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Varvara68 [4.7K]
4 years ago
9

An important thing to consider when responding to a driver in front of you that stops suddenly is:a. the mental state of the oth

er driverb. the position of the car behind youc. whether the other driver acted legally
Physics
1 answer:
Gre4nikov [31]4 years ago
5 0
The correct answer is option A. i.e. An important thing to consider when responding to a driver in front of you that stops suddenly is: the mental state of the other driver.

Our talk or discussion can disrturb the balance of the driver or he can get distracted. So, we must try not to speak much while the driver is driving because by doing this we are putting the life of ourselves in danger. Any distrcaction of driver can cause accident.
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A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Co
Tems11 [23]

(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>
  • The wheels under the wind do not pass through the center line.
  • The position of the front wheel is constant and it is zero mark (origin).
  • The rear wheels are at 21.7 m mark

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}

18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

Learn more about center mass here: brainly.com/question/13499822

7 0
2 years ago
A cart rolling down an incline that is 60 meters long. If the cart has a beginning speed of 2.0 m/s and it
fiasKO [112]

Answer:

4 m/s²

Explanation:

Given:

Δx = 60 m

v₀ = 2.0 m/s

v = 22 m/s

Find: a

v² = v₀² + 2aΔx

(22 m/s)² = (2.0 m/s)² + 2a (60 m)

a = 4 m/s²

3 0
3 years ago
Jake is in chemistry class. He makes a list of the chemicals his instructor described and the properties of each.
koban [17]
Silver: bonds with other atoms because of the weak forces of the valence electrons 
FALSE - The strong forces of the valence electrons is actually the reason why silver bonds with other atoms.

Water: bonds allow for liquid state at room temperature and prevent conduction 
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Carbon: bonds with other atoms through strong shared electrical bonds 
TRUE - Carbon shares covalent bonds with other atoms.

Niobium: bonds allow for a strong conductivity found in stainless steel
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6 0
3 years ago
Read 2 more answers
What is the formula that can be used to find velocity if kinetic energy and mass are known?
viva [34]
The formula for kinetic energy is \frac{1}{2}m\Delta v^2. Thus, the equation for velocity is v=   \sqrt{ \frac{2TotalKineticEnergy}{m} }. 
6 0
3 years ago
Read 2 more answers
A brave child decides to grab onto an already spinning merry‑go‑round. The child is initially at rest and has a mass of 25.5 kg.
Svetllana [295]

Answer:

72.75 kg m^2

Explanation:

initial angular velocity, ω = 35 rpm

final angular velocity, ω' =  19 rpm

mass of child, m = 15.5 kg

distance from the centre, d = 1.55 m

Let the moment of inertia of the merry go round is I.

Use the concept of conservation of angular momentum

I ω = I' ω'

where I' be the moment of inertia of merry go round and child

I x 35 = ( I + md^2) ω'

I x 35 = ( I + 25.5 x 1.55 x 1.55) x 19

35 I = 19 I + 1164

16 I = 1164

I = 72.75 kg m^2

Thus, the moment of inertia of the merry go round is 72.75 kg m^2.

5 0
3 years ago
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