2 years ago

How much force is required to raise a 0.2 kg mass?

Physics

2 answers:

Firlakuza [10]2 years ago

6 0

Answer:

1.96 N

Explanation:

Gnesinka [82]2 years ago

5 0

⇒This is your full answer

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nikklg [1K]

Answer:

The mass goes down

Explanation:

Because mass is the quantity of matter contained in a substance And Volume is the space occupied by a substance. So when the volume is less the mass decreases

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tigry1 [53]

Answer:

a to b is 109 degrees

Explanation:

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3 years ago

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A 90 kg painter is standing on a horizontal wooden scaffolding of length 10 m, which is supported on each end by a rope. The pai

RSB [31]

Explanation:

For equilibrium, $\sum M = 0$ .

So, $8 m \times mg - (10 m) T_{1}$ = 0

$T_{1} = \frac{8 \times mg}{10}$

= $\frac{8 \times 90 \times 9.8}{10}$

= 705.6 N

Also, for equilibrium $\sum F_{y}$ = 0

$T_{1} + T_{2} - mg$ = 0

or, $T_{2} = mg - T_{1}$

= $90 \times 9.8 - 705.6$

= 176.4 N

Thus, we can conclude that the tension in the first rope is 176.4 N.

8 0

3 years ago

Calculate a rate of cooling down of air from 80 C to 5C Show calculation. Give an answer in cubic meters per minute and cfm.

antoniya [11.8K]

Explanation:

Given that,

Rate of cooling of air

Initial temperature= 80°C

Final temperature = 5°C

We need to calculate

Using newton's law of cooling

$\dfrac{dT}{dt}=c(T-T_{0})$

$\dfrac{dT}{dt}=c(\dfrac{T_{1}+T_{2}}{2}-T_{0})$

Where, $dT=T_{1}-T_{2}$

Here, $T =\dfrac{T_{1}+T_{2}}{2}$

$T_{0}$ = 25°C (surrounding temperature)

dt = 1 minute

$\dfrac{dT}{dt}=c(\dfrac{T_{1}+T_{2}}{2}-T_{0})$

Put the value into the formula

$\dfrac{80-5}{1}=c(\dfrac{85}{2}-25)$

$c=\dfrac{75}{17.5}$

$c=4.285\ cubic\ meter/minute$

Hence, This is the required answer.

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2 years ago

Consider a short time span just before and after the spark plug in a gasoline engine ignites the fuel-air mixture and releases 1

Tju [1.3M]

Answer:

Temperature after ignition=7883.205 K

Explanation:

The number of moles is,

n=PV/RT

=(1.18x10^6)(47.9x10^-6)/8.314(325)

= 0.0209 moles

a) In this process volume is constant

Q=U

=nCv.dT

dT= Q/nCv

=1970/(1.5x8.314)(0.0209)

= 7558.205 K

The final temperature is,

= 7558.205+325

= 7883.205 K

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3 years ago

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