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2 years ago

Describe what happens to pressure as the force exerted on a given area increases.

1 answer:

4 0

That's the definition of pressure ... force on a given area.

So when that force increases, it's an increase in pressure.

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9. A sailor pulls a boat along a dock using a rope at an angle of 60.0° with the

aleksklad [387]

Answer: (1) 3.83x10^3 J

Explanation:

(1) Fx=(255N)cos60°

dx=30.0m

w=Fx dx =(255)(cos60°)(30.0m)

6 0

3 years ago

A ventilating fan is operated by 0.5hp electric motor. How much work in joules can the fan do in 3 hours? (I need the answer asa

shtirl [24]

Answer:

Power= 0.5hp = 375W

T = 3hrs = 10800s

Power = Work done/ Time

Work = Power * Time = 375 * 10800 = 4050000J

3 0

2 years ago

What kind of water temperature lowers the dissolved oxygen

ehidna [41]

<h3>Answer</h3>

At a high temperature above 20° oxygen solubility starts to decrease.

<h3>Explanation</h3>

Oxygen, O2 is a very essential component of water as we can see in its chemical formula h2O.

The solubility of oxygen decreases as temperature increases. This means that warmer water will have less dissolved oxygen than does cooler water.

<h3>Other factors that affects oxygen solubility in water</h3>

Salt levels

higher the salt levels in water, lower will be oxygen in it.

Pressure

Water at lower altitudes can hold more dissolved oxygen than water at higher altitudes because dissolved oxygen will increase as pressure increases.

5 0

3 years ago

Read 2 more answers

The sun's energy begins as what form of energy?

blagie [28]

Answer:

nuclear energy.............

4 0

2 years ago

James threw a ball vertically upward with a velocity of 41.67ms-1 and after 2 second David threw a ball vertically upward with a

Reptile [31]

Answer:

When have passed 3.9[s], since James threw the ball.

Explanation:

First, we analyze the ball thrown by James and we will find the final height and velocity by the time two seconds have passed.

We'll use the kinematics equations to find these two unknowns.

$y=y_{0} +v_{0} *t+\frac{1}{2} *g*t^{2} \\where:\\y= elevation [m]\\y_{0}=initial height [m]\\v_{0}= initial velocity [m/s] =41.67[m/s]\\t = time passed [s]\\g= gravity [m/s^2]=9.81[m/s^2]\\Now replacing:\\y=0+41.67 *(2)-\frac{1}{2} *(9.81)*(2)^{2} \\\\y=63.72[m]\\$

Note: The sign for the gravity is minus because it is acting against the movement.

Now we can find the velocity after 2 seconds.

$v_{f} =v_{o} +g*t\\replacing:\\v_{f} =41.67-(9.81)*(2)\\\\v_{f}=22.05[m/s]$

Note: The sign for the gravity is minus because it is acting against the movement.

Now we can take these values calculated as initial values, taking into account that two seconds have already passed. In this way, we can find the time, through the equations of kinematics.

$y=y_{o} +v_{o} *t-\frac{1}{2} *g*t^{2} \\y=63.72 +22.05 *t-\frac{1}{2} *(9.81)*t^{2} \\\\y=63.72 +22.05 *t-4.905*t^{2} \\$

As we can see the equation is based on Time (t).

Now we can establish with the conditions of the ball launched by David a new equation for y (elevation) in function of t, then we match these equations and find time t

$y=y_{o} +v_{o} *t+\frac{1}{2} *g*t^{2} \\where:\\v_{o} =55.56[m/s] = initial velocity\\y_{o} =0[m]\\now replacing\\63.72 +22.05 *t-(4.905)*t^{2} =0 +55.56 *t-(4.905)*t^{2} \\63.72 +22.05 *t =0 +55.56 *t\\63.72 = 33.51*t\\t=1.9[s]$

Then the time when both balls are going to be the same height will be when 2 [s] plus 1.9 [s] have passed after David throws the ball.

Time = 2 + 1.9 = 3.9[s]

4 0

2 years ago