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larisa [96]
3 years ago
12

Laura puts her file cabinet on a small cart with wheels which will occur. A Laura will increase the amount of friction. B Laura

will reduce the force she needs C Laura will decrease the amount of motion D Laura will increases the force she needs
Physics
1 answer:
kow [346]3 years ago
6 0

Answer: b

Explanation:

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Elden [556K]

<u>TRANSISTOR:</u>

The future of transistors based on the ongoing researches suggests that, it would be based on nano technology. This technology is itself still under development and the case is similar in the case of transistors too.

Transistors due to their applications, especially its application as an amplifier has increased its usage and the dependability over them. The hard part often arises with the fabrication techniques which are being tested and developed as we speak in order to increase the efficiency of the transistors and also to reduce their physical size though, these would mean that, the design complications would be higher.

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Two small space probes have been slowed to 10m/s as they approach the moon from the same direction. Probe 1 has a mass of 86kg a
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3 years ago
a body of mass 0.2kg is whirled round a horizontal circle by a string inclined at 30 degrees to the vertical calculate &lt;br /&
Katen [24]

Answer:

a)  T = 2.26 N, b) v = 1.68 m / s

Explanation:

We use Newton's second law

Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string

        sin 30 = \frac{T_x}{T}

        cos 30 = \frac{T_y}{T}

        Tₓ = T sin 30

        T_y = T cos 30

Y axis  

       T_y -W = 0

       T cos 30 = mg                     (1)

X axis

        Tₓ = m a

they relate it is centripetal

        a = v² / r

we substitute

         T sin 30 = m\frac{v^2}{r}            (2)

a) we substitute in 1

         T = \frac{mg }{cos 30}

         T = \frac{ 0.2 \ 9.8}{cos  \ 30}

         T = 2.26 N

b) from equation 2

           v² = \frac{T \ sin 30 \ r}{m}

If we know the length of the string

          sin 30 = r / L

          r = L sin 30

we substitute

          v² = \frac{ T \ sin 30 \ L \ sin 30}{m}

          v² = \frac{TL \ sin^2  30}{m}

For the problem let us take L = 1 m

let's calculate

          v = \sqrt{ \frac{2.26 \ 1 \ sin^230}{0.2} }

          v = 1.68 m / s

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