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2 years ago

Which of the following can significantly change a vehicle’s center of gravity?

Physics

2 answers:

Fittoniya [83]2 years ago

4 0

Answer:

The answer is Loading

Explanation:

The center of gravity is a point inside a mass where the weight of the mass is considered to act on. And the center of gravity of an object can change position if you add more mass on certain areas of the object. So lets say you have a vehicle and the vehicle has the center of gravity right in the middle (this means at mid-height, mid-length and mid-width). Now you put some stuff on the trunk, well guess what? The center of gravity now shifted some towards the trunk of your car. And if you didn´t place the extra mass right in the middle of the trunk, the you also moved the center of gravity to whatever side you put the mass on.

In regards of the other three options:

Force of impact has nothing to do with center of gravity. Although if you want to be very strict, if you crash and the force of impact deforms your car so much that the vehicle has now a totally different shape, then yeah, all the rearrangement of the mass will move the center of gravity from its initial position. But for the scope of this question, the answer is NO.
The antilock brakes or ABS have nothing to do with center of gravity. All they do is prevent your car from sliding by keepin the tires rotating when you slam on the brakes.
Torque has nothing to do with center of gravity. More torque will help you with acceleration of your vehicle.

Mademuasel [1]2 years ago

3 0

loading

Loading determines the weight and/or the center of pressure on a vehicle. Center of gravity refers to the whole weight of the load acting with its orientation. Loading affects the center of gravity as the object is positioned and balanced on the vehicle.

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Many industries are powered via distant power stations. Calculate the current flowing through a 7,300m long 10. copper power lin

Oliga [24]

Answer:

Current, I = 1000 A

Explanation:

It is given that,

Length of the copper wire, l = 7300 m

Resistance of copper line, R = 10 ohms

Magnetic field, B = 0.1 T

$\mu_o=4\pi \times 10^{-7}\ T-m/A$

Resistivity, $\rho=1.72\times 10^{-8}\ \Omega-m$

We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :

$R=\rho \dfrac{l}{A}$

$R=\rho \dfrac{l}{\pi r^2}$

$r=\sqrt{\dfrac{\rho l}{R\pi}}$

$r=\sqrt{\dfrac{1.72\times 10^{-8}\times 7300}{10\pi}}$

r = 0.00199 m

$r=1.99\times 10^{-3}\ m=2\times 10^{-3}\ m$

The magnetic field on a current carrying wire is given by :

$B=\dfrac{\mu_o I}{2\pi r}$

$I=\dfrac{2\pi rB}{\mu_o}$

$I=\dfrac{2\pi \times 0.1\times 2\times 10^{-3}}{4\pi \times 10^{-7}}$

I = 1000 A

So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.

4 0

2 years ago

The suspension cable of a 1,000 kg elevator snaps, sending the elevator moving downward through its shaft. The emergency brakes

tester [92]

Answer:

option (E) 1,000,000 J

Explanation:

Given:

Mass of the suspension cable, m = 1,000 kg

Distance, h = 100 m

Now,

from the work energy theorem

Work done by the gravity = Work done by brake

mgh = Work done by brake

where, g is the acceleration due to the gravity = 10 m/s²

Work done by brake = 1000 × 10 × 100

Work done by brake = 1,000,000 J

this work done is the release of heat in the brakes

Hence, the correct answer is option (E) 1,000,000 J

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2 years ago

The end diastolic volume of a heart is 140 mL Assume that it is a sphere. At end diastole, the intraventricular pressure is 7mmI

Vera_Pavlovna [14]

Answer:

Explanation:

We know that, V = 140 mL = 0.00014 m3

Assume that it is a sphere. so, we have

V = (4/3) \pir3

r3 = (0.00014 m3) (3) / (4) (3.14)

r = \sqrt[3]{}\sqrt[3]{}3\sqrt{}3.34 x 10-5 m3

r = 1.93 x 10-7 m

(a) The wall tension at end diastole will be given as :

using a formula, we have

T = P r / 2 H

where, P = intraventricular pressure at end diastole = 7 mmHg = 933.2 Pa

H = wall thickness at this time = 0.011 m

then, we get

T = (933.2 Pa) (1.93 x 10-7 m) / 2 (0.011 m)

T = 8.18 x 10-3 N

(b) The wall tension at the end of isovolumetric contraction will be given as :

using a formula, we have

T = P r / 2 H

where, P = intraventricular pressure at end of isovolumetric contraction = 80 mmHg = 10665.7 Pa

H = wall thickness at this time = 0.011 m

then, we get

T = (10665.7 Pa) (1.93 x 10-7 m) / 2 (0.011 m)

T = 9.35 x 10-2 N

(d) The wall stress from A and B which will be given as :

we know that, \sigma = T / w

For part A, we have

\sigmaA = (8.18 x 10-3 N) / (0.011 m)

\sigmaA = 0.743 N/m

For part B, we have

\sigmaB = (9.35 x 10-2 N) / (0.011 m)

\sigmaB = 8.5 N/m

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Korvikt [17]

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Westkost [7]

Answer:The anwser is B i Promise ok

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