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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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3 years ago

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How long it takes to reach maximum height in seconds

olganol [36]

The dotted path is the path of the ball. it reaches it's maximum height at the top where vertical, y-velocity = 0
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initial y-velocity = 17.85 m/s

Use one of the kinematic equations with velocity and time. No displacement because we don't want to worry about figuring that out.

v = u - gt
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6 0

3 years ago

The velocity of a ball changes from ‹ 9, −6, 0 › m/s to ‹ 8.96, −6.12, 0 › m/s in 0.02 s, due to the gravitational attraction of

Alenkasestr [34]

Answer:

a) $a=(-2,-6,0)m/s^2$ , with a magnitude of $6.3m/s^2$

b) $\frac{\Delta p}{\Delta t}=(-0.24,-0.72,0)Kgm/s^2$ , with a magnitude of $0.76Kgm/s^2$

c) $F=(-0.24,-0.72,0)N$ , with a magnitude of $0.76N$

Explanation:

We have:

$v_{ix}=9m/s, v_{iy}=-6m/s, v_{iz}=0m/s\\v_{fx}=8.96m/s, v_{fy}=-6.12m/s, v_{fz}=0m/s\\t=0.02s, m=0.12Kg$

We can calculate each component of the acceleration using its definition $a=\frac{\Delta v}{\Delta t}$

$a_x=\frac{v_{fx}-v_{ix}}{t} = \frac{(8.96m/s)-(9m/s)}{0.02s} =-2m/s^2\\a_y=\frac{v_{fy}-v_{iy}}{t} = \frac{(-6.12m/s)-(-6m/s)}{0.02s} =-6m/s^2\\a_y=\frac{v_{fz}-v_{iz}}{t} = \frac{(0m/s)-(0m/s)}{0.02s} =0m/s^2\\$

The rate of change of momentum of the ball is $\frac{\Delta p}{\Delta t} = \frac{\Delta mv}{\Delta t} = \frac{m\Delta v}{\Delta t} = ma$

So for each coordinate:

$\frac{\Delta p_x}{\Delta t}=-0.24Kgm/s^2\\\frac{\Delta p_y}{\Delta t}=-0.72Kgm/s^2\\\frac{\Delta p_z}{\Delta t}=0Kgm/s^2$

And these are equal to the components of the net force since F=ma.

If magnitudes is what is asked:

$a=\sqrt{a_x+a_y+a_z} =6.3m/s^2\\F=ma=\frac{\Delta p}{\Delta t}=0.76N$

3 0

3 years ago

The star Rho1 Cancri is 57 light-years from the earth and has a mass 0.85 times that of our sun. A planet has been detected in a

s344n2d4d5 [400]

Answer:

82780.42123 m/s

14.45 days

Explanation:

m = Mass of the planet

M = Mass of the star = $0.85\times 1.989\times 10^{30}\ kg=1.69065\times 10^{30}\ kg$

r = Radius of orbit of planet = $0.11\times 149.6\times 10^{9}\ m=16.456\times 10^{9}\ m$

v = Orbital speed

The kinetic and potential energy balance is given by

$\frac{GMm}{r^2}=\frac{mv^2}{r}\\\Rightarrow v=\sqrt{\dfrac{Gm}{r}}\\\Rightarrow v=\sqrt{\dfrac{6.67\times 10^{-11}\times 1.69065\times 10^{30}}{16.456\times 10^{9}}}\\\Rightarrow v=82780.42123\ m/s$

The orbital speed of the star is 82780.42123 m/s

The orbital period is given by

$t=\frac{2\pi r}{v}\\\Rightarrow t=\dfrac{2\pi \times 16.456\times 10^{9}}{82780.42123}\\\Rightarrow t=1249040.48419\ seconds=\dfrac{1249040.48419}{24\times 60\times 60}=14.45\ days$

The orbital period is 14.45 days

5 0

3 years ago