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3 years ago

मारवतन गनुहास (What Is MKS SysteConvert 5 solar days into second.)

Physics

1 answer:

mr_godi [17]3 years ago

7 0

Answer:

5 Days to Seconds = 432000

Explanation:

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A roundabout in a fairground requires an input power of 2.5 kW when operating at a constant angular velocity of 0.47 rad s–1 . (

natita [175]

Answer:

Explanation:

Since the roundabout is rotating with uniform velocity ,

input power = frictional power

frictional power = 2.5 kW

frictional torque x angular velocity = 2.5 kW

frictional torque x .47 = 2.5 kW

frictional torque = 2.5 / .47 kN .m

= 5.32 kN . m

= 5 kN.m

b )

When power is switched off , it will decelerate because of frictional torque .

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3 years ago

An object has a kinetic energy of 225 j and a momentum of magnitude 28.3 kg · m/s. (a) find the speed of the object. m/s (b) fin

Irina18 [472]

A. hope this helps ;0

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3 years ago

Assume an axon has an internal diameter of 1μm and a myelin sheath 1μm thick. The internal specific resistance is 100 Ω cm. For

SpyIntel [72]

Answer:

$1.27\times 10^{12}\Omega/m$

Explanation:

We are given that

Diameter=d= $\mu m$

Thickness= $1\mu m$

Radius= $r=\frac{d}{2}=\frac{1}{2}\mu m=0.5\times 10^{-6} m$

Using $1\mu m=10^{-6} m$

Dielectric constant=8

Resistance = $R=2\times 10^5\Omega cm^2$

Internal specific resistance=r=100 ohm cm= $100\times \frac{1}{100}\Omega-m=1\Omega m$

Using 1 m=100 cm

Internal resistance per unit length= $\frac{r}{A}=\frac{1}{\pi r^2}=\frac{1}{3.14\times (0.5\times 10^{-6})^2}=1.27\times 10^{12}\Omega/m$

Using $\pi=3.14$

Internal resistance per unit length= $1.27\times 10^{12}\Omega/m$

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3 years ago

The achievement of lifting a rocket off the ground and into space can be explained by

lawyer [7]

Answer:

The achievement of lifting a rocket off the ground and into space can be explained by Newton's third law of motion. What is required for a rocket to lift off into space? Thrust is required for a rocket to lift off into space, ... An object that travels around another object in space is called a satellite.

Explanation:

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3 years ago

What is the frequency of a clock waveform whose period is 750 microseconds?

Allushta [10]

Use this formula to find your answer...

Determine the frequency of a clock waveform whose period is 2us or (micro) and 0.75ms

frequency (f)=1/( Time period).

Frequency of 2 us clock =1/2*10^-6 =10^6/2 =500000Hz =500 kHz.

Frequency of 0..75ms clock =1/0.75*10^-3 =10^3/0.75 =1333.33Hz =1.33kHz.

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3 years ago

मारवतन गनुहास (What Is MKS SysteConvert 5 solar days into second.)​

मारवतन गनुहास (What Is MKS SysteConvert 5 solar days into second.)