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3 years ago

The tongue weight of a trailer should be what percent of the gross trailer weight rating

Physics

1 answer:

mario62 [17]3 years ago

7 0

Answer:

between 10 and 15 percent

Explanation:

How to put your load

- First load the heavy

The safe trailer starts loading correctly. Uneven weight can affect steering, brakes and swing control.

In general, 60% of the weight of the load should be in the front half of the trailer and 40% in the rear half (unless the manufacturer indicates something different). When you place the load, you want it to be balanced from side to side, keeping the center of gravity near the ground and on the axle of the trailer.

- Hold your load

After balancing the load, you must hold it in place. An untapped load can move when the vehicle is moving and cause trailer instability.

- Trailer weight

To avoid overloading the trailer, look for the recommended weight rating. It is located on the VIN plate in the trailer chassis, usually on the tongue. Confirm the Gross Vehicle Weight Classification (GVWR) before towing.

GVWR: is the total weight that the trailer can support, including its weight. You can also find this number as the Gross Trailer Weight (GTW). The weight of the tongue should be 10-15% of the GTW.

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A sports car is advertised to be able to stop in a distance of 50.0 m from a speed of 80 km. What is its acceleration and how ma

Flauer [41]

Explanation:

Given that,

Initial speed of the sports car, u = 80 km/h = 22.22 m/s

Final speed of the runner, v = 0

Distance covered by the sports car, d = 80 km = 80000 m

Let a is the acceleration of the sports car. It can be calculated using third equation of motion as :

$v^2-u^2=2ad$

$a=\dfrac{v^2-u^2}{2d}$

$a=\dfrac{0-(22.22)^2}{2\times 80000}$

$a=-0.00308\ m/s^2$

Value of g, $g=9.8\ m/s^2$

$a=\dfrac{-0.00308}{9.8}\ m/s^2$

$a=(-0.000314)\ g\ m/s^2$

Hence, this is required solution.

8 0

3 years ago

A linear accelerator uses alternating electric fields to accelerate electrons to close to the speed of light. A small number of

kobusy [5.1K]

Answer:

8.1 x 10^13 electrons passed through the accelerator over 1.8 hours.

Explanation:

The total charge accumulated in 1.8 hours will be:

Total Charge = I x t = (-2.0 nC/s)(1.8 hrs)(3600 s/ 1 hr)

Total Charge = - 12960 nC = - 12.96 x 10^(-6) C

Since, the charge on one electron is e = - 1.6 x 10^(-19) C

Therefore, no. of electrons will be:

No. of electrons = Total Charge/Charge on one electron

No. of electrons = [- 12.96 x 10^(-6) C]/[- 1.6 x 10^(-19) C]

<u>No. of electrons = 8.1 x 10^13 electrons</u>

6 0

3 years ago

A 35 kg child goes down a playground that is inclined at an angle of 27.5 degrees above the horizontal. Find the acceleration if

tigry1 [53]

Answer:

4.16m/s²

Explanation:

According to Newtons second law;

$\sum Fx = ma_x\\Fm - Ff = ma_x\\W sin\theta - \mu R cos \theta = ma_x\\mg sin\theta - \mu mg cos\theta = ma_x\\$

Fm is the moving force

$\mu$ is the coefficient of kinetic friction between the child and the slide

m is the mass

g is the acceleration due to gravity

a is the acceleration of the child

Substitute the given values and get the acceleration as shown;

35(9.8)sin27.5 - 0.415(35)(cos27.5) = 35a

158.38-12.88 = 35a

145.49 = 35a

a = 145.49/35

a = 4.16m/s²

Hence the acceleration of the body is 4.16m/s²

4 0

2 years ago

Cyclotrons are widely used in nuclear medicine for producing short-lived radioactive isotopes. These cyclotrons typically accele

lyudmila [28]

Answer:

Part a)

$v = 3.16 \times 10^7 m/s$

Part b)

r = 0.166 m

Explanation:

Part a)

As we know that the energy of the Hydride ion is given as

$E = 5 MeV$

here we have

$\frac{1}{2}mv^2 = 5\times 10^6(1.6 \times 10^{-19})$

also we know that

$m = 1.6 \times 10^{-27} kg$

now we have

$v = \sqrt{\frac{2 \times 5 \times 10^6(1.6 \times 10^{-19}}{1.6\times 10^{-27}}$

$v = 3.16 \times 10^7 m/s$

Part b)

As we know that magnetic force on the charge is centripetal force

so we have

$qvB = \frac{mv^2}{r}$

so we have

$r = \frac{mv}{qB}$

so we have

$r = \frac{1.6 \times 10^{-27}(3.16 \times 10^7)}{(1.6 \times 10^{-19}) 1.9}$

$r = 0.166 m$

4 0

3 years ago

Compare the energy consumption of two commonly used items in the household. Calculate the energy used by a 1.20 kW toaster oven,

9966 [12]

Answer:

The energy consumption by toaster oven is 432 kJ and that of fluorescent light (CFL) bulb is 435 kJ.

Explanation:

Power of toaster oven, P = 1.2 kW

Time, t = 6 minutes = 360 s

The energy used by toaster oven is given by :

$E=P\times t\\\\E=1.2\times 10^3\times 360\\\\E=432000\ J\\\\E=432\ kJ$

Power of fluorescent light (CFL) bulb, P' = 11 W

Time, t' = 11 h = 39600 s

Energy used by fluorescent light (CFL) bulb is given by :

$E'=P'\times t'\\\\E'=11\times 39600\\\\E'=435600\ J\\\\E'=435\ kJ$

So, the energy consumption by toaster oven is 432 kJ and that of fluorescent light (CFL) bulb is 435 kJ.

5 0

3 years ago