All categories

3 years ago

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

Physics

2 answers:

Fittoniya [83]3 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]3 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.

You might be interested in

How does the mass of an object affect the outcome when an unbalanced force acts on it?

Verdich [7]

Answer:

The magnitude of acceleration is reduced.

Explanation:

Force is defined as push or pull

The force is said to be balance force if the force are equal in size but opposite in direction. ie the object does not move or move with constant speed.

The force are to be unbalanced force if the force cause change in motion. ie the object has force greater than zero and has acceleration.

According to Newton second law of motion , acceleration depends on force acting on the object and mass of object.

F=ma

a= $\frac{F}{m}$

When unbalanced force act on the mass of object it reduces magnitude of acceleration without changing the direction.

6 0

2 years ago

The radius of the aorta is about 1 cm and the blood flowing through it has a speed of about 30 cm/s. Calculate the average speed

puteri [66]

Answer:

The average speed of the blood in the capillaries is 0.047 cm/s.

Explanation:

Given;

radius of the aorta, r₁ = 1 cm

speed of blood, v₁ = 30 cm/s

Area of the aorta, A₁ = πr₁² = π(1)² = 3.142 cm²

Area of the capillaries, A₂ = 2000 cm²

let the average speed of the blood in the capillaries = v₂

Apply continuity equation to determine the average speed of the blood in the capillaries.

A₁v₁ = A₂v₂

v₂ = (A₁v₁) / (A₂)

v₂ = (3.142 x 30) / (2000)

v₂ = 0.047 cm/s

Therefore, the average speed of the blood in the capillaries is 0.047 cm/s.

4 0

2 years ago

The amount by which a machine multiplies input force is known as

fiasKO [112]

Mechanical advantage is the amount by which a machine multiplies input force. It is a measure of the force amplification using a tool or machine system.

4 0

3 years ago

You stand on a merry-go-round which is spinning at f = 0:25 revolutions per second. You are R = 200 cm from the center. (a) Find

ivanzaharov [21]

Answer:

(a) ω = 1.57 rad/s

(b) ac = 4.92 m/s²

Explanation:

(a)

The angular speed of the merry go-round can be found as follows:

ω = 2πf

where,

ω = angular speed = ?

f = frequency = 0.25 rev/s

Therefore,

ω = (2π)(0.25 rev/s)

ω = 1.57 rad/s

(b)

The centripetal acceleration can be found as:

ac = v²/R

but,

v = Rω

Therefore,

ac = (Rω)²/R

ac = Rω²

therefore,

ac = (2 m)(1.57 rad/s)²

ac = 4.92 m/s²

(c)

In order to avoid slipping the centripetal force must not exceed the frictional force between shoes and floor:

Centripetal Force = Frictional Force

m*ac = μs*R = μs*W

m*ac = μs*mg

ac = μs*g

μs = ac/g

μs = (4.92 m/s²)/(9.8 m/s²)

μs = 0.5

7 0

2 years ago

A 1 530-kg automobile has a wheel base (the distance between the axles) of 2.70 m. The automobile's center of mass is on the cen

NeTakaya

Answer:

Force on front axle = 6392.85 N

Force on rear axle = 8616.45 N

Explanation:

As we know that the weight of the car is balanced by the normal force on the front wheel and rear wheels

Now we know that

$F_1 + F_2 = W$

$F_1 + F_2 = (1530\times 9.81)$

$F_1 + F_2 = 15009.3 N$

now we know that distance between the axis is 2.70 m and centre of mass is 1.15 m behind front axle

so we can write torque balance about its center of mass

$F_1(1.15) = F_2(2.70 - 1.15)$

$F_1 = 1.35 F_2$

now from above equation

$F_2 + 1.35F_2 = 15009.3$

now we have

$F_2 = 6392.85 N$

now the other force is given as

$F_1 = 8616.45 N$

4 0

3 years ago