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2 years ago

8

What term refers to a universal fact sometimes based on mathematical equations?

1 answer:

noname [10]2 years ago

3 0

Answer:

Scientific law

Explanation:

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What is the smallest particle of an element that still retains of that element

Thepotemich [5.8K]

Answer:

atom

Explanation:

the first guy who gave link this is for you:

we cannot use links in brainly

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3 years ago

A 260-kg glider is being pulled by a 1,940-kg jet along a horizontal runway with an acceleration of a= 2.20 m/s^2 to the right.

lisov135 [29]

Answer:

a) The magnitude of the thrust provided by the jet's engines is 4840 newtons.

b) The magnitude of the tension in the cable connecting the jet and glider is 572 newtons.

Explanation:

a) By Newton's laws we construct the following equations of equilibrium. Please notice that both the glider and the jet experiments has the same acceleration:

Jet

$\Sigma F = F - T = m_{J}\cdot a$ (1)

Glider

$\Sigma F = T = m_{G}\cdot a$ (2)

Where:

$F$ - Thrust of jet engines, measured in newtons.

$T$ - Tension in the cable connecting the jet and glider, measured in newtons.

$m_{G}$ , $m_{J}$ - Masses of the glider and the jet, measured in kilograms.

$a$ - Acceleration of the glider-jet system, measured in meters per square second.

If we know that $m_{G} = 260\,kg$ , $m_{J} = 1,940\,kg$ and $a = 2.20\,\frac{m}{s^{2}}$ , then the solution of this system of equations:

By (2):

$T = (260\,kg)\cdot \left(2.20\,\frac{m}{s^{2}} \right)$

$T = 572\,N$

By (1):

$F = T+m_{J}\cdot a$

$F = 572\,N+(1,940\,kg)\cdot \left(2.20\,\frac{m}{s^{2}} \right)$

$F = 4840\,N$

The magnitude of the thrust provided by the jet's engines is 4840 newtons.

b) The magnitude of the tension in the cable connecting the jet and glider is 572 newtons.

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3 years ago

The smallest division value of electronic balance

lara [203]

Answer:

0.1g to 0.0000001g hope it helps uu

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3 years ago

A steam engine generating 5 hp is raising a load weighing 2000 lb.How high will thebload be raised in 10 s.

notsponge [240]

Answer:

Explanation:

bi

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4 years ago

You have two springs that are identical except that spring 1 is stiffer than spring 2, (k1 > k2). On which spring is more wor

kicyunya [14]

Answer:

Explanation:

Given

Two springs with spring constant $k_1$ and $k_2$

(a)If they are stretched using the same force

Force $F=k_1x_1=k_2x_2$

where $x_1$ and $x_2$ are the extension in the spring

so $\frac{k_1}{k_2}=\frac{x_2}{x_1}$

therefore $x_2 > x_1$

Work done is given by

$W=F\cdot x$

$W_1=F\cdot x_1$

$W_2=F\cdot x_2$

thus $W_2 > W_1$

(b)If they are stretched the same distance

$\frac{F_1}{F_2}=\frac{k_1}{k_2}$

Thus $F_1 > F_2$

For same extension, work done by Force 1 is more

$W_1 > W_2$

6 0

4 years ago