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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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car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of a. The car
Luba_88 [7]

Answer:

0.572

Explanation:

First examine the force of friction at the slipping point where Ff = µsFN = µsmg.

the mass of the car is unknown,

The only force on the car that is not completely in the vertical direction is friction, so let us consider the sums of forces in the tangential and centerward directions.

First the tangential direction

∑Ft =Fft =mat

And then in the centerward direction ∑Fc =Ffc =mac =mv²t/r

Going back to our constant acceleration equations we see that v²t = v²ti +2at∆x = 2at πr/2

So going backwards and plugging in Ffc =m2atπr/ 2r =πmat

Ff = √(F2ft +F2fc)= matp √(1+π²)

µs = Ff /mg = at /g √(1+π²)=

1.70m/s/2 9.80 m/s² x√(1+π²)= 0.572

7 0
3 years ago
The process that enhances some properties of an object at the expense of other properties is called:
lidiya [134]

Answer:

The answer is: letter a, pop-out effect.

Explanation:

The "pop-out effect" is a phenomenon which allows the person's precognitive processes to detect a<em> visual stimulus that is potentially the most meaningful one</em> in a person's spatial field of attention. The pop-up effect occurs when a person distinguishes one object from the rest.

For example, when a child chooses among pictures in different colors, it is common for the child to point at colored pictures rather than grayscale pictures. This is an example of a pop-out effect. <u>The properties of the colored pictures is more preferred by the child thus, causing him not to choose or mind the grayscale images.</u>

Thus, this explains the answer.

6 0
3 years ago
Calculate the potential energy of a 20-kg sled at 40 meters
irakobra [83]

Answer: 7840N

Explanation:

Given that

Potential energy = ?

Mass of sled = 20-kg

Distance = 40 meters

Acceleration due to gravity = 9.8m/s^2

Recall that potential energy is the energy possessed by a body at rest

i.e potential energy = mass m x acceleration due to gravity g x distance h

P.E = mgh

P.E = 20kg x 9.8m/s^2 x 40m

P.E = 7840N

Thus, the potential energy of the sled is 7840N

5 0
3 years ago
Read 2 more answers
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3
Pachacha [2.7K]

This question is incomplete, the complete question is;

Coulomb's law for the magnitude of the force F between two particles with charges Q and Q' separated by a distance d is

|FI = |QQ'I / d²

where K = 1/4π∈0, and

∈0 = 8.854 × 10⁻¹² C²/(N.m²) is the permittivity of free space.

Consider two point charges located on the x-axis:

one charge, q₁ = -18.5 nC, is located at

x₁ = -1.715m; the charge q₂ = 30.5 nC, is at the origin ( x₂=0 )

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q₃ = 51.0 nC placed between q₁ and q₂ at x₃ = -1.085 m ?

Answer: (Fnet3)x = -3.3287 × 10⁻⁵ N

Explanation:

Given that;

Q₁ = -18.5 nC       Q₃ = 51 nC        Q₂ = 30.5 nC

x₁ = - 1.715m         x₃ = - 1.085m     x₂ = 0

Now

x - component of Net force on charge Q₃ is

(Fnet3)x = -K|Q₁I|Q₃I / r₁3² - -K|Q₂I|Q₃I / r₂3²

(Fnet3)x = -(9×10⁹)(51×10⁻⁹) [ 18.5 / ((-1.085 + 1.715)²) + (30.5 / (-1.085)² ] × 10⁻⁹

(Fnet3)x = -3.3287 × 10⁻⁵ N

6 0
3 years ago
Two in-phase loudspeakers, which emit sound in all directions, are sitting side by side. One of them is moved sideways by 4.0 mm
loris [4]

Answer:

4.4721m

Explanation:

#Use Pythagorean theorem to find the distance between the two speakers:

a^+b^2=c^2\\4^2+2^2=d^2\\\sqrt{20}\ mm=d

There are antinodes 1/4,1/2 and 3/4 of the distance between speakers.

The greatest antinode is 3/4-1/4=1/2

#Distance between consecutive antinodes is:

0.5\times \sqrt{20}=\lambda/2\\\\\lambda=4.4721m

Hence, the maximum possible wavelength of the sound waves is 4.4721m

6 0
3 years ago
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