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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).

You might be interested in
Two spectators at a soccer game in Montjuic Stadium see, and a moment later hear, the ball being kicked on the playing field. Th
Akimi4 [234]

Answer:

a) The distance of spectator A to the player is 79.2 m

b) The distance of spectator B to the player is 43.9 m

c) The distance between the two spectators is 90.6 m

Explanation:

a) Knowing the time it takes the sound to reach both spectators, we can calculate their position relative to the player, using this equation:

x = v * t

where:

x = position of the spectators

v = speed of sound

t = time

Then, the position for spectator A relative to the player is:

x = 343 m/s * 0.231 s = 79.2 m

b)For spectator B:

x = 343 m/s * 0.128 s

x = 43.9 m

The distance of spectator A and B to the player is 79.2 m and 43.9 m respectively.

c) To calculate the distance between the spectators, please see the attached figure. Notice that the distance between the spectators is the hypotenuse of the triangle formed by the sightline of both. We already know the longitude of the two sides. Then, using Pythagoras theorem:

(Distance AB)² = A² + B²

(Distance AB)² = (79.2 m)² + (43.9 m)²

Distance AB = 90. 6 m

6 0
3 years ago
A circular rod with a gage length of 3.5 mm and a diameter of 2.8 cmcm is subjected to an axial load of 68 kN . If the modulus o
Crank

To solve this problem, we will apply the concepts related to the linear deformation of a body given by the relationship between the load applied over a given length, acting by the corresponding area unit and the modulus of elasticity. The mathematical representation of this is given as:

\delta = \frac{PL}{AE}

Where,

P = Axial Load

l = Gage length

A = Cross-sectional Area

E = Modulus of Elasticity

Our values are given as,

l = 3.5m

D = 0.028m

P = 68*10^3 N

E = 200GPa  

A = \frac{\pi}{4}(0.028)^2 \rightarrow 0.0006157m^2

Replacing we have,

\delta = \frac{PL}{AE}

\delta = \frac{( 68*10^3)(3.5)}{(0.0006157)(200*10^9)}

\delta = 0.001932m

\delta = 1.93mm

Therefore the change in length is 1.93mm

7 0
2 years ago
What is the volume of an object if it has a mass of 10 grams and a density of 87 g/ml
belka [17]

Answer:

<h3>The answer is 0.115 mL</h3>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density} \\

From the question

mass = 10 g

density = 87 g/ml

We have

volume =  \frac{10}{87}  \\  = 0.114942528...

We have the final answer as

<h3>0.115 mL</h3>

Hope this helps you

5 0
3 years ago
The diagram below shows the movement of matter in a portion of the water cycle.
Alisiya [41]

Answer:

d) precipitation

Hope it helps you

And if you want to, pls mark it as the brainliest answer

7 0
3 years ago
Although the reactions of the calvin cycle do not depend directly on light. True or False
Harlamova29_29 [7]

Answer:

True

Explanation:

The complete question is:

<em>"Although the reactions of the Calvin cycle do not depend directly on light, they do not usually occur at night. True o False"</em>

<em> </em>The Calvin cycle is also known as the Calvin-Benson cycle or as the CO₂ fixation phase in the photosynthesis process.

The Calvin cycle generates the reactions necessary to fix the carbon in a solid structure for the formation of glucose and, in turn, regenerates the molecules for the continuation of the cycle.

The Calvin cycle is known as the dark phase of photosynthesis, or the carbon fixation phase. It is called the dark phase because this cycle is not dependent on light like other parts that make up the photosynthesis process. But it uses the energy that is produced in the light phase of photosynthesis to fix carbon.

It can be said that it consists of or forms the second stage of photosynthesis, in which the carbon of the carbon dioxide that is absorbed is fixed.

So, the statement is true because the Calvin cycle uses the energy that is produced in the light phase of photosynthesis to fix carbon.

4 0
3 years ago
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