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Hoochie [10]
3 years ago
14

Mercury has a density of 13.6 g/cm , whereas the density of water is 1.00 g/cm. The pressure of the atmosphere is high enough to

push mercury 76.0 cm up
through a vacuum tube. How high can a column of water be pushed up in a vacuum tube by the atmosphere? (Hint: The level of either liquid rises till the atmospheric
pressure is in balance with the pressure by the liquid computed by density xgx height; since the atmospheric pressure is the same in both scenarios, one only
needs to compare the liquid pressures to find the height of water) (Express the height of water in meters and with 3 significant figures in the form XX.X; number only
- no units)
Physics
1 answer:
kenny6666 [7]3 years ago
7 0

10.3

Explanation:

Step 1:

The pressure exerted by any liquid column of height, h density d is given by the formula P = h * d * g

Step 2:

It is given that one atmosphere pressure pushes up 76.0 cm of mercury, we need to calculate the level of water that will be pushed by the same pressure.

Step 3:

Since the pressure pushing up mercury and water is the same

h_{mercury} * d_{mercury} * g = h_{water} * d_{water} * g

 h_{water} = \frac{h_{mercury}*d_{mercury} }{d_{water}}  = (13.6 g/cm * 76 cm)/1 g/cm = 1033.6 cm

Step 4:

Now we need to express the answer in meters.

1 m = 100 cm.

1033.6 cm = 10.336 m

This can be rounded off to 10.3 m

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A box of mass 60 kg is at rest on a horizontal floor that has a static coefficient of friction of 0.6 and a kinetic coefficient
gavmur [86]

Answer:

a) The minimum force required to start moving the box is 352.86 N

b) i) The friction force for the box in motion is 147.025 N

ii) The acceleration of box is 4.21625 m/s²

Explanation:

The parameters of the box at rest and the floor are;

The mass of the box = 60 kg

The static coefficient friction of the floor = 0.6

The kinetic coefficient friction of the floor = 0.25

Frictional force = Normal force × Friction coefficient

For an horizontal floor and the box laying on the floor, we have;

The normal force = The weight of the box = Mass of the box × Acceleration due to gravity, g

The acceleration due to gravity, g = 9.81 m/s²

The weight of the box  = 60 × 9.81 = 588.6 N

a) The static coefficient gives the frictional force observed by the box and which must be surpassed to bring about motion

Therefore;

The minimum force required to start moving the box = The static frictional force = Weight of the box × The static coefficient of friction

The minimum force required to start moving the box = 588.1 × 0.6 = 352.86 N

The minimum force required to start moving the box = 352.86 N

b) i) When an horizontal force of 400 N is applied, the applied force is larger than the static friction force, and the box will be in motion with the kinetic coefficient of friction being the source of friction

The friction force for the box in motion = 588.1 × 0.25 = 147.025 N

ii) The force, F with the box is in motion, is given as follows;

F = Mass of box × Acceleration of box, a = Applied force - Kinematic friction force

F = 60 × a = 400 - 147.025 = 252.975 N

60 × a = 252.975 N

a = 252.975 N/(60 kg) = 4.21625 m/s²

Acceleration of box, a = 4.21625 m/s².

6 0
2 years ago
What does fossil fuels do?
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3 years ago
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A wave travels through a medium because
Amanda [17]

in case you dont want to read the answer is B

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3 years ago
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In which of the following examples does the object have both kinetic and potential energy? Select all that apply.
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3 years ago
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A 0.58 kg mass is moving horizontally with a speed of 6.0 m/s when it strikes a vertical wall. The mass rebounds with a speed of
Sunny_sXe [5.5K]

Answer:

5.8\; {\rm kg\cdot m \cdot s^{-1}}.

Explanation:

If the mass of an object is m and the velocity of that object is v, the linear momentum of that object would be m\, v.

Assume that the initial velocity of the mass is positive (6.0\; {\rm m\cdot s^{-1}}.) However, the direction of the velocity is reversed after the impact. Thus, the sign of the new velocity of the object would be negative- the opposite of that of the initial velocity. The new velocity would be (-4.0\; {\rm m\cdot s^{-1}}).

Thus, the change in the velocity of the mass would be:

\begin{aligned}& (\text{Change in Velocity}) \\ =\; & (\text{Final Velocity}) - (\text{Initial Velocity}) \\ =\; & (-4.0\; {\rm m\cdot s^{-1}}) - (6.0\; {\rm m\cdot s^{-1}}) \\ =\; & (-10\; {\rm m\cdot s^{-1})\end{aligned}.

The change in the linear momentum of the mass would be:

\begin{aligned} & \text{change in momentum} \\ =\; & (\text{mass}) \times (\text{change in velocity}) \\ =\; & 0.58\; {\rm kg} \times (-10\; {\rm m\cdot s^{-1}}) \\  =\; & (-5.8\; {\rm kg \cdot m \cdot s^{-1}})\end{aligned}.

Thus, the magnitude of the change of the linear momentum would be 5.8\; {\rm kg \cdot m \cdot s^{-1}}.

7 0
2 years ago
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