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3 years ago

In which type of reaction is matter created?

Physics

1 answer:

NemiM [27]3 years ago

3 0

Answer:

Matter can not be created or destroyed, it only changes forms.

Explanation:

conservation of matter

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Three positive charges A, B, and C, and a negative charge D are placed in a line as shown in the diagram. All four charges are o

Irina18 [472]

Answer:

a. $q_C$ experiences the greatest net force and $q_B$ experiences the smallest net force

b. Ratio of the greatest to the smallest net force= 9

Explanation:

Electrostatic Forces

Two point-charges q1 and q2, separated a distance d, exert on each other an electrostatic force of magnitude

$\displaystyle F=K\frac{q_1q_2}{d^2}$

If the charges have the same sign, they repel each other, for different signed charges, they attract. That gives us the direction of each force in the space.

Let's assume all the charges of the problem have a magnitude q, and between two consecutive charges, the distance is d. The proposed layout is shown it the image.

The net force on qA is the sum of those exerted by qB, qC, and qD. But note qB and qC repel qA and qD attracts it, so the total force on qA is

$F_{TA}=-F_B-F_C+F_D$

Computing the individual forces we have

$\displaystyle F_B=\frac{K\ q_A\ q_B}{d^2}=K\ \frac{q^2}{d^2}$

$\displaystyle F_C=\frac{K\ q_A\ q_C}{(2d)^2}=\frac{1}{4}\ \ \frac{K\ q^2}{d^2}$

$\displaystyle F_D=\frac{K\ q_A\ q_D}{(3d)^2}=\frac{1}{9}\ \ \frac{K\ q^2}{d^2}$

The total force on qA is:

$\displaystyle F_{TA}=\frac{K\ q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})$

$\displaystyle F_{TA}=-\frac{41}{36}\ \frac{K\ q^2}{d^2}$

$\displaystyle |F_{TA}|=\frac{41}{36}\ \frac{K\ q^2}{d^2}$

Charge qA repels qB to the right, qC repels qB to the left, and qD attracts qB to the right, thus

$\displaystyle F_{TB}=F_A-F_C+F_D$

$\displaystyle F_{TB}=\frac{K\ q^2}{d^2}-\frac{K\ q^2}{d^2}+\frac{K\ q^2}{(2d)^2}$

$\displaystyle F_{TB}=\frac{1}{4}\ \frac{K\ q^2}{d^2}$

$\displaystyle |F_{TB}|=\frac{1}{4}\ \frac{K\ q^2}{d^2}$

Charges qA and qb repel qC to the right, and qD attracts qC to the right, thus

$\displaystyle F_{TC}=F_A+F_B+F_D$

$\displaystyle F_{TC}=\frac{K\ q^2}{(2d)^2}+\frac{K\ q^2}{d^2}+\frac{K\ q^2}{d^2}$

$\displaystyle F_{TC}=\frac{9}{4}\ \frac{K\ q^2}{d^2}$

$\displaystyle |F_{TC}|=\frac{9}{4}\ \frac{K\ q^2}{d^2}$

Charge qA and qB attract qD to the left, and qC atracts qD to the left, thus

$\displaystyle F_{TD}=-F_A-F_B-F_C$

$\displaystyle F_{TD}=-\frac{K\ q2}{(3d)^2}-\frac{K\ q2}{(2d)^2}-\frac{K\ q2}{d^2}$

$\displaystyle F_{TD}=-\frac{49}{36}\ \frac{K\ q^2}{d^2}$

$\displaystyle |F_{TD}|=\frac{49}{36}\ \frac{K\ q^2}{d^2}$

Comparing the relative values of all the forces

$\displaystyle |F_{TC}|>|F_{TD}|>|F_{TA}|>|F_{TB}|$

This means that qc experiences the greatest net force and qB experiences the smallest net force

The ratio of the greatest to the smallest forces is

$\displaystyle \frac{|F_{TC}|}{|F_{TB}|}=\frac{\frac{9}{4}}{\frac{1}{4}}=9$

5 0

3 years ago

A machine part has the shape of a solid uniform sphere of mass 220 g and diameter 4.50 cm . It is spinning about a frictionless

miss Akunina [59]

Answer:

The angular acceleration is 10.10 rad/s².

Explanation:

Given that,

Mass of sphere =220 g

Diameter = 4.50 cm

Friction force = 0.0200 N

Suppose we need to find its angular acceleration.

We need to calculate the angular acceleration

Using formula of torque

$\tau=f\times r$

$I\times\alpha=f\times r$

Here, I = moment of inertia of sphere

$\dfrac{2}{5}mr^2\times\alpha=f\times r$

$\alpha=\dfrac{5\times f}{2mr}$

Put the value into the formula

$\alpha=\dfrac{5\times0.0200}{2\times220\times10^{-3}\times2.25\times10^{-2}}$

$\alpha=10.10\ rad/s^2$

Hence, The angular acceleration is 10.10 rad/s².

7 0

3 years ago

Followed by the previous question: presume that the electron performs a uniform circular motion around the hydrogen nucleus. Wha

Ksivusya [100]

Answer:

$A_c=87.73*10^{21}m/s$

Explanation:

From the question we are told that

$r=5\times 10^{-11}$

$T=1.5 \times 10^{-16}$

Generally the equation for velocity is mathematically given as

$Velocity (v)=\frac{2 \pi r}{t}$

$V=\frac{2 \pi (5*10^{-11})}{1.5*10^{-16}}$

Generally the equation for Centripetal acceleration is mathematically given as

$A_c=\frac{V^2}{r}$

$A_c=(\frac{20.944*10^5)}{r5*10^{-11}}$

$A_c=87.73*10^{21}m/s$

8 0

2 years ago

When a force is applied to an object for an amount of time, it is known as __________.

NikAS [45]

The correct answer to the question is : D) Impulse

EXPLANATION:

Before going to answer this question, first we have to understand impulse.

Impulse of a body is defined as change in momentum or the product of force with time.

Mathematically impulse = F × t = m ( v - u ).

Here, v is the final velocity

u is the initial velocity

F is the force acting on the body for time t.

Hence, the perfect answer of this question is impulse m i.e the force multiplied with time is known as impulse.

7 0

3 years ago

Read 2 more answers

Hurry Please !!!!! Study the diagram Point C identifies the____ of the wave

Nata [24]

Answer: Trough

Explanation: The point labeled C in the wave diagram above is the TROUGH of the wave motion. The trough of a wave motion identifies or signifies the point of least or minimum Displacement by measuring the downward Displacement of the wave. The point A is the CREST which is the opposite of the trough, signifying the point of maximum or upward Displacement of the wave cycle.

Point B is the wave amplitude which signifies the maximum extent of vibration from the equilibrium position of a wave. The point labeled D refers to the wavength of the wave motion which is the distance between successive crest or troughs of a wave motion.

4 0

3 years ago