What kind of medium does ocean waves have

To move a refrigerator of mass 170 kg into a house, a mover puts it on a dolly and covers the steps leading into the house with

Arte-miy333 [17]

Explanation:

The given data is as follows.

mass (m) = 170 kg, Distance (s) = 9.6 m

Height (h) = 3.3 m, Force (F) = 1400 N

First, we will calculate the work performed by her as follows.

W = Fs

= $1400 N \times 9.6 m$

= 13440 J

Hence, minimal work necessary to lift the refrigerator is as follows.

U = mgh

= $170 kg \times 9.8 \times 3.3 m$

= 5497.8 J

Therefore, we can conclude that he performed 5497.8 J of work.

7 0

3 years ago

A heat engine has a maximum possible efficiency of 0.780. If it operates between a deep lake with a constant temperature of-24.8

Volgvan

Answer : The temperature of the hot reservoir (in Kelvins) is 1128.18 K

Explanation :

Efficiency of carnot heat engine : It is the ratio of work done by the system to the system to the amount of heat transferred to the system at the higher temperature.

Formula used for efficiency of the heat engine.

$\eta =1-\frac{T_c}{T_h}$

where,

$\eta$ = efficiency = 0.780

$T_h$ = Temperature of hot reservoir = ?

$T_c$ = Temperature of cold reservoir = $-24.8^oC=273+(-24.8)=248.2K$

Now put all the given values in the above expression, we get:

$\eta =1-\frac{T_c}{T_h}$

$0.780=1-\frac{248.2K}{T_h}$

$T_h=1128.18K$

Therefore, the temperature of the hot reservoir (in Kelvins) is 1128.18 K

8 0

3 years ago

Two forces act on an object. One force has a magnitude of 10 N directed north, and the other force has a magnitude of 2 N direct

xeze [42]

Answer:

8 N North.

Explanation:

Given that,

One force has a magnitude of 10 N directed north, and the other force has a magnitude of 2 N directed south.

We need to find the magnitude of net force acting on the object.

Let North is positive and South is negative.

Net force,

F = 10 N +(-2 N)

= 8 N

So, the magnitude of net force on the object is 8 N and it is in North direction (as it is positive). Hence, the correct option is (d) "8N north".

7 0

3 years ago

Which statement describes the distribution of charge in an atom? (1) A positively charged nucleus is surrounded by one or more n

pantera1 [17]

Answer: option B

Explanation: when a neutral atom loses an electron or gains a positive charge electron, it becomes a positive ion (positively charged) and when an neutral atom gains an electronic charge or losses a positive charge electron, it becomes a negative ion (negatively charged).

4 0

3 years ago

Read 2 more answers

Compare the circular velocity of a particle orbiting in the Encke Division, whose distance from Saturn 133,370 km, to a particle

Ket [755]

Answer:

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

Explanation:

The circular velocity is define as:

$v = \frac{2 \pi r}{T}$

Where r is the radius of the trajectory and T is the orbital period

To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:

$T^{2} = r^{3}$

Where T is orbital period and r is the radius of the trajectory.

Case for the particle in the Encke Division:

$T^{2} = r^{3}$

$T = \sqrt{(133370 Km)^{3}}$

$T = \sqrt{(2.372x10^{15} Km)}$

$T = 4.870x10^{7} Km$

It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( $1.50x10^{8} Km$ )

1 AU is defined as the distance between the earth and the sun.

$\frac{4.870x10^{7} Km}{1.50x10^{8}Km} . 1AU$

$T = 0.324 AU$

But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.

$T = \frac{0.324 AU}{1 AU} . 1 year$

$T = 0.324 year$

That can be expressed in units of days

$T = \frac{0.324 year}{1 year} . 365.25 days$

$T = 118.60 days$

Circular velocity for the particle in the Encke Division:

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (133370 Km)}{(118.60 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$118.60 days .\frac{86400 s}{1 day}$ ⇒ 10247040 s

$133370 Km .\frac{1000 m}{1 Km}$ ⇒ 133370000 m

$v = \frac{2 \pi (133370000 m)}{(10247040 s)}$

$v = 81.778 m/s$

Case for the particle in the D Ring:

For the case of the particle in the D Ring, the same approach used above can be followed

$T^{2} = r^{3}$

$T = \sqrt{(69000 Km)^{3}}$

$T = \sqrt{(3.285x10^{14} Km)}$

$T = 1.812x10^{7} Km$

$\frac{1.812x10^{7} Km}{1.50x10^{8}Km} . 1 AU$

$T = 0.120 AU$

$T = \frac{0.120 AU}{1 AU} . 1 year$

$T = 0.120 year$

$T = \frac{0.120 year}{1 year} . 365.25 days$

$T = 43.83 days$

Circular velocity for the particle in D Ring:

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (69000 Km)}{(43.83 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$43.83 days . \frac{86400 s}{1 day}$ ⇒ 3786912 s

$69000 Km . \frac{1000 m}{ 1 Km}$ ⇒ 69000000 m

$v = \frac{2 \pi (69000000 m)}{(3786912 s)}$

$v = 114.483 m/s$

$\frac{114.483 m/s}{81.778 m/s} = 1.399$

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

7 0

3 years ago