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Sergeeva-Olga [200]

3 years ago

8

how is the process of finding the mass of a gas different from that of finding the mass of a liquid why is the difference necess

ary

1 answer:

PtichkaEL [24]3 years ago

5 0

Because gas and liquid are both different things

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Plzzzzz help me swear will mark u brainiest

photoshop1234 [79]

Answer:

B

Explanation:

4 0

3 years ago

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4. Provide an example of an isolated energy system and explain how it could be changed to create an open energy system. You may

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3 years ago

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A record is spinning at the rate of 25rpm. If a ladybug is sitting 10cm from the center of the record.

marin [14]

A) Angular speed: 0.42 rev/s

B) Frequency: 0.42 Hz

C) Tangential speed: 26.4 cm/s

D) Distance travelled: 528 cm

Explanation:

A)

In this problem, the ladybug is rotating together with the record.

The angular velocity of the ladybug, which is defined as the rate of change of the angular position of the ladybug, in this problem is

$\omega = 25 rpm$

where here it is measured in revolutions per minute.

Keeping in mind that

1 minute = 60 seconds

We can rewrite the angular speed in revolutions per second:

$\omega = 25 \frac{rev}{min} \cdot \frac{1}{60 s/min}=0.42 rev/s$

B)

The relationship between angular speed and frequency of revolution for a rotational motion is given by the equation

$\omega = 2 \pi f$ (1)

where

$\omega$ is the angular speed

f is the frequency of revolution

For the ladybug in this problem,

$\omega=0.42 rev/s$

Keeping in mind that $1 rev = 2\pi rad$ , the angular speed can be rewritten as

$\omega = 0.42 \frac{rev}{s} \cdot 2\pi = 2\pi \cdot 0.42$

And re-arranginf eq.(1), we can find the frequency:

$f=\frac{\omega}{2\pi}=\frac{(2\pi)0.42}{2\pi}=0.42 Hz$

And the frequency is the number of complete revolutions made per second.

C)

For an object in circular motion, the tangential speed is related to the angular speed by the equation

$v=\omega r$

where

$\omega$ is the angular speed

v is the tangential speed

r is the distance of the object from the axis of rotation

For the ladybug here,

$\omega = 2\pi \cdot 0.42 rad/s$ is the angular speed

r = 10 cm = 0.10 m is the distance from the center of the record

So, its tangential speed is

$v=(2\pi \cdot 0.42)(0.10)=0.264 m/s = 26.4 cm/s$

D)

The tangential speed of the ladybug in this motion is constant (because the angular speed is also constant), so we can find the distance travelled using the equation for uniform motion:

$d=vt$

where

v is the tangential speed

t is the time elapsed

Here we have:

v = 26.4 cm/s (tangential speed)

t = 20 s

Therefoe, the distance covered by the ladybug is

$d=(26.4)(20)=528 cm$

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7 0

4 years ago

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A helicopter (m = 3250 kg) is cruising at a speed of 56.9 m/s atan altitude of 185 m. What is the total mechanical energy of the

taurus [48]

Answer:

The mechanical energy of the helicopter is $1.12\times 10^7\ J$ .

Explanation:

It is given that,

Mass of the helicopter, m = 3250 kg

Speed of the helicopter, v = 56.9 m/s

Position of the helicopter, h = 185 m

The energy possessed by an object due to its motion is called its kinetic energy. It is given by :

$E=\dfrac{1}{2}mv^2$

$E=\dfrac{1}{2}\times 3250\times (56.9)^2$

$E=5.26\times 10^6\ J$

The energy possessed by an object due to its position is called its potential energy. It is given by :

$E=mgh$

$E=3250\times 9.8\times 185$

$E=5.89\times 10^6\ J$

The sum of kinetic and potential energy is called mechanical energy of the system. It is given by :

$M=5.26\times 10^6+5.89\times 10^6$

$M=11.15\times 10^6\ J$

or

$M=1.12\times 10^7\ J$

So, the mechanical energy of the helicopter is $1.12\times 10^7\ J$ . Hence, this is the required solution.

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3 years ago

Rank the tensions in the ropes, t1, t2, and t3, from smallest to largest, when the boxes are in motion and there is no friction

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<span>AS T1,T2,T3 are the tensions in the ropes,assuming that there are Three blocks of mass 3m, 2m, and m.T3 is the string between 3m and 2m,T2 is the string between 2m and m ,T1 is the string attached to m thus T1 pulls the whole set of blocks along, so it must be the largest. T2 pulls the last two masses, but T3 only pulls the last mass, so T3 < T2 < T1.</span>

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3 years ago