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3 years ago

Liquid nitrogen boils at 77.2 k. what is this temperature in degrees celsius

1 answer:

6 0

77.2 k is -195.95 C

To convert from Kelvin to Celsius, you simply subtract 273.15 from the temperature. So the expression becomes c = 77.2 - 273.15 = -195.95

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The temperature of a plastic cube is monitored while the cube is pushed 3.4 m across a floor at constant speed by a horizontal f

Nesterboy [21]

Answer:

$\Delta E_{floor} = 51 J$

Explanation:

The work (W) done on the cube to be pushed across the floor is equal to the total thermal energy (ΔE) of the system:

$W = \Delta E_{T} = \Delta E_{cube} + \Delta E_{floor}$ (1)

Also, the work done on the cube by the horizontal force is giving by:

$W = F \cdot d$ (2)

where F: force applied to the cube , d: displacement of the cube

By equaling the equations (1) and (2), we can find the thermal energy of the floor:

$\Delta E_{cube} + \Delta E_{floor} = F \cdot d$

$\Delta E_{floor} = F \cdot d - \Delta E_{cube}$

$\Delta E_{floor} = 20 N \cdot 3.4 m - 17 J$

$\Delta E_{floor} = 51 J$

So, the increase in the thermal energy of the floor is 51 J.

Have a nice day!

7 0

3 years ago

Given a force of 90 N and a acceleration of 30 m/s/s, what is the object’s mass?

vladimir2022 [97]

The object’s mass is 3

7 0

3 years ago

Read 2 more answers

Find Jupiter's mass from the fact that its moon Io orbits every 42.5 hours at an average distance of 422,000 kilometers.

Hitman42 [59]

Answer:

$M=1.9\times 10^{27}\ kg$

Explanation:

Given that,

Orbital period, T = 42.5 hours = 153000 seconds

The average distance of Jupiter's mass, r = 422,000 km

We need to find Jupiter's mass. The formula for the orbital period is given by :

$T^2=\dfrac{4\pi^2}{GM}r^3\\\\T^2GM=4\pi^2 r^3\\\\M=\dfrac{4\pi^2 r^3}{T^2 G}\\\\M=\dfrac{4\pi^2 \times (422000 \times 10^3)^3}{(153000 )^2 \times 6.67\times 10^{-11}}\\\\M=1.9\times 10^{27}\ kg$

So, the mass of Jupiter is $1.9\times 10^{27}\ kg$ .

7 0

3 years ago

A force of 150 N is continuously applied to an object with a mass of 11 kg over a distance of 2.5 m.The object slides over a sur

Ivahew [28]

Answer:

Final velocity will be 7.31 m/sec

Explanation:

We have given applied force F = 150 N

Mass of the object m = 11 KG

Distance traveled by the object s = 2.5 m

Coefficient of kinetic friction $\mu _k=0.3$

Acceleration due to gravity $g=9.8m/sec^2$

So frictional force applied on the object $f=-\mu _kmg=-0.3\times 11\times 9.8=-32.34N$

So net force = 150-32.34 = 117.66 N

As work done us equal to $W=Fs$

So work done = $117.66\times 2.5=294.15j$

Now according to work energy theorem work done is equal to change in kinetic energy

So $\frac{1}{2}mv^2=294.15$

$\frac{1}{2}\times 11\times v^2=294.15$

v =7.31 m/sec

So velocity final velocity will be 7.31 m/sec

4 0

3 years ago

Which statement accurately describes the motion of the object in the graph above over 10 seconds? Group of answer choices The ob

timurjin [86]

Answer:

Object appears to move forward at 1 cm/sec, then the velocity drops to zero for 3 sec and then moves forward at 2 cm/sec (11 - 3) / (10 - 6) = 2 cm/sec

7 0

3 years ago