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4 years ago

What is 2.5 meters in feet?

Physics

2 answers:

Phantasy [73]4 years ago

6 0

The answer is 8 feet 2 27/64 inches.

miss Akunina [59]4 years ago

5 0

If there are about 3.2 feet in a meter then multiply.
2.5 * 3.2 = 8

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does the amount of work to required to move a charge in an electric field depend on the path taken or just the potential differe

Romashka [77]

Answer: No

Explanation:

The work done is not path-dependent but only dependent on the final and initial position of the charge, that is the difference between the end points.

4 0

3 years ago

A cylindrical capacitor is made of two thin-walled concentric cylinders. The inner cylinder has radius , and the outer one a rad

lianna [129]

Answer: $9.6\times 10^{-8} J$

Explanation:

The capacitance of cylindrical capacitor is

C= $\frac{2\pi \varepsilon _0L}{\ln\frac{r_2}{r_1}}$

C= $\frac{2\pi \times 8.854\times 10^{-12}\times 150}{\ln\frac{8}{4}}$

C= $1.2\times 10^{-8} F$

potential Energy stored in capacitor= $\frac{1}{2}CV^2$

= $\frac{1}{2}\times 1.2\times 10^{-8}\times 4^2$

= $9.6\times 10^{-8} J$

6 0

3 years ago

A sine wave with maximum amplitude after 1/4 cycle

frosja888 [35]

Yeah what do you want me to answer for this question?

4 0

4 years ago

Think about the electricity sent to your home from a power plant. How does the voltage of the electricity that leaves the plant

Lana71 [14]

The voltage at the plant is HUGEEEE compared to the one at your house. your house functions at 230V , the plant works at thousands times more voltage

3 0

3 years ago

Read 2 more answers

Three guns are aimed at the center of a circle, and each fires a bullet simultaneously. The directions in which they fire are 12

ikadub [295]

Answer:

The unknown mass of the bullet is $m1=3.751x10^{-3} kg$

Explanation:

According to Newton's laws of motion, when a net external force acts on a body of mass <u><em>m</em></u> , it results in change in momentum of the body and is given by:

$F=\frac{P}{dt}$

Where:

is the linear momentum of the body

As a consequence, when there are no external forces acting on the body the total momentum remains conserved i.e.

Given:

$m_{2}=5.79x10^{-3}kg \\m_{3}=5.79x10^{-3}kg\\v_{2}=v_{3}=392 \frac{m}{s}\\$

For momentum along the y-direction to be zero, it is achieved when the equal masses are moving at angles of

θ1=180°, θ2=60°, θ3=-60°

Therefore, from conservation of momentum along x - direction:

$m_{1}*v_{1}*cos(180)+m_{2}*v_{2}*cos(60)+m_{3}*v_{3}*cos(-60)=0\\$ $m_{1}*605\frac{m}{s}*cos(180)+5.79x10^{-3}kg *392\frac{m}{s}*cos(60)+5.79x10^{-3}kg*392\frac{m}{s}*cos(-60)=0\\$

$-m_{1}*605+5.79x10^{-3}kg*196\frac{m}{s}+5.79x10^{-3}kg*196\frac{m}{s}=0\\m_{1}*605kg= 2.26968\frac{kg*m}{s}\\m_{1}=3.75 x10^{-3} kg$

3 0

3 years ago