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d1i1m1o1n [39]
3 years ago
10

Which has a stronger attractions among its sub microscopic particles of a solid at 25°C or a gas and 25°C

Physics
1 answer:
Pavlova-9 [17]3 years ago
7 0

a gas at 25 C is the answer

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In a research facility, a person lies on a horizontal platform which floats on a film of air. When the person's heart beats, it
anastassius [24]

Answer: 3.48g

Explanation:

here, we will be using conservation of momentum to solve the problem. i.e the total momentum remains unchanged, unless an external force acts on the system. We'll in thus question, there is no external force acting in the system.

Remember, momentum = mass * velocity, then

mass of blood * velocity of blood = combined mass of subject and pallet * velocity of subject and pallet

Velocity of blood = 56.5cm = 0.565m

mass of blood * 0.565 = 54kg * (0.000063/0.160)

mass of blood * 0.565 = 54 * 0.00039375

mass of blood * 0.565 = 0.001969

mass of blood = 0.00348kg

Thus, the mass of blood that leaves the heart is 3.48g

7 0
3 years ago
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What are examples of convection currents?
hram777 [196]

Answer:

I would say all of the above.

Explanation:

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3 years ago
Plastics are used in millions of objects we use every day. Why are plastics NOT considered natural resources?
jeka57 [31]

Answer:

B

Explanation:

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4 0
3 years ago
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If a gas has a gage pressure of 156 kPa, it is absolute pressure is approximately
Art [367]
In the given question, one important information for getting to the actual solution is not given and that is the atmospheric pressure. To find the approximate absolute pressure, it is needed to add the value of atmospheric pressure with the gage pressure.
Atmospheric pressure = 100 kPa
Then
Absolute pressure = 156 + 100 kPa
                             = 256 KPa.
5 0
3 years ago
The parachute on a drag racing car deploys at the end of a run. If the car has a mass of 820 kg and the car is moving 36 m/s, wh
Lelechka [254]

In order to determine the required force to stop the car, proceed as follow:

Calculate the deceleration of the car, by using the following formula:

v^2=v^2_o-2ax

where,

v: final speed = 0m/s (the car stops)

vo: initial speed = 36m/s

x: distance traveled = 980m

a: deceleration of the car= ?

Solve the equation above for a, replace the values of the other parameters and simplify:

\begin{gathered} a=\frac{v^2_o-v^2}{2x} \\ a=\frac{(36\frac{m}{s})^2-(0\frac{m}{s})^2}{2(980m)}=0.66\frac{m}{s^2} \end{gathered}

Next, consider that the formula for the force is:

F=ma

where,

m: mass of the car = 820 kg

a: deceleration of the car = 0.66m/s^2

Replace the previous values and simplify:

F=(820kg)(0.66\frac{m}{s^2})=542.20N

Hence, the required force to stop the car is 542.20N

4 0
1 year ago
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