:\:\:= 2.6×10^6\:\text{m}$

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3 years ago

An angry physics student releases a wrecking ball as shown. The wrecking ball is just about to hit the building at the final tim

daser333 [38]

Answer:

the force between the building and the ball is non-conservative (friction-type force)

Explanation

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When the ball moves it has a kinetic energy and if its height increases or decreases its potential energy also changes, but the sum of being must be equal to the initial work.

When the ball arrives and collides with the building, non-conservative forces, of various kinds; rubbing, breaking, etc. It transforms this energy into a part of heat and another in mechanical energy that the building must absorb, let us destroy its wall

Consequently, the force between the building and the ball is non-conservative (friction-type force

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3 years ago

An object of mass m is traveling in a circle with centripetal force Fc. If the velocity of the object is v, what is the radius o

borishaifa [10]

Hi there!

Recall the equation for centripetal force:
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We can rearrange the equation to solve for 'r'.

Multiply both sides by r:
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Divide both sides by Fc:
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3 years ago

a 45 kg ice skater initially skating at a velocity of 3 m/s speeds up to a velocity of 5 m/s. calculate the difference in the ma

ad-work [718]

Answer: 90 kgm/s

Explanation:

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In this situation the skater has two values of momentum:

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Where:

$V_{1}=3 m/s$

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So, if we want to calculate the difference in the magnitude of the skater's momentum, we have to write the following equation(assuming the mass of the skater remains constant):

$p=p_{2}-p_{1}=m.V_{2}-m.V_{1}$