Please help me, I'm not good in physics.

a large cargo trucks needs to cross a bridge the truck is 30m long, and 3.2 wide, the cargo exerts a force of 54,000 N the bridg

otez555 [7]

Answer:

It isn't safe for the truck to cross the bridge because the pressure exerted by the truck on the bridge is greater than the maximum tolerable pressure for the bridge, 562.5 Pa > 450 Pa.

Explanation:

Pressure is expressed in Force/Area.

So, for the truck, force exerted = 54000 N

Area covered = 30 × 3.2 = 96 m²

Pressure exerted by the truck = 54000/96 = 562.5 Pa

The pressure exerted by the truck on the bridge is greater than the maximum tolerable pressure for the bridge, 562.5 > 450, hence, it isn't safe for the truck to cross the bridge.

5 0

4 years ago

On the same spring day, a station near the equator has a surface temperature of 25°C, 15°C higher than the middle-latitude city

DedPeter [7]

Answer:

The air temperature at the tropopause is - 79 °C

Explanation:

We know that a station near the equator has a surface temperature of 25°C

Vertical soundings reveal an environmental lapse rate of 6.5 °C per kilometer.

The tropopause is encountered at 16 km.

In order to find the air temperature at the tropopause we are going to deduce a linear function for the temperature at the tropopause.

This linear function will have the following structure :

$f(x)=ax+b$

Where '' $a$ '' and '' $b$ '' are real numbers.

Let's write $T(x)$ to denote the temperature '' T '' in function of the distance

'' x '' ⇒

$T(x)=ax+b$

We can reorder the function as :

$T(x)=b+ax$ (I)

Now, at the surface the value of '' $x$ '' is 0 km and the temperature is 25°C so in the function (I) we write :

$T(0)=25=b+a(0)$ ⇒ $b=25$ ⇒

$T(x)=25+ax$ (II)

In (II) the value of '' $a$ '' represents the change in temperature per kilometer.

Because the temperature decrease with the height this number will be negative and also a data from the question ⇒

$T(x)=25-(6.5)x$ (III)

In (III) we deduced the linear equation. The last step is to replace by $x=16$ in (III) ⇒

$T(16)=25-(6.5)(16)=-79$

The air temperature at the tropopause is - 79 °C

6 0

3 years ago

PLEASE SOLVE QUICKLY!!!<br> Solve for A, B, and C from graph

hram777 [196]

A = 59.35cm

B = 196.56g

C = 74.65g

<u>Explanation:</u>

We know,

$x = \frac{L}{\frac{W}{F} +1}$

and L = x+y

1.

Total length, L = 100cm

Weight of Beam, W = 71.8g

Center of mass, x = 49.2cm

Added weight, F = 240g

Position weight placed from fulcrum, y = ?

$L-y = \frac{L}{\frac{W}{F}+1 } \\100 - y = \frac{100}{\frac{71.8}{49.2}+1 } \\100 - y = \frac{100}{1.46+1}\\\\100 - y = \frac{100}{2.46} \\100-y = 40.65\\\\y = 59.35cm$

Therefore, position weight placed from fulcrum is 59.35cm

2.

Total length, L = 100cm

Center of mass, x = 47.8 cm

Added weight, F = 180g

Position weight placed from fulcrum, y = 12.4cm

Weight of Beam, W = ?

$47.8 = \frac{100}{\frac{W}{180}+1 }\\\47.8 = \frac{100}{\frac{W+180}{180} } \\\\47.8 = \frac{100 X 180}{W+180}\\ \\47.8W + 47.8 X 180 = 18000\\47.8W = 18000 - 8604\\W = \frac{9396}{47.8}\\ W = 196.56g$

Therefore, weight of the beam is 196.56g

3.

Total length, L = 100cm

Center of mass, x = 50.8 cm

Position weight placed from fulcrum, y = 9.8cm

Weight of Beam, W = 72.3g

Added weight, F = ?

$50.8 = \frac{100}{\frac{72.3}{F}+1 }\\\ 50.8 = \frac{100}{\frac{72.3+F}{F} } \\\\50.8 = \frac{100 X F}{72.3+F}\\ \\50.8 X 72.3 + 50.8 X F = 100F\\\\3672.84 = 100F-50.8F\\3672.84 = 49.2F\\F = 74.65g$

Therefore, Added weight F is 74.65g

A = 59.35cm

B = 196.56g

C = 74.65g

4 0

3 years ago

A scientist compares two samples of colorless gas present at the beginning and end of a lab process. She wants to determine whet

kompoz [17]

Joseph Priestley was born in Yorkshire, the eldest son of a maker of wool cloth. His mother died after bearing six children in six years. Young Joseph was sent to live with his aunt, Sarah Priestley Keighley, until the age of 19. She often entertained Presbyterian clergy at her home, and Joseph gradually came to prefer their doctrines to the grimmer Calvinism of his father. Before long, he was encouraged to study for the ministry. And study, as it turned out, was something Joseph Priestley did very well.

Aside from what he learned in the local schools, he taught himself Latin, Greek, French, Italian, German and a smattering of Middle Eastern languages, along with mathematics and philosophy. This preparation would have been ideal for study at Oxford or Cambridge, but as a Dissenter (someone who was not a member of the Church of England) Priestley was barred from England's great universities. So he enrolled at Daventry Academy, a celebrated school for Dissenters, and was exempted from a year of classes because of his achievements.

After graduation, he supported himself, as he would for the rest of his life, by teaching, tutoring and preaching. His first full-time teaching position was at the Dissenting Academy in Warrington. (Although obviously brilliant, original, outspoken and, by one report, of "a gay and airy disposition," Priestley had an unpleasant voice and a sort of stammer. That he made a living through lectures and sermons is further evidence of his extraordinary nature.)

https://www.acs.org/content/acs/en/education/whatischemistry/landmarks/josephpriestleyoxygen.html

8 0

3 years ago

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A student decides to give his bicycle a tune up. He flips it upside down (so there's no friction with the ground) and applies a

aleksklad [387]

Answer:

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