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Solnce55 [7]
3 years ago
12

A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from zero to 9.41 m

/s in 4.24 s. What is the magnitude of the linear impulse experienced by a 67.0 kg passenger in the car during this time
Physics
1 answer:
Svetach [21]3 years ago
8 0

Answer:

the impulse experienced by the passenger is 630.47 kg

Explanation:

Given;

initial velocity of the car, u = 0

final velocity of the car, v = 9.41 m/s

time of motion of the car, t = 4.24 s

mass of the passenger in the car, m = 67 kg

The impulse experienced by the passenger is calculated as;

J = ΔP = mv - mu = m(v - u)

           = 67(9.41 - 0)

           = 67 x 9.41

           = 630.47 kg

Therefore, the impulse experienced by the passenger is 630.47 kg

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Answer:

Density of the fuel is 727.3 kilograms per cubic meter.

Specific weight of the fuel is 7127.3 Newtons per cubic meter.

Specific gravity of the fuel is 0,727.

Explanation:

In order to use SI units, we have to convert liters to cubic meters. Knowing that a liter is a cubic decimeter and a cubic decimeter is 1*10^{-3} cubic meters, we know that the tank has 0,055 cubic meters of fuel (because it is full).

Now that we have things in SI units, we calculate density:

p_{fuel}= \frac{mass}{volume} = \frac{40 kg}{0.055 m^{3} } =727.3 \frac{kg }{m^{3} }

Knowing the mass per unit of volume, we can calculate weight per unit of volume thanks to Newton's second law (mass times acceleration, g in this case, equals force (weight)), i.e. specific weight:

y=p*g=727,3 \frac{kg}{m^{3}}*9.8\frac{m }{s^{2}}=7127,3 \frac{N}{m^{3}}

With density we can also calculate how dense the fuel is related to a reference (water), i.e. specific gravity. SG is a dimensionless number that tell us how much denser (SG>1) or lighter per unit of volume (SG<1) a substance is than water. We use water as a reference because it is one of the most used substances in our life, and it is a standard density (1000 kg per cubic meter at 4°C and 1 atm).

SG=\frac{p_{fuel} }{p_{water} } =\frac{727.3 \frac{kg }{m^{3} }}{1000 \frac{kg }{m^{3} }} =0,727

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Students perform a set of experiments by placing a block of mass m against a spring, compressing the spring a distance x along a
Verizon [17]

Increasing the angle of inclination of the plane decreases the velocity of the block as it leaves the spring.

  • The statement that indicates how the relationship between <em>v</em> and <em>x</em> changes is;<u> As </u><u><em>x</em></u><u> increases, </u><u><em>v</em></u><u> increases, but the relationship is no longer linear and the values of </u><u><em>v</em></u><u> will be less for the same value of </u><u><em>x</em></u><u>.</u>

Reasons:

The energy given  to the block by the spring = \mathbf{0.5  \cdot k  \cdot x^2}

According to the principle of conservation of energy, we have;

On a flat plane, energy given to the block = 0.5  \cdot k  \cdot x^2 = kinetic energy of

block = 0.5  \cdot m  \cdot v^2

Therefore;

0.5·k·x² = 0.5·m·v²

Which gives;

x² ∝ v²

x ∝ v

On a plane inclined at an angle θ, we have;

The energy of the spring = \mathbf{0.5  \cdot k  \cdot x^2}

  • The force of the weight of the block on the string, F = m \cdot g  \cdot sin(\theta)

The energy given to the block = 0.5 \cdot k \cdot x^2 - m \cdot g  \cdot sin(\theta) = The kinetic energy of block as it leaves the spring = \mathbf{0.5  \cdot m  \cdot v^2}

Which gives;

0.5 \cdot k \cdot x^2 - m \cdot g  \cdot sin(\theta) = 0.5  \cdot m  \cdot v^2

Which is of the form;

a·x² - b = c·v²

a·x² + c·v² = b

Where;

a, b, and <em>c</em> are constants

The graph of the equation a·x² + c·v² = b  is an ellipse

Therefore;

  • As <em>x</em> increases, <em>v</em> increases, however, the value of <em>v</em> obtained will be lesser than the same value of <em>x</em> as when the block is on a flat plane.

<em>Please find attached a drawing related to the question obtained from a similar question online</em>

<em>The possible question options are;</em>

  • <em>As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x</em>
  • <em>The relationship is no longer linear and v will be more for the same value of x</em>
  • <em>The relationship is still linear, with lesser value of v</em>
  • <em>The relationship is still linear, with higher value of v</em>
  • <em>The relationship is still linear, but vary inversely, such that as x increases, v decreases</em>

<em />

Learn more here:

brainly.com/question/9134528

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