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3 years ago

How does the current values in and out of the battery compare to the sum of currents going through the light bulbs?

1 answer:

6 0

The current values in and out of the battery is going to be equal to the sum of the currents going through the light bulbs. In a parallel circuit, the current is divided as it enters different branches and light bulbs. However, the amount of current leaving the battery is always going to be the same as the amount of current entering the battery. Therefore, while current divides in a parallel circuit, the total amount of current always remains the same.

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A rigid container filled with a gas is placed in ice. What will happen to the pressure of the gas? And what will happen to the v

tankabanditka [31]

The volume of the can remains constant, and as the pressure increases with temperature, the can can't contain the pressure any longer.

7 0

3 years ago

"A parcel moving in a horizontal direction with speed v0 = 13 m/s breaks into two fragments of weights 1.4 N and 1.9 N, respecti

Lady bird [3.3K]

Answer:

the smaller particle moves with speed of 8.706 m/s in the opposite direction to the bigger particle.

Explanation:

Speed of the original particle = 13 m/s

We designate particles as A and B

The final weights of the component particles are

Particle A = 1.4 N

particle B = 1.9 N

The speed of the larger piece (particle B) = 29 m/s

We know that weight is the product of a body's mass and acceleration due to gravity g which is equal to 9.81 m/s^2, therefore, masses of the particles are

particle A = 1.4/9.81 = 0.143 kg

Particle B = 1.9/9.81 = 0.194 kg

The momentum of a body is the product of its mass and its velocity i.e

P = mv

This means that the mass of the particle before splitting is

0.143 kg + 0.194 kg = 0.337 kg

Momentum of the initial whole particle = mv

==> 0.337 x 13 = 4.381 kg-m/s

The bigger particle B remains horizontal, and has a momentum of

mv = 0.194 x 29 = 5.626 kg-m/s

According to the conservation of momentum, the total initial momentum of a system must be equal tot the total final momentum of the system.

Initial total momentum of the system = 4.381 kg-m/s (momentum of original particle before splitting)

Final total momentum of the system = Total momentum of the particles after splitting = 5.626 kg-m/s + ( 0.143 kg x $V_{B}$ )

where $V_{B}$ is the velocity of smaller particle A

final total momentum of the system = 5.626 + 0.143 $V_{B}$

Equating the two momenta of the system, we'll have

4.381 = 5.626 + 0.143 $V_{B}$

4.381 - 5.626 = 0.143 $V_{B}$

-1.245 = 0.143 $V_{B}$

$V_{B}$ = -1.245/0.143 = -8.706 m/s

The negative sign indicates that the smaller particle moves in the opposite direction to the bigger particle

5 0

3 years ago

A potter spins his wheel at 0.98 rev/s. The wheel has a mass of 4.2 kg and a radius of 0.35 m. He drops a chunk of clay of 2.9 k

Bad White [126]

Answer:

$v_{f,w} = 1.791\,\frac{m}{s}$ , $v_{f,c} = 0.972\,\frac{m}{s}$

Explanation:

The situation can be modelled by applying the Principle of Angular Momentum Conservation:

$I_{w} \cdot \omega_{o} = (I_{w} + I_{c})\cdot \omega_{f}$

The final angular speed is:

$\omega_{f} = \frac{I_{w}}{I_{w}+I_{c}}\cdot \omega_{o}$

$\omega_{f} = \left(\frac{\frac{1}{2}\cdot (4.2\,kg)\cdot (0.35\,m)^{2} }{\frac{1}{2}\cdot (4.2\,kg)\cdot (0.35\,m)^{2} + \frac{1}{2}\cdot (2.9\,kg)\cdot (0.19\,m)^{2}}\right)\cdot (0.98\,\frac{rev}{s} )\cdot \left(\frac{2\pi\,rad}{1\,rev} \right)$

$\omega_{f} \approx 5.116\,\frac{rad}{s}$

The tangential velocities of the wheel and the clay are, respectively:

$v_{f, w} = (0.35\,m)\cdot (5.116\,\frac{rad}{s} )$

$v_{f,w} = 1.791\,\frac{m}{s}$

$v_{f, c} = (0.19\,m) \cdot (5.116\,\frac{rad}{s} )$

$v_{f,c} = 0.972\,\frac{m}{s}$

5 0

3 years ago

Which of the following statements are true? Check all that apply. A. Atoms are always moving. B. The faster atoms are moving, th

Alik [6]

A, d are the true ones :))))))))))

6 0

4 years ago

Read 2 more answers

Before using a string in a comparison, you can use either the To Upper method or the To Lower method to convert the string to up

diamong [38]

Answer:

True, check attachment for code

Explanation:

To convert java strings of text to upper or lower case, we can use and inbuilt methods To Uppercase and To lower case.

The first two lines of code will set up a String variable to hold the text "text to change", and then we print it out.

The third line sets of a second String variable called result.

The fourth line is where the conversion is done.

We can compare the string

We can compare one string to another. (When comparing, Java will use the hexadecimal values rather than the letters themselves.) For example, if we wanted to compare the word "Fat" with the word "App" to see which should come first, you can use an inbuilt string method called compareTo.

Check attachment for the code

3 0

3 years ago