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Assoli18 [71]
3 years ago
5

A car starts from rest and travels for t1 seconds with a uniform acceleration a1. The driver then applies the brakes, causing a

uniform acceleration a2. If the brakes are applied for t2 seconds, determine the following. Answers are in terms of the variables a1, a2, t1, and t2.
THIS IS NOT A MULTIPLE-CHOICE QUESTION - I NEED THE EQUATION FOR EACH SECTION

a) How fast is the car going just before the beginning of the braking period?

b) How far does the car go before the driver begins to brake?

c) Using the answers to parts (a) and (b) as the initial velocity and position for the motion of the car during braking, what total distance does the car travel?
Physics
1 answer:
aksik [14]3 years ago
6 0

Answer:

Here, we are required to determine how fast the car is moving, and how far the car goes before the braking period.

A. Vf = acceleration, a1 × time, t1

B. Therefore, D1 = a1 × (t1)².

C. D = {a1 × (t1)² + a2 × (t2)²}.

A. At uniform acceleration, the rate of change of velocity is constant with time.

However, since the car starts from rest, it's initial velocity, Vi = 0.

Therefore, the velocity (speed) at which the car is moving before the braking period, is,

Vf = acceleration, a1 × time, t1

B. To determine how far

the car has moved before the braking period is, D1 = Velocity, Vf × time, t1.

However, velocity, Vf = a1 × t1.

Therefore, D1 = a1 × (t1)².

This is so because the velocity, Vf is the velocity of travel from rest and lasts over time, t1 at which point the braking period commenced.

C. During the braking period, there's deceleration (i.e decay in speed) which returns the velocity of the car back to zero, ultimately bringing the car to a halt.

During the braking period, the total distance covered is also,

D2 = a2 × (t2)².

Therefore, the total distance covered during the motion is, D1 + D2

Total distance, D = D1 + D2.

D = {a1 × (t1)² + a2 × (t2)²}.

Read more:

brainly.com/question/13664094

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dem82 [27]

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4 0
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Scientists have concluded that the uppermost part of the mantle is partially-molten. Which observation helped them reach this co
Olenka [21]

Answer:

 P and S waves slow down when they reach this layer. The asthenosphere, also known as the magma chamber, is the uppermost component of the mantle. This layer is partially molten and is a ductile zone in a tectonically poor state.

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seismic waves travel more quickly through denser materials and therefore generally travel more quickly with the depth it moves more slowly through a liquid than a solid. Molten areas within the Earth slow down P waves and stop S waves because their shearing motion cannot be transmitted through a liquid. Partially molten areas may slow down the P waves and attenuate or weaken S waves.

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5 0
3 years ago
Two long, parallel wires are separated by a distance of 3.20 cm. The force per unit length that each wire exerts on the other is
AURORKA [14]

Answer:

(a)The current in the second wire = 9.43 A

(b) The two current are in the opposite direction.

Explanation:

The force per unit length (F/l) on a current carrying conductor = μ₀I₁I₂/2πr

Making I₂ The subject of the equation,

I₂ = [2πr(F/l)]/μ₀I₁.................... Equation 1

Where I₁ = Current in the first wire, I₂ = current in the second wire, μ₀ = permeability of vacuum, r = distance of separation between the two wires.

<em>Given: F/l = 3.30 × 10⁻⁵ N/m, I₁ = 0.56 A, r = 3.20 cm = 0.032 m.</em>

<em>Constant: </em>μ₀= 4π × 10⁻⁷ Am⁻¹.

Substituting these values into equation 1

I₂ = (2π × 0.032 × <em>3.30 × 10⁻⁵ )/(</em>4π × 10⁻⁷×0.56)

I₂ = (0.032 × 3.3 × 100)/1.12

I₂ = 10.56/1.12

I₂ = 9.43 A

Therefore the current in the second wire = 9.43 A

(b) The two current are in the opposite direction, because the force on each wire repel each other.

7 0
3 years ago
When a force of 450N pushes on a 20kg box as
ehidna [41]

Answer:

ma+mgsinh0​+f=F∴(25)(0.75)+(25)(10)sinh0​+μk​N=F∴18.75+(250)(0.6h)+μk​(mgcosh0​=F⟹18.75+150+μk​((25)(10)(0.76))=500∴168.75+μk​(190)=500⟹μk​(190)=331.25⟹μk​=1.74

Explanation:

7 0
3 years ago
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