You are playing baseball and hit a single, running to first base. Your distance and displacement are

(20 points) You are at the center of a boat and have been rowing the boat for a long time. You weigh only 80 kg and your 120 kg

valina [46]

Answer:

Explanation:

From the given information:

Let the first weight be $m_ 1$ = 80 kg

The weight of the buddy be $m_2$ = 120 kg

The weight of Bubba be $m_3$ = 60 kg

Also, since you and Budda are a distance of 4m to each other, then the length to which both meet buddy will be:

$x_1 = x_3 = \dfrac{4}{2} \\ \\ = 2$

The length of the boat be $x_2$ = 4 m

∴

We can find the center of mass of the system by using the formula:

$X_{CM} = \dfrac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3} \\ \\ X_{CM} = \dfrac{(80 \times 2)+(120\times4)+(60\times2)}{80+120+60} \\ \\ X_{CM} = \dfrac{160+480+120}{260} \\ \\ \mathbf{X_{CM} = 2.923}$

4 0

2 years ago

g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

weeeeeb [17]

Complete Question

A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

Answer:

The value $v_2 = 4 \sqrt{10} \ m/s$

Explanation:

From the question we are told that

The charge on the first sphere is $q_1 = 2\mu C = 2*10^{-6} \ C$

The charge on the second sphere is $q_2 = 8 \mu C = 8*10^{-6} \ C$

The mass of the second charge is $m = 1.50 \ g = 1.50 *10^{-3} \ kg$

The distance apart is $d = 0.4 \ m$

The speed of the second sphere is $v_1 = 20 \ ms^{-1}$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.8 \ m$ is mathematically represented

$Q = KE + U$

Here KE is the kinetic energy which is mathematically represented as

$KE = \frac{1 }{2} m (v_1)^2$

substituting value

$KE = \frac{1 }{2} * ( 1.50 *10^{-3}) (20 )^2$

$KE = 0.3 \ J$

And U is the potential energy which is mathematically represented as

$U = \frac{k * q_1 * q_2 }{d }$

substituting values

$U = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.8 }$

$U = 0.18 \ J$

So

$Q = 0.3 + 0.18$

$Q = 0.48 \ J$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.4 \ m$ is mathematically represented

$Q_f = KE_f + U_f$

Here $KE_f$ is the kinetic energy which is mathematically represented as

$KE_f = \frac{1 }{2} m (v_2^2$

substituting value

$KE_f = \frac{1 }{2} * ( 1.50 *10^{-3}) (v_2 )^2$

$KE_f = 7.50 *10^{ -4} (v_2 )^2$

And $U_f$ is the potential energy which is mathematically represented as

$U_f = \frac{k * q_1 * q_2 }{d }$

substituting values

$U_f = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.4 }$

$U_f = 0.36 \ J$

From the law of energy conservation

$Q = Q_f$

So

$0.48 = 0.36 +(7.50 *10^{-4} v_2^2)$

$v_2 = 4 \sqrt{10} \ m/s$

6 0

2 years ago

A student touches sphere x and moves it close to, but not touching sphere y. What are the natures of the charges left on the two

e-lub [12.9K]

No charge I know this because

7 0

2 years ago

Three bullets are fired simultaneously by three guns aimed toward the center of a circle where they mash into a stationary lump.

earnstyle [38]

Answer:

2,87 * $10^{-3}$

Explanation:

When the bullets meet at the center and collide, since momentum is a vectoral quantity, their momentum vectors even up and are sumof zero. Formula of momentum is P = m.v , where m is mass and v is velocity. Let’s name the first two bullets as x,y and the one which mass is unknown as z. Then calculate momentum of x and y:

Px= 5,30 * $10^{-3}$ * 301 = 1,5953 kg*m/s

Py= 5,30 * $10^{-3}$ * 301 = 1,5953 kg*m/s

The angle between x and y bullets is 120°, and we know that if the angle between two equal magnitude vectors is 120°, the magnitude of the resultant vector will be equal to first two and placed in exact middle of two vectors. So we can say total momentum of x and y (Px+Py) equals to 1,5953 kg*m/s as well (Shown in the figure).

For z bullet to equalize the total momentum of x and y bullets, it needs to have the same amount of momentum in the opposite way.

Pz = 1,5953 = m * 554

m = 2,87 * $10^{-3}$ kg

8 0

2 years ago

There are several ways to model a compound. One type of model is shown. 4 C's are connected in a line by 3 black lines. The C on

stealth61 [152]

Answer:

The chemical formula for the molecule is $C_4H_9O_2$

Explanation :

Molecular formula : It is the chemical formula which depicts the actual number of atoms of each element present in the compound.

Structural formula : It is a formula in which the lines are used between the bonded atoms and the atoms are also shown in the structural formula.

In the structural formula, the lines are used between the atoms.

As per given information of compound we conclude that, there are 4 carbon atoms, 9 hydrogen atoms and 2 oxygen atoms.

Thus, the chemical formula for the given molecule will be, $C_4H_9O_2$

7 0

2 years ago