I wouldn't consider it construction
We have,
- Mass of both students is 60 kg
- Surface area of heels of one student say A is 0.45 cm² = 0.45/10000 m²
- Surface area of heals of another student say B is 74 cm² = 74/10000 m²
And we know that,
Here, area is provided but the force applied will be due to the acceleration of gravity so,
- F = ma
- F = 60 × 9.8
- F = 588
Now, let us calculate the pressure applied by both students;
Pressure(Student A)
- P = 588/0.45 × 10000
- P = 58800 × 10000/45
- P = 1306.6 × 10000
- P = 1306000.0 pascal
Pressure(Student B)
- P = 588/74 × 10000
- P = 5880000/74
- P = 7837.83
From the above two solutions we can say that, Pressure(Student A) > Pressure(student B)
And thus we can also say that,
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In order to find the our own velocity with respect to land,we need to apply the theory of relative velocity.
Now consider the velocity of the ship traveling towards the north with respect to land as A.Consider our own velocity headed northwards as B.
The relative velocity is the velocity that the body A would appear to an observer on the body B and vice versa.
In this case the relative velocity would be arrived by summing up our velocity with the velocity of the ship as the object (I) is travelling in the ship.
Relative velocity = Velocity of Body A+ Velocity of Body B.
Velocity of the ship traveling towards the north with respect to land(A)= 13.0m/s. (Given)
Our own velocity headed northwards(B)= 2.8 m/s.
Relative velocity = Velocity of Body A+ Velocity of Body B.
Relative velocity= 13.0 + 2.8 = 15.8m/s.
Thus our own velocity with respect to the land is 15.8 m/s.
Answer:
<em>The range is 35.35 m</em>
Explanation:
<u>Projectile Motion</u>
It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.
Being vo the initial speed of the object, θ the initial launch angle, and the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:
The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:
The range is 35.35 m