What is the velocity of a 0.8kg ball that has a momentum of 3 kg*m/s?

When the speed of your car is doubled, by what factor does its kinetic energy increase?

Tju [1.3M]

Answer:

a. 4

Explanation:

Hi there!

The equation of kinetic energy (KE) is the following:

KE = 1/2 · m · v²

Where:

m = mass of the car.

v = speed of the car.

Let´s see how would be the equation if the velocity is doubled (2 · v)

KE2 = 1/2 · m · (2 · v)²

Distributing the exponent:

KE2 = 1/2 · m · 2² · v²

KE2 = 1/2 · m · 4 · v²

KE2 = 4 (1/2 · m · v²)

KE2 = 4KE

Doubling the velocity increased the kinetic energy by 4.

7 0

3 years ago

determine the force of gravitational attraction between a 78kg boy sitting 2 meters away from a 65kg girl.

bixtya [17]

Answer:

$F=8.45\times 10^{-8}\ N$

Explanation:

Given that,

Mass of a boy is 78 kg

Mass of a girl is 65 kg

We need to find the force of gravitational attraction between them if they are 2 m away.

The formula for the gravitational force is given by :

$F=\dfrac{Gm_1m_2}{r^2}\\\\F=\dfrac{6.67\times 10^{-11}\times 78\times 65}{(2)^2}\\\\=8.45\times 10^{-8}\ N$

So, the force between them is $8.45\times 10^{-8}\ N$ .

8 0

2 years ago

In a vacuum, two particles have charges of q1 and q2, where q1 = +3.11 μC. They are separated by a distance of 0.241 m, and part

lesya [120]

Answer:

Charge of particle 2, $q_2=-7.13\ \mu C$

Explanation:

Given that,

Charge 1, $q_1=3.11\ \mu C=3.11\times 10^{-6}\ C$

The distance between charges, r = 0.241 m

Force experienced by particle 1, F = 3.44 N

We need to find the magnitude of electric charge 2. It can be calculated using formula of electrostatic force. It is given by :

$F=k\dfrac{q_1q_2}{r^2}$

$q_2=\dfrac{Fr^2}{kq_1}$

$q_2=\dfrac{3.44\times (0.241)^2}{9\times 10^9\times 3.11\times 10^{-6}}$

$q_2=7.13\times 10^{-6}\ C$

or

$q_2=7.13\ \mu C$

So, the magnitude of electric charge 2 is $q_2=7.13\ \mu C$ . Since, the force is attractive then the magnitude of charge 2 must be negative.

5 0

3 years ago

The half-life of Iodine-131 is 8.0252 days. If 14.2 grams of I-131 is released in Japan and takes 31.8 days to travel across the

MakcuM [25]

Answer:

Explanation:

Half-life problems are modeled as exponential equations. The half-life formula is $P=P_o\left (\dfrac{1}{2} \right)^{\frac{t}{k}}$ where $P_o$ is the initial amount, $k$ is the length of the half-life, $t$ is the amount of time that has elapsed since the initial measurement was taken, and $P$ is the amount that remains at time $t$ .

$P=14.2\left (\dfrac{1}{2} \right)^{\frac{t}{8.0252}}$

<u>Deriving the half-life formula</u>

If one forgets the half-life formula, one can derive an equivalent equation by recalling the basic an exponential equation, $y=a b^{t}$ , where $t$ is still the amount of time, and $y$ is the amount remaining at time $t$ . The constants a and b can be solved for as follows:

Knowing that amount initially is 14.2g, we let this be time zero:

$y=a b^{t}$

$(14.2)=ab^{(0)}$

$14.2=a *1$

$14.2=a$

So, $a=14.2$ , which represents out initial amount of the substance, and our equation becomes: $y=14.2 b^{t}$

Knowing that the "half-life" is 8.0252 days (note that the unit here is "days", so times for all future uses of this equation must be in "days"), we know that the amount remaining after that time will be one-half of what we started with:

$\left(\frac{1}{2} *14.2 \right)=14.2 b^{(8.0252)}$

$\dfrac{7.1}{14.2}=\dfrac{14.2 b^{8.0252}}{14.2}$

$0.5=b^{8.0252}$

$\sqrt[8.0252]{\frac{1}{2}}=\sqrt[8.0252]{b^{8.0252}}$

$\sqrt[8.0252]{\frac{1}{2}}=b$

Recalling exponent properties, one could find that $\left ( \frac{1}{2} \right )^{\frac{1}{8.0252}}=b$ , which will give the equation identical to the half-life formula. However, recalling this trivia about exponent properties is not necessary to solve this problem. One can just evaluate the radical in a calculator:

$b=0.9172535661...$

Using this decimal approximation has advantages (don't have to remember the half-life formula & don't have to remember as many exponent properties), but one minor disadvantage (need to keep more decimal places to reduce rounding error).

So, our general equation derived from the basic exponential function is:

$y=14.2* (0.9172535661)^t$ or $y=14.2*(0.5)^{\frac{t}{8.0252}}$ where y represents the amount remaining at time t.