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3 years ago

1875 kg car is traveling at 22 m/s. It’s momentum is?

Physics

2 answers:

Irina-Kira [14]3 years ago

8 0

Momentum is Mass × Velocity
= 1875×22 = 41250 kg m/sec

Troyanec [42]3 years ago

8 0

84.22 I pretty sure because you divide them

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What are the forces that act on the ball?

scoundrel [369]

Answer:

This slide shows the three forces that act on a baseball in flight. The forces are the weight, drag, and lift. Lift and drag are actually two components of a single aerodynamic force acting on the ball. Drag acts in a direction opposite to the motion, and lift acts perpendicular to the motion

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2 years ago

A wave is 8 meters long and has a frequency of 3 Hz. Find speed

Olenka [21]

Answer:

The speed is 24 $\frac{meter}{s}$

Explanation:

A wave is a disturbance that propagates through a certain medium or in a vacuum, with transport of energy but without transport of matter.

The wavelength is the minimum distance between two successive points of the wave that are in the same state of vibration. It is expressed in units of length (m).

Frequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).

The speed of propagation is the speed with which the wave propagates in the middle, that is, the magnitude that measures the speed at which the wave disturbance propagates along its displacement. Relate wavelength (λ) and frequency (f) inversely proportionally using the following equation:

v = f * λ.

In this case, λ= 8 meter and f= 3 Hz

Then:

v= 3 Hz* 8 meter

So:

v= 24 $\frac{meter}{s}$

The speed is 24 $\frac{meter}{s}$

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3 years ago

What change in entropy occurs when a 0.15 kg ice cube at -18 °C is transformed into steam at 120 °c 4.

Studentka2010 [4]

Answer: The change in entropy of the given process is 1324.8 J/K

Explanation:

The processes involved in the given problem are:

$1.)H_2O(s)(-18^oC,255K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\3.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\4.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\5.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(120^oC,393K)$

Pressure is taken as constant.

To calculate the entropy change for same phase at different temperature, we use the equation:

$\Delta S=m\times C_{p,m}\times \ln (\frac{T_2}{T_1})$ .......(1)

where,

$\Delta S$ = Entropy change

$C_{p,m}$ = specific heat capacity of medium

m = mass of ice = 0.15 kg = 150 g (Conversion factor: 1 kg = 1000 g)

$T_2$ = final temperature

$T_1$ = initial temperature

To calculate the entropy change for different phase at same temperature, we use the equation:

$\Delta S=m\times \frac{\Delta H_{f,v}}{T}$ .......(2)

where,

$\Delta S$ = Entropy change

m = mass of ice

$\Delta H_{f,v}$ = enthalpy of fusion of vaporization

T = temperature of the system

Calculating the entropy change for each process:

For process 1:

We are given:

$m=150g\\C_{p,s}=2.06J/gK\\T_1=255K\\T_2=273K$

Putting values in equation 1, we get:

$\Delta S_1=150g\times 2.06J/g.K\times \ln(\frac{273K}{255K})\\\\\Delta S_1=21.1J/K$

For process 2:

We are given:

$m=150g\\\Delta H_{fusion}=334.16J/g\\T=273K$

Putting values in equation 2, we get:

$\Delta S_2=\frac{150g\times 334.16J/g}{273K}\\\\\Delta S_2=183.6J/K$

For process 3:

We are given:

$m=150g\\C_{p,l}=4.184J/gK\\T_1=273K\\T_2=373K$

Putting values in equation 1, we get:

$\Delta S_3=150g\times 4.184J/g.K\times \ln(\frac{373K}{273K})\\\\\Delta S_3=195.9J/K$

For process 4:

We are given:

$m=150g\\\Delta H_{vaporization}=2259J/g\\T=373K$

Putting values in equation 2, we get:

$\Delta S_2=\frac{150g\times 2259J/g}{373K}\\\\\Delta S_2=908.4J/K$

For process 5:

We are given:

$m=150g\\C_{p,g}=2.02J/gK\\T_1=373K\\T_2=393K$

Putting values in equation 1, we get:

$\Delta S_5=150g\times 2.02J/g.K\times \ln(\frac{393K}{373K})\\\\\Delta S_5=15.8J/K$

Total entropy change for the process = $\Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4+\Delta S_5$

Total entropy change for the process = $[21.1+183.6+195.9+908.4+15.8]J/K=1324.8J/K$

Hence, the change in entropy of the given process is 1324.8 J/K

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3 years ago

Which could be an example of a balanced force?

Rus_ich [418]

Answer:

An asteroid moving at a constant speed through space.

Explanation:

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2 years ago

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