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2 years ago

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A car company wants to ensure its newest model can stop in less than 450 ft when traveling at 60 mph. If we assume constant dece

leration, find the value of acceleration (in ft/sec2) that accomplishes this.

1 answer:

seraphim [82]2 years ago

4 0

Answer:

The value of acceleration that accomplishes this is 8.61 ft/s² .

Explanation:

Given;

maximum distance to be traveled by the car when the brake is applied, d = 450 ft

initial velocity of the car, u = 60 mph = (1.467 x 60) = 88.02 ft/s

final velocity of the car when it stops, v = 0

Apply the following kinematic equation to solve for the deceleration of the car.

v² = u² + 2as

0 = 88.02² + (2 x 450)a

-900a = 7747.5204

a = -7747.5204 / 900

a = -8.61 ft/s²

|a| = 8.61 ft/s²

Therefore, the value of acceleration that accomplishes this is 8.61 ft/s² .

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Two bicycle tires are set rolling with the same initial speed of 3.30 m/s along a long, straight road, and the distance each tra

vredina [299]

Answer:

At low pressure- $\mu_{k}=0.02315$

At high pressure- $\mu_{k}=0.00445$

Explanation:

Initial speed, $V_{i}=3.3 m/s$

Final speed, $V_{f}=3.3/2= 1.65 m/s$

Net horizontal force due to rolling friction $F_{net}=\mu_{k}$ mg where m is mass, g is acceleration due to gravity, $\mu_{k}$ is coefficient of rolling friction

From kinematic relation, $V_{f}^{2}- V_{i}^{2}=2ad$

For each tire,

$V_{f}^{2}- V_{i}^{2}=2\mu_{k}gd$

Making $\mu_{k}$ the subject

$\mu_{k}=\frac {V_{f}^{2}- V_{i}^{2}}{2gd}$

Under low pressure of 40 Psi, d=18 m

$\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*18}=-0.02315$

Therefore, $\mu_{k}=0.02315$

At a pressure of 105 Psi, d=93.7

$\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*93.7}=-0.00445$

Therefore, $\mu_{k}=0.00445$

4 0

3 years ago

Emilio pushes a 100 kg freshman with 200 N of force. How much is the freshman accelerated?

ladessa [460]

Explanation:

F = MA

200 = 100 * A

A = 200/100

A = 2m/sec^2

<h3><em>hope </em><em>it </em><em>helps </em><em>you </em></h3>

8 0

3 years ago

What is the pressure at a depth of 15 cm brine of density 1.2/cm³?

Anna [14]

P = density × gravity acceleration × height

P = 1200 × 9.81 × 15/100

P = 1765.8

6 0

2 years ago

Read 2 more answers

A scientist discovered an elementary particle paired with another particle. It has a positive charge equal to two–thirds of the

julia-pushkina [17]

W boson has +1e or - 1e charge, Z boson has 0 charge.

Leptons have +1e, -1e or 0 charge.

Photons have 0 charge.

Only quarks have a charge of +2/3e or -1/3e of an electron charge.

To be exact, only up-type quarks (Up, Down and Top quarks) have a +2/3e or two thirds of an electron charge.

So the correct answer is D) Quark.

6 0

3 years ago

A carbon resistor is to be used as a thermometer. On a winter day when the temperature is 4.00°C, the resistance of the carbon r

dusya [7]

Answer:

28 degree C

Explanation:

We are given that

$T_1=4.00^{\circ}$

$R_1=217.7 \Ohm$

$R_2=215.1\Ohm$

$\alpha=-5.00\times 10^{-4}C^{-1}$

We have to find the temperature on a spring day when resistance is 215.1 ohm.

We know that

$\alpha(T_2-T_1)=\frac{R_2}{R_2}-1$

Using the formula

$-5.00\times 10^{-4}(T_2-4)=\frac{215.1}{217.7}-1$

$-5\times 10^{-4}(T_2-4)=0.988-1=-0.012$

$T_2-4=\frac{0.012}{5\times 10^{-4}}=24$

$T_2=24+4=28^{\circ}C$

Hence, the temperature on a spring day 28 degree C.

7 0

3 years ago