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Basile [38]
3 years ago
5

Two identical trucks have mass 5500 kg when empty, and the maximum permissible load for each is 8000 kg. The first truck, carryi

ng a 3900 kg, is at rest. The second truck plows into it at 64 km/h, and the pair moves away at 44 km/h. As an expert witnes, you're asked to determine whether the second truck was overloaded. What do you report? Yes the truck is overloaded, or no, the truck is not overloaded?
Physics
1 answer:
Alexeev081 [22]3 years ago
4 0

Answer:

no, the truck is not overloaded

Explanation:

The computation is shown below;

Let us assume the mass of the loan in the second truck be M

So, the equation is as follows

{(Mass + M) × second truck × 1000 ÷ 3,600} = {(Mass + M + mass + first truck) × Pair moves away  × 1,000 ÷ 3,600}

{(5500 + M) × 64 × 1,000 ÷ 3,600 = {(5,500 + M + 5,500 + 3,900) × 44 × 1,000 ÷ 3,600}

(5500 + M) × 64 = (14,900 + M) × 44

352,000 + 64 M = 655,600 + 44 M

After solving this

M = 15,180 kg

Therefore the second truck is not overloaded

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Students have a small steel ball and a large steel ball, and they have a short ramp and a tall ramp.
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A uniform disk, a thin hoop, and a uniform sphere, all with the same mass and same outer radius, are each free to rotate about a
olga nikolaevna [1]

Answer:

4 hoop, disk, sphere

Explanation:

Because

We are given data that

Hoop, disk, sphere have Same mass and radius

So let

And Initial angular velocity, = 0

The Force on each be F

And Time = t

Also let

Radius of each = r

So let's find the inertia shall we!!

I1 = m r² /2

= 0.5 mr² the his is for dis

I2 = m r² for hoop

And

Moment of inertia of sphere wiil be

I3 = (2/5) mr²

= 0.4 mr²

So

ωf = ωi + α t

= 0 + ( τ / I ) t

= ( F r / I ) t

So we can see that

ωf is inversely proportional to moment of inertia.

And so we take the

Order of I ( least to greatest ) :

I3 (sphere) , I1 (disk) , I2 (hoop) , ,

Order of ωf: ( least to greatest)

That of omega xf is the reverse of inertial so

hoop, disk, sphere

Option - 4

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Find the cost of excavating a space 84 ft long, 42 ft wide, and 9 ft deep at a cost of $39/yd3. (simplify your answer completely
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The cost of excavating a space of 84 ft long, 42 ft wide, and 9 ft deep is $45864

Information about the problem:

  • Space long= 84 ft
  • Space wide= 42 ft
  • Space deep= 9 ft
  • Cost by yard3 = $39/yd3
  • Total cost= ?

To solve this problem, we have to state the equation using the information of the problem:

Calculating the volume of the total space:

space volume = space long * space wide * space deep

space volume = 84 ft * 42 ft * 9 ft

space volume = 31752 ft3

By converting the volume from ft3 to yd3, we have:

31752 ft3 * (0,037037 yd3 / 1 ft3) = 1176 yd3

Calculating the cost of excavating the volume space:

Total cost = space volume * cost by yard3

Total cost = 1176 yd3 * $39/yd3

Total cost = $45864

<h3>What is volume?</h3>

It is the space occupied by a body, it is calculated by multiplying its dimensions, for example: length, height and width.

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4 0
1 year ago
Which method will correctly determine whether the forces on an object are balanced or unbalanced?
MaRussiya [10]

For this case, the first thing you should do is define a reference system.

Once the system is defined, we must follow the following steps:

1) Do the sum of forces in a horizontal direction

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Answer:

Add the forces in the horizontal and vertical directions separately.

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