2. An auditorium has 58 seats in the first row, 62 seats in the second row, 66 seats in the third row, and so

Illustrated here are three different<br> of carbon. They vary by the number of

Basile [38]

D) isotopes; neutrons

Explanation:

The diagram illustrates the three different isotopes of carbon. They vary by the number of neutrons.

Isotopy is the existence of two or more atoms of the same element having the same atomic number but different mass numbers due to the differences in the number of neutrons in their various nuclei.

Isotopes of elements have the same electronic configuration
They share the same chemical properties
They differ in masses due the number of neutrons they contain.

Learn more:

Isotopes brainly.com/question/1915462

#learnwithBrainly

5 0

4 years ago

What are three ways in which heat is transferred

Sedaia [141]

Conduction, Convection, Radiation.

Hope this helps!

-Payshence

6 0

3 years ago

which relationship best represents the relationship between the magnitude of the centripetal acceleration and the speed of an ob

pashok25 [27]

the object's speed and centripetal acceleration in a circle with a constant radius is the exponential curve going up.

acceleration is the rate of change in both speed and direction of velocity over time. When something moves faster or slower in a straight line, it is said to have been accelerated. Because the direction is constantly shifting, motion on a circle accelerates even when the speed is constant. Both effects add to the acceleration for all other types of motion.

It is a vector quantity because acceleration has both a magnitude and a direction. Another vector quantity is velocity. The velocity vector change over a given period of time, divided by that period of time, is the definition of acceleration. The upper limit of the ratio of velocity change provides instantaneous acceleration (at a specific time and place).

Learn more about acceleration here:

brainly.com/question/591649

#SPJ4

5 0

1 year ago

If the third charge (–| q 3 |) is placed at point P, but not held fixed, it will experience a force and accelerate away from the

Mazyrski [523]

Complete Question

The complete question is shown on the first uploaded image

Answer:

A

The potential of this system is $U=6.75*10^{-7}J$

B

The electric potential at point p is $V_p= -900V$

C

The work required is $W= 9*10^{-7}J$

D

The speed of the charge is $v=600m/s$

Explanation:

A sketch to explain the question is shown on the second uploaded image

Generally the potential energy for a system of two charges is mathematically represented as

$U = \frac{kq_1 q_2}{d}$

where k is the electrostatic constant with a value of $k = 9*10^9 N m^2 /C^2$

q is the charge with a value of $q = 1*10^{-9}C$

d is the distance given as $d =5m$

Now we are given that $q_1 = q$ and $q_2 = 3q$ and

Now substituting values

$U = \frac{9*10^9 *1*10^{-9} * 3*10^{-9}}{5}$

$U=6.75*10^{-7}J$

The electric potential at point P is mathematically obtained with the formula

$V_p = V_{-q} + V_{-3q}$

I.e the potential at $q_1$ plus the potential at $q_2$

Now potential at $q_1$ is mathematically represented as

$V_{-q} = \frac{-kq}{s}$

and the potential at $q_2$ is mathematically represented as

$V_{-3q} = \frac{-3kq}{s}$

Now substituting into formula for potential at P

$V_p = \frac{-kq}{s} + \frac{-3kq}{s} = -\frac{4kq}{s}$

$= \frac{4*9*10^9 *1*10^{-9}}{4*10^{-2}}$

$V_p= -900V$

The Workdone to bring the third negative charge is mathematically evaluated as

$W =\Delta U = \frac{kq_1q_3}{s} + \frac{kq_2q_3}{s}$

$= \frac{kq*q}{s} + \frac{kq*3q}{s}$

$= \frac{4kq^2}{s}$

$= \frac{4* 9*10^9 * (1*10^{-9})^2}{4*10^{-2}}$

$W= 9*10^{-7}J$

From the Question are told that the charge $q_3$ would a force and an acceleration which implies that all its potential energy would be converted to kinetic energy.This can be mathematically represented as

$\Delta U = W = \frac{1}{2} m_{q_3} v^2$

$9*10^{-7} = \frac{1}{2} m_{q_3} v^2$

Where $m_{q_3} = 5.0*10^{-12}kg$

Now making v the subject we have

$v = \sqrt{\frac{9*10^{-12}}{5*10^{-12}*0.5} }$

$v=600m/s$

8 0

3 years ago

A 1130-kg car is held in place by a light cable on a smooth (frictionless) ramp. The cable makes an angle of 31.0° above the sur

zubka84 [21]

Answer:

T = 5163.89 N

Explanation:

Newton's first law:

∑F =0 Formula (1)

∑F : algebraic sum of the forces in Newton (N)

We define the x-axis in the direction parallel to the movement of the car on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the car

W: Weight of the car : In vertical direction

FN : Normal force : perpendicular to the ramp

T :Tension force: at angle of 31.0° above the surface of the ramp

Calculated of the Weight of the car (W)

W = m*g m: mass g:acceleration due to gravity

W = 1130-kg* 9.8 m/s² = 11074 N

x-y weight components

Wx = 11074 N*sin 25.0° = 4680.07 N

Wy = 11074 N*cos 25.0° = 10036.45 N

x-y Tension components

Tx = T*cos 25.0°

Ty = T*sin 25.0°

Newton's first law:

∑Fx =0 Formula (1)

Tx-Wx = 0

T*cos 25.0° - 4680.07 = 0

T*cos 25.0° = 4680.07

T = 4680.07 / cos 25.0°

T = 5163.89 N

4 0

3 years ago