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MrMuchimi
3 years ago
8

Show that the units N/kg can be written using only units of meters (m) and second (s). Is this a unit of mass ,acceleration or f

orce.
Physics
1 answer:
nikdorinn [45]3 years ago
3 0
<span>When the difference between two results is larger than the estimates error, the result is</span>
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What is power?
hammer [34]
We can define power as the rate of doing work, it is the work done in unit time. The SI unit of power is Watt (W) which is joules per second (J/s). Sometimes the power of motor vehicles and other machines are given in terms of Horsepower (hp) which is approximately equal to 745.7 watts.


Power is the rate at which a force is applied to an object for example.current wire
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3 years ago
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The transfer of energy that occurs when a force is applied over a distance is
astra-53 [7]
The transfer of energy that occurs when a force is applied over a distance is WORK.
3 0
3 years ago
Numbered 8 nowwwwwww
amm1812

Torque =  r x F

|F| =  mg =  60 * 10 N = 600 N ( assuming g ~ 10m/s^2)

distance of fulcrum = torque / Force = 90/600 m = .15 m.

7 0
3 years ago
What word can be used to describe the compression of a longitudal wave
NemiM [27]

Answer:

Mechanical longitudinal waves are also called compressional or compression waves, because they produce compression and rarefaction when traveling through a medium, and pressure waves, because they produce increases and decreases in pressure.

Explanation:

8 0
3 years ago
A motor is connected to the smaller 0.2 m radius pulley and cable, lifting 150 kg mass. Neglecting the mass/weight of the cable
-Dominant- [34]

Answer:

324.3Nm

Explanation:

The torque is given by the equation

\tau=r\ X\ F

in this case the vectors r and F are perpendicular between them, thus:

\tau=rF

The forces acting on the mass are:

T-Mg=Ma    (1)

where T is the tension of the cable, M is the mass and a is the acceleration.

Furthermore, we have that the acceleration is:

a=\frac{v-v_0}{t}=\frac{10m/s-0m/s}{10s}=1\frac{m}{s^2}

By replacing in (1) we can obtain:

T=Ma-Mg=M(a-g)=(150kg)(1\frac{m}{s^2}+9.8\fac{m}{s^2})=1620N

The force T produces the torque on the pulley, hence:

\tau=rT=(0.2m)(1620N)=324Nm

the answer is 324.4Nm

hope this helps!

7 0
3 years ago
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