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3 years ago

Discuss velocity by providing at least one example and explain why velocity is either a scalar or vector quantity.

2 answers:

5 0

Velocity is speed plus direction, so an example of velocity would be a vehicle traveling at 75mph north. Velocity is a vector quantity because it describes both magnitude and direction.

Fed [463]3 years ago

3 0

Answer: Velocity is speed and the direction where the speed is pointing.

If you have a map, with the y axis = north and the x axis = east. For you to describe your velocity, you must describe how fast are moving to north and how fast are you moving to east. So your velocity will be Vnorth + Veast, and this is a vector of the form (Vnorth, Veast). with total speed of $\sqrt{Vnorth^{2} + Veast^{2} }$ .

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Active immunity is passed on from a mother to her baby.

viva [34]

True.The immune system in babies. Antibodies are passed from mother to baby through the placenta during the last three months of pregnancy.

3 0

3 years ago

A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region

zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

$K_{1} = K_{2} + W_{f}$

Where:

$K_{1}$ , $K_{2}$ are the initial and final translational kinetic energies of the tobbogan, measured in joules.

$W_{f}$ - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

$f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Distance travelled by the toboggan in the rough region, measured in meters.

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

$v_{1}$ , $v_{2}$ - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

$f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}$

If $m = 375\,kg$ , $v_{1} = 4.50\,\frac{m}{s}$ , $v_{2} = 1.20\,\frac{m}{s}$ and $\Delta s = 5.40 \,m$ , then:

$f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}$

$f = 653.125\,N$

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

$\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%$

$\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%$

$\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%$

$\%K_{loss} = 92.889\,\%$

The rough region reduced the kinetic energy of the toboggan in 92.889 %.

c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:

$\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%$

$\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%$

$\%v_{loss} = 73.333\,\%$

The speed of the toboggan is reduced in 73.333 %.

5 0

3 years ago

What are 16 stages of a stars life

Alla [95]

Https://www.slideshare.net/mobile/jan_parker/life-cycle-of-stars-3196871
the answer to your question!

8 0

3 years ago

Calculate the electrical energy use of the microwave in 120 V circuit has 7.5 A current flow, if it used for 0.05 hours

Elden [556K]

Answer:

162 KJ

Explanation:

The electrical energy can be calculated using the formula

E = V×I×t

Where, V= voltage = 120 V

I is current in ampere = 7.5 A and t is time in seconds

1 hour = 3600 sec

Therefore, 0.05 hour = 3600×0.05 = 180 sec

Substitute the values in above equation find electrical power

E = 120×7.5×180

= 162000 J

= 162 KJ

Therefore, the electrical energy use of the microwave = 162 KJ

3 0

3 years ago

5–111. A box having a weight of 8 lb is moving around in a circle of radius rA = 2 ft with a speed of (vA)1 = 5 ft>s while co

Elis [28]

Answer:

a) vB = 10.77 ft/s

b) W = 11.30 lb*ft

Explanation:

a) W = 8 lb ⇒ m = W/g = 8 lb/32.2 ft/s² = 0.2484 slug

vA lin = 5 ft/s

rA = 2 ft

v rad = 4 ft/s

vB = ?

rB = 1 ft

W = ?

We can apply The law of conservation of angular momentum

Lin = Lfin

m*vA*rA = m*vB*rB ⇒ vB = vA*rA / rB

⇒ vB = (5 ft/s)*(2 ft) / (1 ft) = 10 ft/s (tangential speed)

then we get

vB = √(vB tang² + vB rad²) ⇒ vB = √((10 ft/s)² + (4 ft/s)²)

⇒ vB = 10.77 ft/s

b) W = ΔK = KB - KA = 0.5*m*vB² - 0.5*m*vA²

⇒ W = 0.5*m*(vB² - vA²) = 0.5*0.2484 slug*((10.77 ft/s)²-(5 ft/s)²)

⇒ W = 11.30 lb*ft

6 0

3 years ago