Ask question

Ask question

All categories

denis-greek [22]

2 years ago

14

A pressure cooker is a pot whose lid can be tightly sealed to prevent gas from entering or escaping. even without knowing how bi

g the pressure cooker is, or what altitude it is being used at, we can make predictions about how much force the lid will experience under different conditions.

1 answer:

ankoles [38]2 years ago

5 0

Even without knowing the size of the pressure cooker or the altitude at which the pressure cooker will be used, we can easily determine the force on the lid by experience because just as boiling water, the amount of pressure inside the vessel (in this case, the cooker) increases as the temperature increases.

You might be interested in

An electron moving at right angles to a 0.1 T magnetic field experiences an acceleration of 6 × 1015 m.s-2. What is the speed of

GaryK [48]

Explanation:

It is given that,

Magnetic field, B = 0.1 T

Acceleration, $a=6\times 10^{15}\ m/s^2$

Charge on electron, $q=1.6\times 10^{-19}\ C$

Mass of electron, $m=9.1\times 10^{-31}\ kg$

(a) The force acting on the electron when it is accelerated is, F = ma

The force acting on the electron when it is in magnetic field, $F=qvB\ sin\theta$

Here, $\theta=90$

So, $ma=qvB$

Where

v is the velocity of the electron

B is the magnetic field

$v=\dfrac{ma}{qB}$

$v=\dfrac{9.1\times 10^{-31}\ kg\times 6\times 10^{15}\ m/s^2}{1.6\times 10^{-19}\ C\times 0.1\ T}$

v = 341250 m/s

or

$v=3.41\times 10^5\ m/s$

So, the speed of the electron is $3.41\times 10^5\ m/s$

(b) In 1 ns, the speed of the electron remains the same as the force is perpendicular to the cross product of velocity and the magnetic field.

7 0

2 years ago

On a ring road, 12 trams are spaced at regular intervals and travel at a constant speed. How many trams need to be added to the

Sphinxa [80]

3 trams must be added

Explanation:

In this problem, there are 12 trams along the ring road, spaced at regular intervals.

Calling L the length of the ring road, this means that the space between two consecutive trams is

$d=\frac{L}{12}$ (1)

In this problem, we want to add n trams such that the interval between the trams will decrease by 1/5; therefore the distance will become

$d'=(1-\frac{1}{5})d=\frac{4}{5}d$

And the number of trams will become

$12+n$

So eq.(1) will become

$\frac{4}{5}d=\frac{L}{n+12}$ (2)

And substituting eq.(1) into eq.(2), we find:

$\frac{4}{5}(\frac{L}{12})=\frac{L}{n+12}\\\rightarrow n+12=15\\\rightarrow n = 3$

Learn more about distance and speed:

brainly.com/question/8893949

#LearnwithBrainly

4 0

2 years ago

"The predictions of Einstein’s Theory of General Relativity were tested on a double pulsar system in January of 2004. His equati

Rasek [7]

Answer:

99.95%

Explanation:

A double pulsar system named PSR J0737-3039A/B in Puppis constellation was discovered in the year 2003. Pulsars are second most densest object in the universe after black holes and they emit radio waves at regular intervals. This pair presented a great and natural setup to test the Theory of General Relativity presented by Einstein in 1915. In this theory Einstein had presented a set of equations on how the space-time fabric will be curved because of the very dense objects such as Neutron stars. It also predicted how the gravitational waves are created because of stars orbiting each other.

A team of astrophysicists led by Michael Kramer, conducted a study on how these gravitational waves will impact the time in which the radio waves emitted by pulsars will reach Earth. The result of the study proved the theory of General Relativity to be accurate up to 99.95%.

8 0

2 years ago

Air at 3 104 kg/s and 27 C enters a rectangular duct that is 1m long and 4mm 16 mm on a side. A uniform heat flux of 600 W/m2 is

ad-work [718]

Answer:

$T_{out}=27.0000077$ ºC

Explanation:

First, let's write the energy balance over the duct:

$H_{out}=H_{in}+Q$

It says that the energy that goes out from the duct (which is in enthalpy of the mass flow) must be equals to the energy that enters in the same way plus the heat that is added to the air. Decompose the enthalpies to the mass flow and specific enthalpies:

$m*h_{out}=m*h_{in}+Q\\m*(h_{out}-h_{in})=Q$

The enthalpy change can be calculated as Cp multiplied by the difference of temperature because it is supposed that the pressure drop is not significant.

$m*Cp(T_{out}-T_{in})=Q$

So, let's isolate $T_{out}$ :

$T_{out}-T_{in}=\frac{Q}{m*Cp}\\T_{out}=T_{in}+\frac{Q}{m*Cp}$

The Cp of the air at 27ºC is 1007 $\frac{J}{kgK}$ (Taken from Keenan, Chao, Keyes, “Gas Tables”, Wiley, 1985.); and the only two unknown are $T_{out}$ and Q.

Q can be found knowing that the heat flux is 600W/m2, which is a rate of heat to transfer area; so if we know the transfer area, we could know the heat added.

The heat transfer area is the inner surface area of the duct, which can be found as the perimeter of the cross section multiplied by the length of the duct:

Perimeter:

$P=2*H+2*A=2*0.004m+2*0.016m=0.04m$

Surface area:

$A=P*L=0.04m*1m=0.04m^2$

Then, the heat Q is:

$600\frac{W}{m^2} *0.04m^2=24W$

Finally, find the exit temperature:

$T_{out}=T_{in}+\frac{Q}{m*Cp}\\T_{out}=27+\frac{24W}{3104\frac{kg}{s} *1007\frac{J}{kgK} }\\T_{out}=27.0000077$

$T_{out}$ =27.0000077 ºC

The temperature change so little because:

The mass flow is so big compared to the heat flux.
The transfer area is so little, a bigger length would be required.

3 0

1 year ago

Two blocks, with masses m and 3m, are attached to the ends of a string with negligible mass that passes over a pulley, as shown

olasank [31]

Answer:

a) v = √ g x , b) W = 2 m g d , c) a = ½ g

Explanation:

a) For this exercise we use Newton's second law, suppose that the block of mass m moves up

T-W₁ = m a

W₃ - T = M a

w₃ - w₁ = (m + M) a

a = (3m - m) / (m + 3m) g

a = 2/4 g

a = ½ g

the speed of the blocks is

v² = v₀² + 2 ½ g x

v = √ g x

b) Work is a scalar, therefore an additive quantity

light block s

W₁ = -W d = - mg d

3m heavy block

W₂ = W d = 3m g d

the total work is

W = W₁ + W₂

W = 2 m g d

c) in the center of mass all external forces are applied, they relate it is

a = ½ g

8 0

1 year ago

Read 2 more answers