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charle [14.2K]
3 years ago
9

A car is going south on I-69 at 33 m/s (74 mph). The car has good brakes so its maximum braking acceleration is – 8.5 m/s^2 . Tr

affic has come to a stop in all lanes 120 m ahead. What is the maximum reaction time the driver can have and still be able to stop in the available distance? Hint: During the driver’s reaction time, the car continues to have the initial velocity of 33 m/s.
Physics
1 answer:
dmitriy555 [2]3 years ago
4 0

Answer:

1.69515 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-33^2}{2\times -8.5}\\\Rightarrow s=64.06\ m

The distance between the traffic and the car after braking is 120-64.06 = 55.94 m

Time = Distance / Speed

\text{Time}=\frac{55.94}{33}\\\Rightarrow \text{Time}=1.69515\ seconds

The reaction time cannot be more than 1.69515 seconds

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SVEN [57.7K]

Let's explain the following electrical terms.

• Electric current, can be said to be the flow of electric charge. These charges are through a conductor by moving electrons.

• Ammeter, is an instrument used to measure the amount of electric current in a circuit.

The name ammeter was derived from the unit of electric current (Amperes).

• Voltmeter, can be said to be an instrument used to measure the voltage(potential difference) in a circuit. It measures the electric potential between two points in an electrical circuit. It can also be called voltage meter.

• Electric circuit, can be defined as the conductive path for the flow of electric current.

It allows electric charge carriers to flow continuously.

• Resistance ,can be said to be the property of an electrical conductor which resists the flow of electric current. It is measured in ohms.

• Ohms law, states that the potential difference (V) between two points is directly proportional to the electric current across two points.

It is deonted as: V = I x R

Voltage = Current x Resistance

• Electric potential ,can be said to be the amount of electric potential energy at a point.

3 0
1 year ago
URGENT. Please help.
exis [7]

1). c ... 2). d ... 3). a ... 4). d ... 5). c ... 6). a

7). b-mass ... c-m/s ... d-Newton's 1st ... e-Newton's 2nd

6 0
3 years ago
Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.57 A out of the jun
AlekseyPX

Answer:

a. 1.56 × 10¹⁸ electrons per second

b. The electrons in wire 3 flow into the junction.

Explanation:

Here is the complete question

Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.65 A out of the junction. (a) How many electrons per second move past a point in wire 3? (b) In which direction do the electrons move in wire 3 -- into or out of the junction?

Solution

(a) How many electrons per second move past a point in wire 3?

Using Kirchhoff's current law, at the junction, i₁ + i₂ + i₃ = 0 where i₁ = current in wire 1 = 0.40 A, i₂ = current in wire 2 = 0.65 A and  i₃ = = current in wire 3,

So, i₃ = -(i₁ + i₂)

taking current flowing into the junction as positive and those leaving as negative, i₁ = + 0.40 A and i₂ = -0.65 A

So, i₃ = -(i₁ + i₂)

i₃ = -(0.40 A + (-0.65 A))

i₃ = -(0.40 A - 0.65 A)

i₃ = -(-0.25 A)

i₃ = 0.25 A

Since i₃ = 0.25 C/s and we have e = 1.602 × 10⁻¹⁹ C per electron, then the number of electrons flowing in wire 3 per second is i₃/e = 0.25 C/s ÷ 1.602 × 10⁻¹⁹ C per electron = 0.1561  × 10¹⁹ electrons per second = 1.561  × 10¹⁸ electrons per second ≅ 1.56 × 10¹⁸ electrons per second

(b) In which direction do the electrons move -- into or out of the junction?

Given that i₃ = + 0.25 A and that positive flows into the junction, thus, the electrons in wire 3 flow into the junction.

8 0
3 years ago
Point masses m1 m2 are placed at opposite ends
tiny-mole [99]

(a) x = \frac{m_2L}{m_1+m_2}

<u>Explanation:</u>

Given:

Moment of Inertia of m₁ about the axis, I₁ = m₁x²

Moment of Inertia of m₂ about the axis. I₂ = m₂ (L - x)²

Kinetic energy is rotational.

Total kinetic energy is E = \frac{1}{2} I_1w_0^2 + \frac{1}{2}I_2w_0^2 = \frac{1}{2} w_0^2(m_1x^2 + m_2(L-x)^2)

Work done is change in kinetic energy.

To minimize E, differentiate wrt x and equate to zero.

m_1x - m_2(L-x) = 0\\\\x = \frac{m_2L}{m_1+m_2}

Alternatively, work done is minimum when the axis passes through the center of mass.

Center of mass is at \frac{m_2L}{m_1 + m_2}

7 0
3 years ago
An 1100-kg car traveling at 27.0 m/s starts to slow down and comes to a complete stop in 578 m. What is the magnitude of the ave
Trava [24]

Answer:

693.685 N

Explanation:

t = Time taken

u = Initial velocity = 27 m/s

v = Final velocity

s = Displacement = 578 m

a = Acceleration

m = Mass

Equation of motion

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-27^2}{2\times 578}\\\Rightarrow a=-0.63062\ m/s^2

Force

F = ma

F=1100\times -0.63062\\\Rightarrow F=-693.6851\ N

Magnitude of braking force required to stop the car is 693.685 N

7 0
3 years ago
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