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3 years ago

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Two vehicles approach an intersection: a truck moving eastbound at 16.0 m/s and an SUV moving southbound at 20.0 m/s. Suppose th

at the x-axis is directed eastward and the y-axis is directed northward. Find the magnitude of the velocity vector of the truck relative to the SUV.

1 answer:

Alex Ar [27]3 years ago

8 0

Answer:

The value of magnitude of the velocity vector of the truck relative to the SUV $V_{xy}= 25.62 \frac{m}{sec}$

Explanation:

Velocity of the truck $V_{x}$ = 16 $\frac{m}{s}$

Velocity of the SUV $V_{y}$ = - 20 $\frac{m}{s}$

The magnitude of the velocity vector of the truck relative to the SUV is

$V_{xy} =\sqrt{ V_{x} ^{2} + V_{y}^{2}}$

Put the value of $V_{x}$ & $V_{y}$ in the above equation we get

$V_{xy} = \sqrt{(16)^{2} + (-20)^{2}}$

$V_{xy} = \sqrt{656}$

$V_{xy}= 25.62 \frac{m}{sec}$

This is the value of magnitude of the velocity vector of the truck relative to the SUV.

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Calculate the energy (in eV/atom) for vacancy formation in some metal, M, given that the equilibrium number of vacancies at 296o

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Explanation:

The given data is as follows.

Temperature of metal = $296^{o}C$ = (296 + 273) K

= 569 K

Density of the metal = 8.85 $g/cm^{3}$ = $8.85 \times 10^{-6} g/m^{3}$ (as $1 cm^{3} = 10^{-6} m^{3}$ )

Atomic mass = 51.40 g/mol

Vacancies = $9.19 \times 10^{23} m^{-3}$

Formula to calculate the number of atomic sites is as follows.

n = $\frac{\rho \times N_{A}}{\text{atomic weight}}$

= $\frac{8.85 \times 10^{-6} \times 6.022 \times 10^{23}}{51.40 g/mol}$

= $1.036 \times 10^{17} atom/m^{3}$

Now, we will calculate the energy as follows.

E = $-KT \times ln (\frac{\text{no. of vacancies}}{\text{no. of atomic sites}})$

where, K = $8.62 \times 10^{-5}$

E = $-8.62 \times 10^{-5} \times 569 K \times ln (\frac{9.19 \times 10^{23}}{1.036 \times 10^{17} atom/m^{3}})$

= $78.46 eV/atom$

Therefore, we can conclude that energy (in eV/atom) for vacancy formation in given metal, M, is $78.46 eV/atom$ .

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A 54.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 126 m/s from the top of a cliff

Nikolay [14]

Answer:

a) The initial total mechanical energy of the projectile is 498556.296 joules.

b) The work done on the projectile by air friction is 125960.4 joules.

c) The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.

Explanation:

a) The system Earth-projectile is represented by the Principle of Energy Conservation, the initial total mechanical energy ( $E$ ) of the project is equal to the sum of gravitational potential energy ( $U_{g}$ ) and translational kinetic energy ( $K$ ), all measured in joules:

$E = U_{g} + K$ (Eq. 1)

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

$E = m\cdot g\cdot y + \frac{1}{2}\cdot m\cdot v^{2}$ (Eq. 1b)

Where:

$m$ - Mass of the projectile, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$y$ - Initial height of the projectile above ground, measured in meters.

$v$ - Initial speed of the projectile, measured in meters per second.

If we know that $m = 54\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y = 132\,m$ and $v = 126\,\frac{m}{s}$ , the initial mechanical energy of the earth-projectile system is:

$E = (54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (132\,m)+\frac{1}{2}\cdot (54\,kg)\cdot \left(126\,\frac{m}{s} \right)^{2}$

$E = 498556.296\,J$

The initial total mechanical energy of the projectile is 498556.296 joules.

b) According to this statement, air friction diminishes the total mechanical energy of the projectile by the Work-Energy Theorem. That is:

$W_{loss} = E_{o}-E_{1}$ (Eq. 2)

Where:

$E_{o}$ - Initial total mechanical energy, measured in joules.

$E_{1}$ - FInal total mechanical energy, measured in joules.

$W_{loss}$ - Work losses due to air friction, measured in joules.

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

$W_{loss} = E_{o}-K_{1}-U_{g,1}$

$W_{loss} = E_{o} -\frac{1}{2}\cdot m\cdot v_{1}^{2}-m\cdot g\cdot y_{1}$ (Eq. 2b)

Where:

$m$ - Mass of the projectile, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$y_{1}$ - Maximum height of the projectile above ground, measured in meters.

$v_{1}$ - Current speed of the projectile, measured in meters per second.

If we know that $E_{o} = 498556.296\,J$ , $m = 54\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y_{1} = 297\,m$ and $v_{1} = 89.3\,\frac{m}{s}$ , the work losses due to air friction are:

$W_{loss} = 498556.296\,J -\frac{1}{2}\cdot (54\,kg)\cdot \left(89.3\,\frac{m}{s} \right)^{2} -(54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (297\,m)$

$W_{loss} = 125960.4\,J$

The work done on the projectile by air friction is 125960.4 joules.

c) From the Principle of Energy Conservation and Work-Energy Theorem, we construct the following model to calculate speed of the projectile before it hits the ground:

$E_{1} = U_{g,2}+K_{2}+1.5\cdot W_{loss}$ (Eq. 3)

$K_{2} = E_{1}-U_{g,2}-1.5\cdot W_{loss}$

Where:

$E_{1}$ - Total mechanical energy of the projectile at maximum height, measured in joules.

$U_{g,2}$ - Potential gravitational energy of the projectile, measured in joules.

$K_{2}$ - Kinetic energy of the projectile, measured in joules.

$W_{loss}$ - Work losses due to air friction during the upward movement, measured in joules.

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

$\frac{1}{2}\cdot m \cdot v_{2}^{2} = E_{1}-m\cdot g\cdot y_{2}-1.5\cdot W_{loss}$ (Eq. 3b)

$m\cdot v_{2}^{2} = 2\cdot E_{1}-2\cdot m \cdot g \cdot y_{2}-3\cdot W_{loss}$

$v_{2}^{2} = 2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m}$

$v_{2} = \sqrt{2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m} }$

If we know that $E_{1} = 372595.896\,J$ , $m = 54\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y_{2} =0\,m$ and $W_{loss} = 125960.4\,J$ , the final speed of the projectile is:

$v_{2} =\sqrt{2\cdot \left(\frac{372595.896\,J}{54\,kg}\right)-2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0\,m)-3\cdot \left(\frac{125960.4\,J}{54\,kg}\right) }$

$v_{2} \approx 82.475\,\frac{m}{s}$

The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.

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3 years ago