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2 months ago

The magnitude of the electric current is inversely proportional to the _____ of the circuits in R2D2.

1 answer:

4 0

The essential principle for this question is Ohm’s Law: V=IR, I=V/R, R=V/I. Therefore, the answer is (3) Resistance, as it is inversely related to Current (I=V/R).

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An advertiser who calls a basement apartment a “garden apartment” is using the rhetorical device called

serg [3582]

I would choose B or D, based on the location. If it’s situated in South Kensington, London, then D might be appropriate. Conversely, if it’s in an underprivileged area, I'd opt for B.

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2 months ago

Derive an equation for the acceleration of block 3 for any arbitrary values of m3 and m2. Express your answer in terms of m3, m2

Ostrovityanka [3204]

The complete question is;

Block 1 sits on the floor with block 2 resting atop it. Block 3, which is stationary on a frictionless table, is attached to block 2 via a string that passes over a pulley depicted in the illustration below. Both the string and pulley have negligible mass.

Once block 1 is taken away without impacting block 2.

Derive an equation for the acceleration of block 3 considering arbitrary values for m3 and m2. Express your answer in terms of m3, m2, and relevant physical constants as needed.

Answer:

a = (m2)g/(m3 + m2)

Explanation:

Examining the attached illustration, by analyzing the free body diagram for block 3 and utilizing Newton's first law of motion, we reach the following formula;

T = (m3)a - - - (eq 1)

where;

T is the tension in the string

a is acceleration

m3 is the mass of block 3

Simultaneously, doing the same for Block 2, the free body diagram yields the equation; (m2)g - T = (m2)a

Rearranging for T results in;

T = (m2)g - (m2)a - - - (eq 2)

where;

g represents acceleration due to gravity

T is the tension in the string

a is acceleration

m2 is the mass of block 2

To deduce the acceleration, we will substitute (m3)a in place of T in eq 2.

Thus;

(m3)a = (m2)g - (m2)a

(m3)a + (m2)a = (m2)g

a(m3 + m2) = (m2)g

a = (m2)g/(m3 + m2)

3 0

1 month ago

What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t

Yuliya22 [3333]

Details that are not provided: the problem figure is included.

We can address the exercise by applying Poiseuille's law. This law indicates that for a fluid flowing in a laminar manner within a confined pipe,

$\Delta P = \frac{8 \mu L Q}{\pi r^4}$

where:

$\Delta P$ represents the pressure difference across the two ends

$\mu$ denotes the viscosity of the fluid

L signifies the length of the pipe

$Q=Av$ indicates the volumetric flow rate, where $A=\pi r^2$ is the cross-sectional area of the tube and $v$ refers to the fluid's velocity

r stands for the pipe's radius.

This law can be utilized for the needle, allowing us to compute the pressure difference between point P and the needle's end. In this scenario, we have:

$\mu=0.001 Pa/s$ is the dynamic viscosity of water at $20^{\circ}$

L=4.0 cm=0.04 m

$Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s$

and r=1 mm=0.001 m

Substituting these values into the formula yields:

$\Delta P = 3200 Pa$

This pressure difference specifies the value between point P and the needle's termination. As the end of the needle is under atmospheric pressure, the gauge pressure at point P, relative to atmospheric pressure, is exactly 3200 Pa.

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3 months ago

A center lane with solid and broken yellow lines that is used by vehicles making left turns in both directions is called a:

Sav [3153]

A left turn lane is designated for vehicles making left turns. Vehicles must completely enter this lane and wait for a clear path before proceeding with the turn. It is not permissible for a vehicle to be partially in and out of the lane, as this obstructs traffic and creates hazardous conditions.

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2 months ago