Closer lines of__ indicate a stronger__ field. What are the missing words??

13. Which compound is most reactive?<br> ethene<br> ethyne<br> ethane

katen-ka-za [31]

Answer:

Ethyne consists of a triple bond that is most easier to be broken as compared to that of ethyne with a double bond and benzene single bonds. Therefore ethyne is most reactive among the given options.

Explanation:

4 0

2 years ago

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Find the potential energy of a 40kg cart that is on a hill that is 10 meters high

Vilka [71]

Explanation:

P.E. = mgh

PE = 40 × 10 × 10

PE = 4000 Joule

3 0

2 years ago

A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne

kati45 [8]

Answer:

The average emf that will be induced in the wire loop during the extraction process is 37.9 V

Explanation:

The average emf induced can be calculated from the formula

$Emf = -N\frac{\Delta \phi}{\Delta t}$

Where $N$ is the number of turns

$\Delta \phi$ is the change in magnetic flux

$\Delta t$ is the time interval

The change in magnetic flux is given by

$\Delta \phi = \phi _{f} - \phi _{i}$

Where $\phi _{f}$ is the final magnetic flux

and $\phi _{i}$ is the initial magnetic flux

Magnetic flux is given by the formula

$\phi = BAcos(\theta)$

Where $B$ is the magnetic field

$A$ is the area

and $\theta$ is the angle between the magnetic field and the area.

Initially, the magnetic field and the area are pointed in the same direction, that is, $\theta = 0^{o}$

From the question,

B = 1.5 T

and radius = 15.0 cm = 0.15 m

Since it is a circular loop of wire, the area is given by

$A = \pi r^{2}$

∴ $A = \pi (0.15)^{2}$

$A = 0.0225\pi$

∴ $\phi_{i} = (1.5)(0.0225\pi)cos(0^{o} )$

$\phi_{i} = (1.5)(0.0225\pi)$

( NOTE: $cos (0^{o}) = 1$ )

$\phi_{i} = 0.03375\pi$ Wb

For $\phi_{f}$

The field pointed upwards, that is $\theta = 90^{o}$ . Since $cos (90^{o}) = 0$

Then

$\phi_{f} = 0$

Hence,

$\Delta \phi = 0- 0.03375\pi$

$\Delta \phi = - 0.03375\pi$

From the question

$\Delta t = 2.8 ms = 2.8 \times 10^{-3} s$

Here, $N$ = 1

Hence,

$Emf = -N\frac{\Delta \phi}{\Delta t}$ becomes

$Emf = -(1)\frac{-0.03375\pi}{2.8 \times 10^{-3} }$

$Emf = 37.9 V$

Hence, the average emf that will be induced in the wire loop during the extraction process is 37.9 V.

5 0

3 years ago

Energy is released during the fission of pu-239 atoms as a result of the

pav-90 [236]

<h2>Answer: process of converting matter into energy</h2><h2></h2>

Nuclear fission consists of dividing a heavy nucleus into two or more lighter or smaller nuclei, by means of the bombardment with neutrons to make it unstable. In this process that takes place in the atomic nucleus, neutrons, gamma rays and <u>large amounts of energy are emitted. </u>

Then, with this division a great release of energy occurs and the emission of two or three neutrons, other particles and gamma rays.

This means fission is a process in which energy is released by the separation of the components of the nucleous of the atom.

In other words:

<h2>Matter is converted to energy .</h2>

6 0

3 years ago

a 230 kg roller coaster reaches the top of the steepest hill with a speed of 6.2 km/h. It then descends the hill, which is at an

igor_vitrenko [27]

Answer: 81.619 kJ

Explanation:

Given

Mass of roller coaster is $m=230\ kg$

It reaches the steepest hill with speed of $u=6.2\ km/h\ or \ 1.72\ m/s$

Hill to bottom is 51 m long with inclination of $45^{\circ}$

Height of the hill is $h=51\sin 45^{\circ}=36.06\ m$

Conserving energy to get kinetic energy at bottom

Energy at top=Energy at bottom

$\Rightarrow K_t+U_t=K_b+U_b\\\Rightarrow \dfrac{1}{2}mu^2+mgh=K_b+0\\\\\Rightarrow K_b=0.5\times 230\times 1.72^2+230\times 9.8\times 36.06\\\Rightarrow K_b=340.216+81,279.24\\\Rightarrow K_b=81,619.456\ J\\\Rightarrow K_b=81.619\ kJ$

8 0

3 years ago