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2 years ago

14

A rabbit runs 4.4 m across a lawn, stops, then runs 2.2 m back in the opposite direction. What is the rabbit’s displacement from

its starting point?

1 answer:

Alla [95]2 years ago

6 0

Answer:

it is 2.2 m

Explanation:

because he goes back 2.2 m so 4.4 minus 2.2 equals 2.2

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Corey runs a 100-meter race. 7 seconds after the race started Corey is 45 meters from the starting line and reaches his max spee

RUDIKE [14]

Answer:

Corey's max speed is $7 \frac{m}{s}$
the distance Corey's covers in z seconds is $7 \frac{m}{s} * z \ s$
$d (z) = 45 m + 7 \frac{m}{s} * z$
$d (x) = 45 m + 7 \frac{m}{s} * (x-7 s)$

Explanation:

<h3>Corey's max speed</h3>

For constant speed, we know:

$v=\frac{distance}{time}$

The distance between the 80 meters and the 45 meters is:

$distance = 80 m - 45 m = 35 m$

and the time it took to reach the 80 meter will be:

$time = 12 s - 7 s = 5 s$

So, Corey's max speed is

$v_{max}=\frac{35 m}{5 s} = 7 \frac{m}{s}$

<h3>How far runs Corey</h3>

As the velocity of Corey's is $v_{max}$ , the distance Corey's covers in z seconds is

$distance = v_{max} * z \ s$

$distance = 7 \frac{m}{s} * z \ s$

<h3>What is Corey's distance from the starting line</h3>

At time 7 + z seconds the distance will be the 45 meters he covers in the first part of the race plus the distance he traveled at constant speed. this is:

$d (z) = 45 m + v_{max} * z$

$d (z) = 45 m +7 \frac{m}{s} * z$

At time x ( x greater or equal to 7 seconds) the distance will be the 45 meters he covers in the first part of the race plus the distance he traveled at constant speed. this is:

$d (x) = 45 m + v_{max} * (x-7 s)$

$d (x) = 45 m + 7 \frac{m}{s} * (x-7 s)$

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2 years ago

A gamma ray with an energy of 3.40 × 10-14 joules strikes a photographic plate. We know that Planck's constant is 6.63 × 10-34 j

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The correct answer is: "A. 5.13 x 10^19 hertz"

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2 years ago

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A runner whose mass is 54 kg accelerates from a stop to a speed of 7 m/s in 3 seconds. (A good sprinter can run 100 meters in ab

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Answer:

a. $F=126N$

b. $E_K=1323J$

Explanation:

Given:

$m=54kg$

$v=7 m/s$

$t= 3s$

The runner force average to find given the equations

a.

$F=m*a$

$a=\frac{v}{t}$

$F=m*\frac{v}{t}=54kg*\frac{7m/s}{3s}$

$F=126N$

b.

Work done by the system by this force so

$W=F*d$

$W=E_K$

$E_K=\frac{1}{2}*m*v^2$

$E_K=\frac{1}{2}*54kg*(7m/s)^2$

$E_K=1323J$

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