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3 years ago

Please help!!

Physics

1 answer:

dmitriy555 [2]3 years ago

6 0

Answer:

the bag DOES NOT BREAK, F = 59 N

Explanation:

For this exercise, let's use Newton's second law, let's set a reference frame with the positive direction up

F -W = m a

the mass of the groceries is m = 5 kg, the weight is

W = m g

indicate the acceleration a = 2 m / s², let's find the force

F = mg + ma

F = m (g + a)

let's calculate

F = 5 (9.8 + 2)

F = 59 N

F <65 N

as the force with which it pulls the bag less than the breaking force the bag DOES NOT BREAK

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A computer is purchased for $2816 and depreciates at a constant rate to $0 in 8 years. Find a formula for the value, V , of the

marusya05 [52]

Answer:

The formula its $f(t) \ = \ - \ 352 \ \frac{\$ }{years} \ t \ + \ \$ \ 2816$
After 5 years, the computer value its $ 1056

Explanation:

<h3>Obtaining the formula</h3>

We wish to find a formula that

Starts at 2816. $f(0 \ years) \ = \ \$ \ 2816$
Reach 0 at 8 years. $f( 8 \ years) \ = \ \$ \ 0$
Depreciates at a constant rate. m

We can cover all this requisites with a straight-line equation. (an straigh-line its the only curve that has a constant rate of change) :

$f(t) \ = \ m\ t \ + \ b$ ,

where m its the slope of the line and b give the place where the line intercepts the <em>y</em> axis.

So, we can use this formula with the data from our problem. For the first condition:

$f ( 0 \ years ) = m \ (0 \ years) + b = \$ \ 2816$

$b = \$ \ 2816$

So, b = $ 2816.

Now, for the second condition:

$f ( 8 \ years ) = m \ (8 \ years) + \$ \ 2816 = \$ \ 0$

$m \ (8 \ years) = \ - \$ \ 2816$

$m = \frac{\ - \$ \ 2816}{8 \ years}$

$m = \ - \ 352 \frac{\$ }{years}$

So, our formula, finally, its:

$f(t) \ = \ - \ 352 \ \frac{\$ }{years} \ t \ + \ \$ \ 2816$

<h3>After 5 years</h3>

Now, we just use <em>t = 5 years</em> in our formula

$f(5 \ years) \ = \ - \ 352 \ \frac{\$ }{years} \ 5 \ years \ + \ \$ \ 2816$

$f(5 \ years) \ = \ - \$ \ 1760 + \ \$ \ 2816$

$f(5 \ years) \ = $ \ 1056$

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4 years ago

The Tambora volcano on the island of Sumbawa, Indonesia has been known to throw ash into the air with a speed of 625 m/s during

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About 1-3 km

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4 years ago

The equation T^2=A^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A

kobusy [5.1K]

Answer:

D. 2^(3/2)

Explanation:

Given that

T² = A³

Let the mean distance between the sun and planet Y be x

Therefore,

T(Y)² = x³

T(Y) = x^(3/2)

Let the mean distance between the sun and planet X be x/2

Therefore,

T(Y)² = (x/2)³

T(Y) = (x/2)^(3/2)

The factor of increase from planet X to planet Y is:

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4 years ago

A surface completely surrounds a 3.3 × 10-6 C charge. Find the electric flux through this surface when the surface is (a) a sphe

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Answer:

Electric flux in a) , b) and c) is same which is 0.373 × 10 ⁶ N m²/C

Explanation:

given,

surface charge (q) = 3.3 × 10⁻⁶ C

to calculate electric flux = ?

a) radius = 0.76 m

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electric flux = $\dfrac{3.3 \times 10^{-6}}{8.85 \times 10^{-12} }$

flux = 0.373 × 10 ⁶ N m²/C

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4 years ago

The rotational inertia of a thin rod about one end is 1/3 ML2. What is the rotational inertia of the same rod about a point loca

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Answer:

The value is $I = 0.0932 ML ^2$

Explanation:

From the question we are told that

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Generally the mass of the portion of the rod from the axis of rotation considered to the end of the rod is $0.4 M$

Generally the length of the rod from the its beginning to the axis of rotation consider is

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Generally the mass of the portion of the rod from the its beginning to the axis of rotation consider is

$m = 1- 0.4 M = 0.6 M$

Generally the rotational inertia about the axis of rotation consider for the first portion of the rod is

$I_{R1} = \frac{1}{3} (0.6 M )(0.6L)^2$

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Generally the rotational inertia about the axis of rotation consider for the second portion of the rod is

$I_{R2} = \frac{1}{3} (0.6 M )(0.6L)^2$

=> $I_{R2} = \frac{1}{3} (0.4 M )(0.4L)^2$

=> $I_{R2} = \frac{1}{3} (0.4 M )L^2 0.4^2$

Generally by the principle of superposition that rotational inertia of the rod at the considered axis of rotation is

$I = \frac{1}{3} (0.6 M )L^2 0.6^2 + \frac{1}{3} (0.4 M )L^2 0.4^2$

=> $I = \frac{1}{3} ML ^2 [0.6 * (0.6)^2 + 0.4 * (0.4)^2 ]$

=> $I = 0.0932 ML ^2$

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3 years ago