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3 years ago

7

a soccer player is running at 6m/s. he then stumbles over an opponents foot falling, and rolling to a stop. this took 4 seconds.

what was his acceleration?

2 answers:

Natali [406]3 years ago

8 0

$a = \frac{(vf - vi)}{t} \\ a = \frac{0 - 6}{4} \\ a = - 1.5 \frac{m}{ {s}^{2} }$
top one is the formula
mid- i plug in the numbers
bottom- answer

marishachu [46]3 years ago

5 0

Answer:

The acceleration of the soccer player is $1.5\ m/s^2$ and it is decelerating.

Explanation:

Given that,

Initial speed of a soccer player, u = 6 m/s

Finally, it stops, v = 0

Time, t = 4 s

We need to find the acceleration of a soccer player. We know that the rate of change of velocity is called acceleration of an object. It is given by :

$a=\dfrac{v-u}{t}\\\\a=\dfrac{0-6}{4}\\\\a=-1.5\ m/s^2$

So, the acceleration of the soccer player is $1.5\ m/s^2$ and it is decelerating.

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The maximum distance from the Earth to the Sun (at aphelion) is 1.521 1011 m, and the distance of closest approach (at perihelio

LUCKY_DIMON [66]

Answer:

29274.93096 m/s

$2.73966\times 10^{33}\ J$

$-5.39323\times 10^{33}\ J$

$2.56249\times 10^{33}\ J$

$-5.21594\times 10^{33}$

Explanation:

$r_p$ = Distance at perihelion = $1.471\times 10^{11}\ m$

$r_a$ = Distance at aphelion = $1.521\times 10^{11}\ m$

$v_p$ = Velocity at perihelion = $3.027\times 10^{4}\ m/s$

$v_a$ = Velocity at aphelion

m = Mass of the Earth = 5.98 × 10²⁴ kg

M = Mass of Sun = $1.9889\times 10^{30}\ kg$

Here, the angular momentum is conserved

$L_p=L_a\\\Rightarrow r_pv_p=r_av_a\\\Rightarrow v_a=\frac{r_pv_p}{r_a}\\\Rightarrow v_a=\frac{1.471\times 10^{11}\times 3.027\times 10^{4}}{1.521\times 10^{11}}\\\Rightarrow v_a=29274.93096\ m/s$

Earth's orbital speed at aphelion is 29274.93096 m/s

Kinetic energy is given by

$K=\frac{1}{2}mv_p^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(3.027\times 10^{4})^2\\\Rightarrow K=2.73966\times 10^{33}\ J$

Kinetic energy at perihelion is $2.73966\times 10^{33}\ J$

Potential energy is given by

$P=-\frac{GMm}{r_p}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.471\times 10^{11}}\\\Rightarrow P=-5.39323\times 10^{33}$

Potential energy at perihelion is $-5.39323\times 10^{33}\ J$

$K=\frac{1}{2}mv_a^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(29274.93096)^2\\\Rightarrow K=2.56249\times 10^{33}\ J$

Kinetic energy at aphelion is $2.56249\times 10^{33}\ J$

Potential energy is given by

$P=-\frac{GMm}{r_a}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.521\times 10^{11}}\\\Rightarrow P=-5.21594\times 10^{33}$

Potential energy at aphelion is $-5.21594\times 10^{33}\ J$

6 0

2 years ago

How much work, in N*m, is done when a 10.0 N force moves an object 2.5 m?

Alik [6]

W = F * d
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So the answer you want is the third one down.

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tester [92]

Answer:

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Explanation:

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If you are given the mass of an object in pounds, the time in seconds, and the distance in feet, what must you do before you can

Alexandra [31]

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Explanation:

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You are given :

mass- in pounds

for distance - in feet

before you do any calculation first you have to convert pounds in kilograms

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