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3 years ago

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A car is traveling along a straight road at a velocity of +30.0 m/s when its engine cuts out. For the next 1.79 seconds, the car

slows down, and its average acceleration is . For the next 4.03 seconds, the car slows down further, and its average acceleration is . The velocity of the car at the end of the 5.82-second period is +18.4 m/s. The ratio of the average acceleration values is = 1.53. Find the velocity of the car at the end of the initial 1.79-second interval.

1 answer:

Tanzania [10]3 years ago

8 0

Answer:

first value+2nd +3rd

Explanation:

thug life and there

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A 3.2 kg particle starts from rest at x = 0 and moves under the influence of a single force Fx = 4 + 15.7 x − 1.5 x 2 , where Fx

garik1379 [7]

Answer:

Explanation:

Work: This can be defined as the product of force and distance. The unit of work is Joules (J). it can be expressed mathematically as

W = F×d

or

W = $\int\limits^b_a {Fx} \, dx$ .................................. Equation 1

Where b = upper limit, a = lower limit, Fx = expression of force.

<em>Given: a = 0 , b = 1.3 m, Fx = 4 + 15.7x - 1.5x²</em>

Substituting these values into equation 1

<em>W = $\int\limits^a_b {(4 + 15.7x - 1.5x^{2} )dx} \,$ </em>

W = ᵇ[4x + 15.7x²/2-1.5x³/3 +C]ₐ

Work = upper limit - lower limit

Work = ᵃ[4x + 15.7x²/2 - 1.5x³/3 +C] - [4x + 15.7x²/2 + 1.5x³/3 +C]ᵇ............... Equation 2

Substituting the values of a and b into equation 2

Work = [4(1.3) + 15.7(1.3)²/2-1.5(1.3)³/3 + C] - [0 +C]

Work = [5.2 + 26.53 -3.29 + C] - C

Work = 28.44 J

Work done by the force = 28.44 J.

8 0

3 years ago

A car with a mass of 1500 kg is pulled by a rope that is horizontal to the ground. The tension in the rope is 2000 N and a frict

Tanzania [10]

Answer:

Explanation:

Assuming the ground is level as well.

F = ma

a = F/m

a = (2000 - 350) / 1500

a = 1.1 m/s²

7 0

3 years ago

2013 Indianapolis 500 champion Tony Kanaan holds his hand out of his IndyCar while driving through still air with standard atmos

Citrus2011 [14]

Answer:

(a). The pressure is 14.76 psi.

(b). The pressure is 15.59 psi.

(c). The pressure is 15.68 psi.

All answer are reasonable.

Explanation:

Given that,

Speed v₁= 60 mph

Speed v₂ = 225 mph

Speed v₃ = 235 mph

(a). We need to calculate the maximum pressure on his hand

Using equation of pressure

$P_{1}=P+\dfrac{1}{2}\rho v^2+\rho gh$

there, no vertical movement

So, on neglect of height term

$P_{1}=P+\dfrac{1}{2}\rho v_{1}^2$

Where, P= atmospheric pressure

$\rho$ = air density

v = speed

Put the value in the equation

$P_{1}=14.7\times144+\dfrac{1}{2}\times(0.002376\times(60\times1.4667)^2)$

$P_{1}=2126.0\ lb/ft^2$

$P_{1}=\dfrac{2126.0}{144}$

$P_{1}= 14.76\ psi$

(b). Speed v₂ = 225 mph

We need to calculate the maximum pressure on his hand

Using equation of pressure

$P_{2}=P+\dfrac{1}{2}\rho v_{2}^2$

Put the value in the equation

$P_{2}=14.7\times144+\dfrac{1}{2}\times(0.002376\times(225\times1.4667)^2)$

$P_{2}=2246.17\ lb/ft^2$

$P_{2}=\dfrac{2246.17}{144}$

$P_{2}= 15.59\ psi$

(c). Speed v₃ = 235 mph

We need to calculate the maximum pressure on his hand

Using equation of pressure

$P_{3}=P+\dfrac{1}{2}\rho v_{3}^2$

Put the value in the equation

$P_{3}=14.7\times144+\dfrac{1}{2}\times(0.002376\times(235\times1.4667)^2)$

$P_{3}=2257.93\ lb/ft^2$

$P_{3}=\dfrac{2257.93}{144}$

$P_{3}= 15.68\ psi$

According to bernoulli's equation,

If the car increases the velocity the the pressure on the surface of the driver's hand increases.

The pressure from P₁ to P₃ are all near the value of one atmosphere.

So, the pressure difference of one atmosphere is not enough to break the driver's hand.

Hence, (a). The pressure is 14.76 psi.

(b). The pressure is 15.59 psi.

(c). The pressure is 15.68 psi.

All answer are reasonable.

5 0

3 years ago

List examples of how the Bill of Rights protects you:<br> .<br> .<br> .<br> .<br> .

Naya [18.7K]

guaranteeing freedom of speech, press, assembly, and exercise of religion

4 0

3 years ago

Two resistors have resistances R(smaller) and R(larger), where R(smaller) < R(larger). When the resistors are connected in se

ASHA 777 [7]

Answer:

R = 9.85 ohm , r = 0.85 ohm

Explanation:

Let the two resistances by r and R.

when they are connected in series:

V = 12 V

i = 1.12 A

The equivalent resistance when they are connected in series is

Rs = r + R

So, By using Ohm's law

V = i Rs

Rs = V / i = 12 / 1.12 = 10.7 ohm

R + r = 10.7 ohm .... (1)

When they are connected in parallel:

V = 12 V

i = 9.39 A

The equivalent resistance when they are connected in parallel

$R_{p}=\frac{R+r}{rR}$

So, By using Ohm's law

V = i Rp

Rp = V / i = 12 / 9.39 = 1.28 ohm

$\frac{R+r}{rR}=1.28$ .... (2)

by substituting the value of R + r from equation (1) in equation (2), we get

r R = 8.36 ..... (3)

$R-r = \sqrt{\left ( R+r \right )^{2}-4rR}$

$R-r = \sqrt{\left ( 10.7 \right )^{2}-4\times 8.36}=9$ ..... (4)

By solvng equation (1) and (4), we get

R = 9.85 ohm , r = 0.85 ohm

8 0

3 years ago