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stellarik [79]
3 years ago
15

On a straight, level, two-lane road, two cars moving in opposite directions approach and pass each other. Car A is in the eastbo

und lane moving with a speed of 11 m/s, and Car B is in the westbound lane moving with a speed of 31 m/s.
(a) What is the magnitude and direction of the velocity of car A as seen by the driver of car B? magnitude m/s direction
(b) What is the magnitude and direction of the velocity of car B as seen by the driver of car A? magnitude m/s direction
Physics
1 answer:
ludmilkaskok [199]3 years ago
5 0

Answer:

a) 42 m/s, positive direction (to the east), b) 42 m/s, negative direction (to the west).

Explanation:

a) Let consider that Car A is moving at positive direction. Then, the relative velocity of Car A as seen by the driver of Car B is:

\vec v_{A/B} = \vec v_{A} - \vec v_{B}\\\vec v_{A/B} = 11 \frac{m}{s} \cdot i + 31 \frac{m}{s} \cdot i\\\vec v_{A/B} = 42 \frac{m}{s} \cdot i

42 m/s, positive direction (to the east).

b) The relative velocity of Car B as seen by the drive of Car A is:

\vec v_{B/A} = \vec v_{B} - \vec v_{A}\\\vec v_{B/A} = -31 \frac{m}{s} \cdot i - 11 \frac{m}{s} \cdot i\\\vec v_{B/A} = - 42 \frac{m}{s} \cdot i

42 m/s, negative direction (to the west).

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A screw having 50% efficiency is driven by a rod and 25 cm. The pitch of the screw is 1/10cm Calculate velocity ratio and mechan
neonofarm [45]

(a) The velocity ratio of the screw is 1570.8.

(b) The mechanical advantage of the screw is 785.39.

<h3>Velocity ratio of the screw</h3>

The velocity ratio of the screw is calculated as follows;

V.R = 2πr/P

where;

  • P is the pitch = 1/10 cm = 0.1 cm = 0.001 m
  • r is radius = 25 cm = 0.25 m

V.R = (2π x 0.25)/(0.001)

V.R = 1570.8

<h3>Mechanical advantage of the screw</h3>

E = MA/VR x 100%

0.5 = MA/1570.8

MA = 785.39

Learn more about mechanical advantage here: brainly.com/question/18345299

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4 0
2 years ago
If it takes 200 joules of energy to lift a bucket of water 3 meters in 2 seconds, how much power would be required to do the sam
Harlamova29_29 [7]

200 joules of work energy are involved.  That's all we need to know to answer the question.  Once we know that 200 joules of work energy are involved, we don't care what was lifted, or how far, or how long it took, or how many people worked on it, or how much they were paid, or what was the distribution of their gender identities, or the ethnic diversity among the team. or what day each of them celebrates as their sabbath.  Any other information besides the 200 joules is only there to distract us, and see whether we're paying attention.

Power = (work or energy) / (time to do the work or move the energy)

Power = (200 joules) / (5 seconds)

<em>Power = 40 watts</em>

3 0
3 years ago
Two blocks a and b ($m_a&gt;m_b$) are pushed for a certain distance along a frictionless surface. how does the magnitude of the
Yuki888 [10]

Answer:

the magnitude of the work done by the two blocks is the same.

Explanation:

The work done by block a on block b is given by:

W_a = F_a d

where Fa is the force exerted by block a on block b, and d is the distance they cover.

The work done by block b on block a is given by:

W_b = F_b d

where Fb is the force exerted by block b on block a, and d is still the distance they cover.

For Newton's third law, the force exerted by block a on block b is equal to the force exerted by block b on block a, therefore

F_a = F_b

and so

W_a=W_b

3 0
3 years ago
Do longitude lines run horizontally (east-west) or vertically (north-south)?
nlexa [21]
Longitude- Horizontal (East West)
Latitude- Vertical (North South)
3 0
3 years ago
A magic medallion is suspended from a string inside a compartment of Hogwarts Express which is running straight westwards on hor
Serggg [28]

Answer:

a = 1 m/s²  and

Explanation:

The first two parts can be seen in attachment

We use Newton's second law on each axis

Y axis

      Ty - W = 0        

      Ty = w

X axis

     Tx = m a

With trigonometry we find the components of tension

    Sin θ = Ty / T

    Ty = T sin θ

    Cos θ = Tx / T

    Tx = T cos θ

We calculate the acceleration with kinematics

   Vf = Vo + a t

   a = (Vf -Vo) / t

   a = (20 -10) / 10

   a = 1 m/s²

We substitute in Newton's equations

     

  T Sin θ = mg

  T cos θ = ma

We divide the two equations

  Tan θ = g / a

  θ = tan⁻¹ (g / a)

  θ = tan⁻¹ (9.8 / 1)

  θ = 84º

We see that in the expression of the angle the mass does not appear therefore you should not change the angle

4 0
3 years ago
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