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4 years ago

when you squeeze the rubber bulb of a perfume atomizer how do you change the air pressure at the top of the tube

Physics

1 answer:

LuckyWell [14K]4 years ago

8 0

You push air and perfume out of the tube, creating a small vaccum that pulls more perfume into the tube

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A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal for

nadezda [96]

Answer:

a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) The final speed of the block of ice is 9.8 meters per second.

Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block ( $W$ ), measured in newtons, is:

$W = m\cdot g$ (1)

Where:

$m$ - Mass of the block of ice, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $m = 14\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , the magnitudes of the weight and normal force of the block of ice are, respectively:

$N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$N = W = 137.298\,N$

And the pull force is:

$F_{pull} = 98\,N$

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

$m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f}$ (2)

Where:

$v_{o}$ , $v_{f}$ - Initial and final speeds of the block, measured in meters per second.

$\Sigma F$ - Horizontal net force, measured in newtons.

$\Delta t$ - Impact time, measured in seconds.

Now we clear the final speed in (2):

$v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}$

If we know that $v_{o} = 0\,\frac{m}{s}$ , $m = 14\,kg$ , $\Sigma F = 98\,N$ and $\Delta t = 1.40\,s$ , then final speed of the ice block is:

$v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}$

$v_{f} = 9.8\,\frac{m}{s}$

The final speed of the block of ice is 9.8 meters per second.

6 0

3 years ago

Carbon dioxide enters an adiabatic nozzle steadily at 1 MPa, 518 oC, and mass flow rate of 5,322 kg/h and exits the system at 96

Orlov [11]

Answer:

The velocity at the nozzle at inlet $V_{1}$ = 3584 $\frac{m}{sec}$

Explanation:

Pressure at inlet $P_{1}$ = 1 × $10^{6}$ Pa

Temperature at inlet $T_{1}$ = 518 ° c = 791 K

Mass flow rate = $\frac{5322}{60}$ $\frac{kg}{sec}$ = 88.7

Gas constant for carbon die oxide is R = 189 $\frac{J}{kg k}$

Mass flow rate inside the nozzle is given by the formula = $\frac{P_{1} }{R T_{1} }$ × $A_{1}$ × $V_{1}$

⇒ $P_{1}$ = = 1 × $10^{6}$ Pa

⇒ R $T_{1}$ = 791 × 189 = 149499 $\frac{J}{kg}$

⇒ $A_{1}$ = 0.0037 $m^{2}$

Put all the above values in above formula we get,

⇒ 88.7 = $\frac{10^{6} }{149499}$ × 0.0037 × $V_{1}$

⇒ $V_{1}$ = 3584 $\frac{m}{sec}$

This is the velocity at the nozzle at inlet.

3 0

3 years ago

Potassium is a crucial element for the healthy operation of the human

Dovator [93]

The second one if it’s on edge

8 0

3 years ago

What is newton's 3rd law of physics

ValentinkaMS [17]

Answer:

His third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal and opposite force on object A.

4 0

3 years ago

Anyone know what type of circuit this is?

Oksana_A [137]

Answer:

Explanation:

BECAUSE ITS GOING ON AND ON IF ITS NOT CORRECT I WILL VOTE YOU BRAINLEST ON MY QUESTION

3 0

3 years ago