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2 years ago

6

A cat's displacement is 15 meters to the right in 7.0 seconds. If, at the start of the 7.0 seconds, the cat was moving at a velo

city of 2.0 m/s left what was its final velocity?

1 answer:

kogti [31]2 years ago

3 0

Answer:

6.3 m/s

Explanation:

From the given information:

The displacement (x) = 15 m

time (t) = 7.0 s

initial velocity = -2.0 m/s (since it is moving in the opposite direction)

We need to determine the acceleration then find the final velocity.

By applying the kinematics equation:

$x = ut + \dfrac{1}{2}at^2$

$15 = (-2.0)(7.0) + \dfrac{1}{2}a(7.0)^2\\ \\ 15 = -14.0 + \dfrac{49}{2}a \\ \\ 29= 24.5a \\ \\ a= \dfrac{29}{24.5} \\ \\ a = 1.184 \ m/s^2$

Now, to determine the final velocity by using the equation:

v = u + at

v = -2 + 1.184(7.0)

v = 6.288 m/s

v ≅ 6.3 m/s

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How many days of a diet of 1910 large calories are equivalent to the gravitational energy change from sea level to the top of Mo

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Answer:

0.58 days is needed

Explanation:

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2 years ago

Drop a rock from a 5-m height and it accelerates at 9.81 m/s^2 and strikes the ground 1s later. Drop the sane rock from a height

KiRa [710]

Answer:

A. The same amount

Explanation:

The acceleration at which objects free falls to the ground on Earth is constant and its value is always $9.81 m/s^2$ , regardless of their mass (in this problem we neglect air resistance).

So, it doesn't matter if the two rocks are different or they are launched from different heights: their acceleration will be exactly the same.

This can be proved this way: first of all, the force of gravity exerted on every object is equal to the weight of the object,

$F=mg$

where m is the mass of the object and g the acceleration of gravity.

However, we also know for Newton's second law that

$F=ma$

where a is the acceleration of the object.

Combining the two equations,

$ma=mg\\a=g$

So, the acceleration of an object in free fall is exactly the acceleration of gravity.

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2 years ago

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of -0.9537 N when separated by

scoray [572]

Answer:

<em>The initial charges on the spheres are </em> $6.796\ 10^{-6}\ c$ and $-3.898\ 10^{-6}\ c$

Explanation:

<u>Electrostatic Force </u>

Two charges q1 and q2 separated a distance d exert a force on each other which magnitude is computed by the known Coulomb's formula

$\displaystyle F=\frac{K\ q_1\ q_2}{d^2}$

We are given the distance between two unknown charges d=50 cm = 0.5 m and the attractive force of -0.9537 N. This means both charges are opposite signs.

With these conditions we set the equation

$\displaystyle F_1=\frac{K\ q_1\ q_2}{0.5^2}=-0.9537$

Rearranging

$\displaystyle q_1\ q_2=\frac{-0.9537(0.5)^2}{k}$

Solving for q1.q2

$\displaystyle q_1\ q_2=-2.6492.10^{-11}\ c^2\ \ ......[1]$

The second part of the problem states the spheres are later connected by a conducting wire which is removed, and then, the spheres repel each other with an electrostatic force of 0.0756 N.

The conducting wire makes the charges on both spheres to balance, i.e. free electrons of the negative charge pass to the positive charge and they finally have the same charge:

$\displaystyle q=\frac{q_1+q_2}{2}$

Using this second condition:

$\displaystyle F_2=\frac{K\ q^2}{0.5^2}=\frac{K(q_1+q_2)^2}{(4)0.5^2}=0.0756$

$\displaystyle q_1+q_2=2.8983\ 10^{-6}\ C$

Solving for q2

$\displaystyle q_2=2.8983\ 10^{-6}\ C-q_1$

Replacing in [1]

$\displaystyle q_1(2.8983\ 10^{-6}-q_1)=-2.64917.10^{-11}$

Rearranging, we have a second-degree equation for q1.

$\displaystyle q_1^2-2.8983.10^{-6}q_1-2.64917.10^{-11}=0$

Solving, we have two possible solutions

$\displaystyle q_1=6,796.10^{-6}\ c$

$\displaystyle q_1=-3.898.10^{-6}\ c$

Which yields to two solutions for q2

$\displaystyle q_2=-3.898.10^{-6}\ c$

$\displaystyle q_2=6.796.10^{-6}\ c$

Regardless of their order, the initial charges on the spheres are $6.796\ 10^{-6}\ c$ and $-3.898\ 10^{-6}\ c$

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2 years ago