A 25 kg mass is accelerated by a force at a rate of 5 m/s2. What is the magnitude of the force that accelerates the man?

Why is it important to understand your strength and weakness<br><br>

S_A_V [24]

Answer: be a better person

Explanation:

6 0

3 years ago

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Which process requires more energy: completely vaporizing 1 kg of saturated liquid water at 1 atm pressure or completely vaporiz

Lilit [14]

The correct answer would be the first option. The process that would need more energy would be vaporizing 1 kg of saturated liquid water at a pressure of 1 atmosphere. This can be seen from the latent heat of vaporization of each system. For the saturated water at 1 atm, the latent heat is equal to 40.7 kJ per mole while, at 8 atm, the latent heat is equal to 36.4 kJ per mole. The latent heat of vaporization is the amount of heat needed in order to vaporize a specific amount of substance without any change in the temperature. As we can observe, more energy is needed by the liquid water at 1 atm.

7 0

3 years ago

At the equator, the radius of the Earth is approximately 6370 km. A plane flies at a very low altitude at a constant speed of v

Anna007 [38]

To solve this problem we will apply the concepts related to the kinematic equations of linear motion. For this purpose we will define the speed as the distance traveled in a given period of time. Here the distance is equivalent to the orbit traveled around the earth, that is, a circle. Approaching the height of the aircraft with the radius of the earth, we will have the following data,

$R= 6370*10^3 m$

$v = 219m/s$

$a = 17m/s^2$

The circumference of the earth would be

$\phi = 2\pi R$

Velocity is defined as,

$v = \frac{x}{t}$

$t = \frac{x}{v}$

Here $x = \phi$ , then

$t = \frac{\phi}{v} = \frac{2\pi (6370*10^3)}{219}$

$t = 1.82*10^5s$

Therefore will take $1.82*10^5$ s or 506 hours, 19 minutes, 17 seconds

3 0

2 years ago

You go to the hardware store to buy a new 50 ft garden hose. You find you can choose between hoses of ½ inch and 5/8 inch inner

omeli [17]

To solve this problem it is necessary to consider two concepts. The first of these is the flow rate that can be defined as the volumetric quantity that a channel travels in a given time. The flow rate can also be calculated from the Area and speed, that is,

Q = V*A

Where,

A= Cross-sectional Area

V = Velocity

The second concept related to the calculation of this problem is continuity, which is defined as the proportion that exists between the input channel and the output channel. It is understood as well as the geometric section of entry and exit, defined as,

$Q_1 = Q_2$