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Bond [772]
3 years ago
5

When you approach a yield sign, while trying to enter or merge onto another road, traffic already on that road has the right of

way. true or false.?
Physics
2 answers:
vovangra [49]3 years ago
7 0

Answer:

True.

Explanation:

The road signs that appear on the public road have several designs and each one has a specific meaning and intends to alert or inform the driver about different things.  It is important to pay attention to the signs because they avoid accidents and injuries.

The yield signal is an example of important signals that must be respected. However it is important to remember that when you approach a yield signal, when trying to enter or enter another road, traffic already on the road has the right of way. For this reason, you should expect all the cars that are already in the traffic to pass and when you do not have any cars in sight you can complete your maneuver.

kogti [31]3 years ago
5 0
TRUE. When you approach a yield sign, while trying to enter or merge onto another road, traffic already on that road has the right of way.
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Required information Problem 16.048 - DEPENDENT MULTI-PART PROBLEM - ASSIGN ALL PARTS NOTE: This is a multi-part question. Once
GaryK [48]

NOTE: The diagram is attached to this solution

Answer:

The acceleration of point A = 14.64 ft/s²

Explanation:

By proper analysis of the diagram, acceleration of point A will be: (Check the free body diagram attached)

a_{A}  = \bar{a} + \frac{\alpha L}{2}

Weight, W = mg

g = 32.2 ft/s²

m = W/g

p = m \bar{a}\\p = w \bar{a} /g

\bar{a} = pg/w\\\bar{a} = 0.25g/2.2\\\bar{a} =3.66

\frac{pL}{2} = I \alpha

but I = \frac{wL^{2} }{12g}

\frac{pL}{2} = \frac{\alpha wL^{2} }{12g}\\\alpha = \frac{6 g p}{wL}\\\alpha = \frac{6*g*0.25}{2.2L} \\\alpha = 21.96/L

a_{A}  = 3.66 + \frac{(21.96/L ) * L}{2}\\a_{A}  = 3.66 + 10.98\\a_{A}  = 14. 64 ft/s^{2}

6 0
3 years ago
A 3.00-kg pendulum is 28.84 m long. what is its period on earth?
GrogVix [38]
T=2 \pi  \sqrt{ \frac{l}{g} }
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T=10.7731s.
7 0
3 years ago
Vector A has a magnitude of 16 m and makes an angle of 44° with the positive x axis. Vector B also has a magnitude of 13 m and i
marshall27 [118]

Answer with explanation:

The given vectors in are reduced to their componednt form as shown

For vector A it can be written as

\overrightarrow{v}_{a}=16cos(44^{o})\widehat{i}+16sin(44^{o})\widehat{j}

Similarly vector B can be written as

\overrightarrow{v}_{b}=-13\widehat{i}

Hence The sum and difference is calculated as

\overrightarrow{v}_{a}+\overrightarrow{v}_{b}=16cos(44^{o})\widehat{i}+16sin(44^{o})\widehat{j}+(-13\widehat{i})\\\\\overrightarrow{v}_{a}+\overrightarrow{v}_{b}=(16cos(44^{o})-13)\widehat{i}+16sin(44^{o})\widehat{j}\\\\\therefore \overrightarrow{v}_{a}+\overrightarrow{v}_{b}=-1.49\widehat{i}+11.11\widehat{j}\\\\\therefore |\overrightarrow{v}_{a}+\overrightarrow{v}_{b}|=\sqrt{(-1.49)^{2}+11.11^{2}}=11.21m

The direction is given by

\theta =tan^{-1}\frac{r_{y}}{r_{x}}\\\\\theta =tan^{-1}\frac{11.11}{-1.49}=97.64^{o}with positive x axis.

Similarly

\overrightarrow{v}_{a}-\overrightarrow{v}_{b}=16cos(44^{o})\widehat{i}+16sin(44^{o})\widehat{j}-(-13\widehat{i})\\\\\overrightarrow{v}_{a}-\overrightarrow{v}_{b}=(16cos(44^{o})+13)\widehat{i}+16sin(44^{o})\widehat{j}\\\\\therefore \overrightarrow{v}_{a}-\overrightarrow{v}_{b}=24.51\widehat{i}+11.11\widehat{j}\\\\\therefore |\overrightarrow{v}_{a}-\overrightarrow{v}_{b}|=\sqrt{(24.51)^{2}+11.11^{2}}=26.91m

The direction is given by

\theta =tan^{-1}\frac{r_{y}}{r_{x}}\\\\\theta =tan^{-1}\frac{11.11}{24.51}=24.38^{o}with positive x axis.

3 0
3 years ago
A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.
Bas_tet [7]

Answer:

Explanation:

Total momentum of the system before the collision

.5 x 3 - 1.5 x 1.5 = -0.75 kg m/s towards the left

If v be the velocity of the stuck pucks

momentum after the collision = 2 v

Applying conservation of momentum

2 v = -  .75

v =  - .375 m /s

Let after the collision v be the velocity of .5 kg puck

total momentum after the collision

.5 v + 1.5 x .231 = .5v +.3465

Applying conservation of momentum law

.5 v +.3465 = - .75

v = - 2.193 m/s

2 ) To verify whether the collision is elastic or not , we verify whether the kinetic energy is conserved or not.

Kinetic energy before the collision

= 2.25 + 1.6875

=3.9375 J

kinetic energy after the collision

= .04 + 1.2 =1.24 J

So kinetic energy is not conserved . Hence collision is not elastic.

3 ) Change in the momentum of .5 kg

1.5 - (-1.0965 )

= 2.5965

Average force applied = change in momentum / time

= 2.5965 / 25 x 10⁻³

= 103.86 N

5 0
3 years ago
Given a force of 56 N and an acceleration of 7 m/s2<br><br> , what is the mass?
melisa1 [442]

Answer:

Your mass is 60 kg.

Explanation:

5 0
3 years ago
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