What is an ip Number
1 answer:
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Answer:
Impulse =14937.9 N
tangential force =14937.9 N
Explanation:
Given that
Mass of car m= 800 kg
initial velocity u=0
Final velocity v=390 km/hr
Final velocity v=108.3 m/s
So change in linear momentum P= m x v
P= 800 x 108.3
P=86640 kg.m/s
We know that impulse force F= P/t
So F= 86640/5.8 N
F=14937.9 N
Impulse force F= 14937.9 N
We know that
v=u + at
108.3 = 0 + a x 5.8
So tangential force F= m x a
F=18.66 x 800
F=14937.9 N
Answer: Advertising acts in a method similar to a fee. People who watch TV broadcasts must watch ADs. TV stations turn this into money by selling airtime to advertisers.
Explanation:
A non-rival good is a good whose consumption by one person does not reduce the remaining quantity available. An example is a street light.
For non-excludable goods, it is impossible to prevent everyone from enjoying the benefits of the good. An example is a lighthouse. This is where the free rider problem comes in.
A free rider is someone enjoying the benefits of a good without paying for it. When a good is both non-rival and non-excludable, it is convenient for consumers to enjoy the benefit without paying for it.
If TV broadcasts are both non-rival and non-excludable, everybody can choose to become a free rider. Advertising can solve this problem by converting free riders to potential buyers of goods or services advertised during broadcasts. This way, stations can generate revenue by selling airtime.
Answer:
B A and C
Explanation:
Given:
Specimen σ σ
A +450 -150
B +300 -300
C +500 -200
Solution:
Compute the mean stress
σ = (σ
+ σ
)/2
σ = (450 + (-150)) / 2
= (450 - 150) / 2
= 300/2
σ = 150 MPa
σ = (300 + (-300))/2
= (300 - 300) / 2
= 0/2
σ = 0 MPa
σ = (500 + (-200))/2
= (500 - 200) / 2
= 300/2
σ = 150 MPa
Compute stress amplitude:
σ = (σ
- σ
)/2
σ = (450 - (-150)) / 2
= (450 + 150) / 2
= 600/2
σ = 300 MPa
σ = (300- (-300)) / 2
= (300 + 300) / 2
= 600/2
σ = 300 MPa
σ = (500 - (-200))/2
= (500 + 200) / 2
= 700 / 2
σ = 350 MPa
From the above results it is concluded that the longest fatigue lifetime is of specimen B because it has the minimum mean stress.
Next, the specimen A has the fatigue lifetime which is shorter than B but longer than specimen C.
In the last comes specimen C which has the shortest fatigue lifetime because it has the higher mean stress and highest stress amplitude.