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jarptica [38.1K]
2 years ago
8

What is the law of physics 10 points if you tell me the answer and your name

Engineering
1 answer:
levacccp [35]2 years ago
7 0
<h2>꧁༒Answer༒꧂</h2>

<h2>F=m•a</h2>

f = force

m = mass \: of \: an \: object

a = acceleration

Newton's second law is one of the most important in all of physics. For a body whose mass m is constant, it can be written in the form F = ma, where F (force) and a (acceleration) are both vector quantities. If a body has a net force acting on it, it is accelerated in accordance with the equation

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c   Waste stack

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You want a potof water to boil at 105 celcius. How heavy a
ankoles [38]

Answer:

35.7 kg lid we put

Explanation:

given data

temperature = 105 celcius

diameter = 15 cm

Patm = 101 kPa

to find out

How heavy a  lid should you put

solution

we know Psaturated from table for temperature is 105 celcius is

Psat = 120.8 kPa

so

area will be here

area = \frac{\pi }{4} d^2    ..................1

here d is diameter

put the value in equation 1

area = \frac{\pi }{4} 0.15^2

area = 0.01767 m²

so net force is

Fnet = ( Psat - Patm ) × area

Fnet = ( 120.8 - 101 ) × 0.01767

Fnet = 0.3498 KN = 350 N

we know

Fnet = mg

mass = \frac{Fnet}{g}

mass  = \frac{350}{9.8}

mass = 35.7 kg

so 35.7 kg lid we put

3 0
3 years ago
Bessy's calf weighed 99.19 pounds when calved on March 7th. Her calf gained an average of 1.85 pounds per day for 266 days. What
Zepler [3.9K]

Answer:

591.3

Explanation:

99.19 + (1.85 × 266) = 591.29

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4 0
4 years ago
In 1945, the United States tested the world’s first atomic bomb in what was called the Trinity test. Following the test, images
Zarrin [17]

Answer:

r=K A^{1/5} \rho^{-1/5} t^{2/5}

A= \frac{r^5 \rho}{t^2}

A=1.033x10^{21} ergs *\frac{Kg TNT}{4x10^{10} erg}=2.58x10^{10} Kg TNT

Explanation:

Notation

In order to do the dimensional analysis we need to take in count that we need to conditions:

a) The energy A is released in a small place

b) The shock follows a spherical pattern

We can assume that the size of the explosion r is a function of the time t, and depends of A (energy), the time (t) and the density of the air is constant \rho_{air}.

And now we can solve the dimensional problem. We assume that L is for the distance T for the time and M for the mass.

[r]=L with r representing the radius

[A]= \frac{ML^2}{T^2} A represent the energy and is defined as the mass times the velocity square, and the velocity is defined as \frac{L}{T}

[t]=T represent the time

[\rho]=\frac{M}{L^3} represent the density.

Solution to the problem 

And if we analyze the function for r we got this:

[r]=L=[A]^x [\rho]^y [t]^z

And if we replpace the formulas for each on we got:

[r]=L =(\frac{ML^2}{T^2})^x (\frac{M}{L^3})^y (T)^z

And using algebra properties we can express this like that:

[r]=L=M^{x+y} L^{2x-3y} T^{-2x+z}

And on this case we can use the exponents to solve the values of x, y and z. We have the following system.

x+y =0 , 2x-3y=1, -2x+z=0

We can solve for x like this x=-y and replacing into quation 2 we got:

2(-y)-3y = 1

-5y = 1

y= -\frac{1}{5}

And then we can solve for x and we got:

x = -y = -(-\frac{1}{5})=\frac{1}{5}

And if we solve for z we got:

z=2x =2 \frac{1}{5}=\frac{2}{5}

And now we can express the radius in terms of the dimensional analysis like this:

r=K A^{1/5} \rho^{-1/5} t^{2/5}

And K represent a constant in order to make the porportional relation and equality.

The problem says that we can assume the constant K=1.

And if we solve for the energy we got:

A^{1/5}=\frac{r}{t^{2/5} \rho^{-1/5}}

A= \frac{r^5 \rho}{t^2}

And now we can replace the values given. On this case t =0.025 s, the radius r =140 m, and the density is a constant assumed \rho =1.2 kg/m^2, and replacing we got:

A=\frac{140^5 1.2 kg/m^3}{(0.025 s)^2}=1.033x10^{14} \frac{kg m^2}{s^2}

And we can convert this into ergs we got:

A= 1.033x10^{14} \frac{kgm^2}{s^2} * \frac{1 x10^7 egrs}{1 \frac{kgm^2}{s^2}}=1.033x10^{21} ergs

And then we know that 1 g of TNT have 4x10^4 erg

And we got:

A=1.033x10^{21} ergs *\frac{Kg TNT}{4x10^{10} erg}=2.58x10^{10} Kg TNT

3 0
3 years ago
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