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3 years ago

Work is being done in which of these situations? all motions are at a constant velocity

Physics

1 answer:

fredd [130]3 years ago

7 0

Where the force is not perpendicular to the path of motion

are you missing the the situations ?

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The local church is hosting a carnival which includes a bumper car ride. Bumper car A and its driver have a mass of 300 kg; bump

lukranit [14]

Answer:

a. 20 s

b. 0 m/s

c. right

d.no its inelastic because when the car b was at rest and a was coming in at it, since b had no force what so ever car a swept it away with it moving to the right

Explanation:

im not sure though

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3 years ago

A(n ___ is used to determine where an implanted electrode is to be placed in a brain in 3-dimensions.

Fynjy0 [20]

A stereotaxic atlas is used to determine where an implanted electrode is to be placed in a brain in 3-dimensions. A stereotaxic atlas is used for doing a stereotactic surgery in order to cure the disorder occurred in the brain. This tool increase the accuracy, reliability, and the measurement of the brain because it gives the accurate brain map<span>.</span>

5 0

3 years ago

How do I solve this problem

PilotLPTM [1.2K]

Answer:

it is light

Explanation:

the arrow that says light is on the glass it must be near from tungsten

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3 years ago

Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m

alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = $0.407\times 10^{-3}\ m$

Energy stored in the capacitor (U) = $7.87\ nJ=7.87\times 10^{-9}\ J$

Assuming the medium to be air.

So, permittivity of space (ε) = $8.854\times 10^{-12}\ F/m$

Area of the square plates is given as:

$A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2$

Capacitance of the capacitor is given as:

$C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F$

Now, we know that, the energy stored in a parallel plate capacitor is given as:

$U=\frac{CE^2d^2}{2}$

Rewriting in terms of 'E', we get:

$E=\sqrt{\frac{2U}{Cd^2}}$

Now, plug in the given values and solve for 'E'. This gives,

$E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C$

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0

4 years ago

What type of relationship exists between the length of a wire and the resistance, if all other factors remain the same? A. Resis

ZanzabumX [31]

<h3>Answer</h3>

(A) Resistance is directly related to length.

<h3>Explanation</h3>

Formula for resistance

R = p(length) / A

where R = resistance

p = resistivity(material of wire)

A = cross sectional area

So it can be seen that resistance depends upon 3 factors that are length of wire , resistivity of wire and the cross sectional area of the wire.

If two of the factors, resistivity and cross sectional area, are kept constant then the resistance is directly proportional to the length of wire.

<h3> R ∝ length</h3>

This means that the resistance of the wire increases with the increase in length of the wire and decreases with the decrease of length of the wire.

5 0

3 years ago