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sveticcg [70]
2 years ago
14

Which trailer has more downward pressure where it attaches to the car?

Physics
1 answer:
VARVARA [1.3K]2 years ago
4 0

The one that is loaded worst. The overall weight is not important; tongue weight is a matter of loading. Our 12,000 lb snow cat trailer, which has stops to position the cat properly, has under 100 lbs tongue weight. Excessive tongue weight is a Bad Thing because it reduces weight on the towing vehicle's front wheels, leading to instability.

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7) A heavy dart and a light dart are launched horizontally on a frictionless table by identical ideal springs. Both springs were
Elenna [48]

Answer:

A, B, and E

Explanation:

The springs are identical, and are compressed the same amount, so they have the same initial elastic potential energy. (E is true)

Energy is conserved, so the darts have the same amount of kinetic energy. (A is true, C is false)

The lighter dart has the same energy as the heavier dart.  Since it has less mass, it must have a greater speed. (B is true, D is false)

6 0
2 years ago
A 4kg block is sliding on a horizontal frictionless floor at a speed of 2.5ms and runs into a horizontal spring. The spring has
alexgriva [62]

Answer:

Explanation:

Given

mas of block m=4\ kg

speed of block v=2.5\ m/s

spring constant k=30\ N-m

As the mass collides with the spring its kinetic energy is converted to the Elastic Potential energy of the spring

\frac{1}{2}mv^2=\frac{1}{2}kx^2

x=v\sqrt{\frac{m}{k}}

x=2.5\times \sqrt{\frac{4}{30}}

x=0.912\ m

8 0
2 years ago
Read 2 more answers
Joan and Mike's teacher show them a picture of an atom and ask them to identify it using the periodic table. Joan says the atom
zhenek [66]

Well we can't see the picture that the teacher has, and we have to figure out what's in the picture from some clues in the answer choices.  The picture seems to show an atom with 3 protons and 3 neutrons in the nucleus, and 3 electrons whizzing around the nucleus.

If that's what's in the picture, then Mike is correct (first choice), because the atomic number shows the number of protons in all atoms. There are 3 protons and the atomic number of lithium is 3.

Second choice . . . False, because electrons are not involved in the atomic mass.

Third and fourth choices . . . both false; sadly, Joan is woefully unclear on the concepts.

3 0
2 years ago
A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s
Bezzdna [24]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring?

0.57 m

0.64 m  

0.80 m  

1.25 m

Data:

E_{pe}\:(elastic\:potential\:energy) = 5184\:J

K\:(constant) = 16200\:N/m

x\:(displacement) =\:?

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

E_{pe} = \dfrac{k*x^2}{2}

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:  

E_{pe} = \dfrac{k*x^2}{2}

5184 = \dfrac{16200*x^2}{2}

5184*2 = 16200*x^2

10368 = 16200\:x^2

16200\:x^2 = 10368

x^{2} = \dfrac{10368}{16200}

x^{2} = 0.64

x = \sqrt{0.64}

\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark

Answer:  

The displacement of the spring = 0.8 m (or 0.80 m)

_________________________________________

I Hope this helps, greetings ... Dexteright02! =)

8 0
2 years ago
Read 2 more answers
Eac of the two Straight Parallel Lines Each of two very long, straight, parallel lines carries a positive charge of 24.00 m C/m.
Cloud [144]

Answer:

The magnitude of the electric field at a point equidistant from the lines is 4.08\times10^{5}\ N/C

Explanation:

Given that,

Positive charge = 24.00  μC/m

Distance = 4.10 m

We need to calculate the angle

Using formula of angle

\theta=\sin^{-1}(\dfrac{\dfrac{d}{2}}{2d})

\theta=\sin^{-1}(\dfrac{1}{4})

\theta=14.47^{\circ}

We need to calculate the magnitude of the electric field at a point equidistant from the lines

Using formula of electric field

E=\dfrac{2k\lambda}{r}\times2\cos\theat

Put the value into the formula

E=\dfrac{2\times9\times10^{9}\times24.00\times2\times10^{-6}\cos14.47}{2.05}

E=408094.00\ N/C

E=4.08\times10^{5}\ N/C

Hence, The magnitude of the electric field at a point equidistant from the lines is 4.08\times10^{5}\ N/C

6 0
2 years ago
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