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katrin2010 [14]

2 years ago

12

Daria was swimming in a friend’s pool yesterday, when she saw that a fly had landed in the water about 5 feet away from her. She

began screaming, jumped out of the pool as fast as she could, and refused to get back in for the rest of the day. This is not the first time that her fear of bugs has interfered with a nice afternoon outside, and she has been less and less willing to participate in such activities recently. Daria probably suffers from _________.

1 answer:

jasenka [17]2 years ago

5 0

Answer:

Daria probably suffers from Entomophobia.

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: Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at

ser-zykov [4K]

Complete Question

Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A=0.0155 m. The air pressure in Denver is PD = 79000 Pa. and in New Orleans is PNo = 100250 Pa. Assume the lid is weightless.

Part (a) Write an expression for the force FNo required to remove the container lid in New Orleans.

Part (b) Calculate the force FNo required to lift off the container lid in New Orleans, in newtons.

Part (c) Calculate the force Fp required to lift off the container lid in Denver, in newtons.

Part (d) is more force required to lift the lid in Denver (higher altitude, lower pressure) or New Orleans (lower altitude, higher pressure)?

Answer:

a

The expression is $F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

b

$F_{No}= 7771.125 \ N$

c

$F_p = 2.2*10^{6} N$

d

From the value obtained we can say the that the force required to open the lid is higher at Denver

Explanation:

The altitude of container in Denver is $d_D = 6000 \ ft = 6000 * 0.3048 = 1828.8m$

The surface area of the container lid is $A = 0.0155m^2$

The altitude of container in New Orleans is sea-level

The air pressure in Denver is $P_D = 79000 \ Pa$

The air pressure in new Orleans is $P_{ro} = 100250 \ Pa$

Generally force is mathematically represented as

$F_{No} = \Delta P A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

So the $\Delta P$ which is the difference in pressure within and outside the container is

$\Delta P = P_{No} - \frac{P_{sea}}{2}$

Therefore

$F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

Now substituting values

$F_{No} = 0.0155 [100250 - \frac{101000}{2}]$

$F_{No}= 7771.125 \ N$

The force to remove the lid in Denver is

$F_p = \Delta P_d A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

At sea level the air pressure in Denver is mathematically represented as

$P_D = \rho g h$

=> $g = \frac{P_D}{\rho h}$

Let height at sea level is h = 1

The air pressure at height $d_D$

$P_d__{D}} = \rho gd_D$

=> $g = \frac{P_d_D}{\rho d_D}$

Equating the both

$\frac{P_D}{\rho h} = \frac{P_d_D}{\rho d_D}$

$P_d_D = P_D * d_D$

Substituting value

$P_d__{D}} = 1828.2 * 79000$

$P_d__{D}} = 1.445*10^{8} Pa$

So

$\Delta P_d = P_{d} _D - \frac{P_{sea}}{2}$

=> $\Delta P_d = 1.445 *10^{8} - \frac{101000}{2}$

$\Delta P_d = 1.44*10^{8}Pa$

So

$F_p = \Delta P_d A$

$= 1.44*10^8 * 0.0155$

$F_p = 2.2*10^{6} N$

6 0

2 years ago

Water is boiled in a pan on a stove at sea level. During 10 min of boiling, it is observed that 200 g of water has evaporated. D

Ymorist [56]

Answer : The rate of heat transfer to the water is, 37.92 kJ/min

Explanation : Given,

Time = 10 min

Mass of water = 200 g

Latent heat of fusion of water = 334 J/g

Latent heat of vaporization of water = 2230 J/g

Now we have to calculate the rate of heat transfer to the water.

$Q=\frac{m\times (L_v-L_f)}{t}$

Now put all the given values in the above formula, we get:

$Q=\frac{200g\times (2230-334)J/g}{10min}$

$Q=37920J/min=37.92kJ/min$

Thus, the rate of heat transfer to the water is, 37.92 kJ/min

4 0

2 years ago

You have a suspect on the run in a dry, wooded area; they’ve ditched the gun but are hiding from you. how long do you have to fi

erma4kov [3.2K]

Answer:

C. a few days

Explanation:

The evidence that we will have to investigate is the gunshot residue on the body of the suspect. When a person fires a gun there will be a deposition of material from the components of the chemicals present in the gun. This deposit is called gunshot residue and consists of particles of the propellant or explosive primer.

Experimentation has shown that the period of time that is person will test possible for gunshot residue is 5.27 days.

7 0

1 year ago

Please help! will give brainlest!!!!!!!!!!!!

eimsori [14]

The force of friction is 19.1 N

Explanation:

According to Newton's second law, the net force acting on the bag is equal to the product between its mass and its acceleration:

$\sum F = ma$

where

$\sum F$ is the net force

m is the mass

a is the acceleration

The bag is moving at constant speed, so its acceleration is zero:

$a=0$

Therefore the net force is zero as well:

$\sum F = 0$

Here we are interested only in the forces acting along the horizontal direction, therefore the net force is given by:

$\sum F = F cos \theta - F_f = 0$

where

$F cos \theta$ is the horizontal component of the applied force, with

F = 22.5 N

$\theta=32.0^{\circ}$

$F_f$ is the force of friction

And solving for $F_f$ , we find

$F_f =Fcos \theta=(22.5)(cos 32.0^{\circ})=19.1 N$

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

7 0

2 years ago

A dog runs 3 miles and 4 miles west in 6 hours. What’s the dogs total distance and displacement ?

Inga [223]

Complete question:

A dog runs 3 miles east and 4 miles west in 6 hours. What’s the dogs total distance and displacement ?

Answer:

The total distance covered by the dog is 7 miles

The displacement of the dog is 1 mile west

Explanation:

Given;

initial position of the dog = 3 miles east

final position of the dog = 4 miles west

time of motion, t = 6 hours

The total distance covered by the dog is given as;

Total distance = 3 miles + 4 miles = 7 miles

The displacement of the dog is given as;

displacement = final position of the dog - initial position of the dog

displacement = 4 miles west - 3 miles east = 1 mile west

6 0

2 years ago