Home / Assignment Help / A model used for the yield y of an agricultural crop as a function of the nitrogen level n in the soil (measured in appropriate units) is y = kn 1 + n2 where k is a positive constant. what nitrogen level gives the best yield?

A model used for the yield y of an agricultural crop as a function of the nitrogen level n in the soil (measured in appropriate units) is y = kn 1 + n2 where k is a positive constant. what nitrogen level gives the best yield?

14. E 2.0 m/s^2

Initially, a 20 N force is applied to the block, so it has an acceleration of 3.9 m/s^2. According to Newton’s law, the mass of the block is:

m=frac{F}{a}=frac{20 N}{3.9 m/s^2}=5.13 kg

In the second situation, a force of 10 N is applied to the block. Since the mass is still the same, the acceleration now is:

a=frac{F}{m}=frac{10 N}{5.13 kg}=1.95 m/s^2

So, approximately 2.0 m/s^2.

15. C. 866 N at 78.1° counterclockwise to the x-axis

Resultant along the x- and y-axis:

R_x = (985 N)(cos 31^{circ})-(788 N)(sin 32^{circ})-(411 N)(cos 53^{circ})=179.4 N

R_y = (985 N)(sin 31^{circ})+(788 N)(cos 32^{circ})-(411 N)(sin 53^{circ})=847.3 N

Magnitude and direction of the resultant:

R=sqrt{R_x^2+R_y^2}=sqrt{(179.4 N)^2+(847.3 N)^2}=866.0 N

theta=arctan(frac{R_y}{R_x})=arctan(frac{847.3 N}{179.4 N})=78.1^{circ}

16. D. 1720 N

Since the sphere is suspended, it is in equilibrium, therefore the tension in the chain is equal to the weight of the sphere attached to it, therefore:

T=mg=(175 kg)(9.81 m/s^2)=1720 N

17. C. A

This arrangement generates the largest tension in the chain, because in all other arrangements the weight of the object is split between the two chains, while in this case all the weight is hold by one chain, therefore the tension in this case is larger.

18. Straight path

The gravity “holds” the planets keeping them in a circular orbit. If we remove gravity, the planets would continue in a straight path with constant speed, because now there are no more forces acting on it, so by inertia they will continue their uniform motion with constant speed.

19. B. 1.6 × 104 N

First we can find the deceleration of the car by using the SUVAT equation:

v^2 -u^2 =2aS

where v=0 m/s, u=2 m/s, and S=15 cm=0.15 m. Re-arranging, we have

a=frac{-u^2}{2S}=frac{-(2 m/s)^2}{2(0.15 m)}=-13.3 m/s^2

And now we can calculate the average force exerted on the car, by using Newton’s second law:

F=ma=(1200 kg)(-13.3 m/s^2)=-15960 N=-1.6 cdot 10^4 N

(the negative sign means that the force’s direction is opposite to the motion of the car)

20. A. magnetism

Magnetism is part of the electromagnetic force, which is one of the fundamental forces which act also through empty space. All the other forces need some object in order to act.

21. E. 45 N

The magnitude of the force in link A is equal to the weight of the rod plus the weight of the lower block, therefore:

W=(m_1 + m_2)g=(0.6 kg+4.0 kg)(9.8 m/s^2)=45 N

22. C. on Earth at sea level

The weight of the bowling ball is given by: W=mg, where m is the mass of the ball and g is the acceleration due to gravity. The value of g increases when moving from the Earth’s center to the Earth’s surface, then decreases when moving far from the surface, so the point where g is greatest is at sea level, where it is 9.81 m/s^2. On the surface of the Moon, g is much smaller (about 1/6 of the value on Earth).

23. A. 0.5 m/s2

The acceleration of the block is given by Newton’s second law:

a=frac{F}{m}=frac{20 N}{40 kg}=0.5 m/s^2

24. D. Both forces are equal in magnitude but opposite in direction.

According to Newton’s third law: if an object A exerts a force on an object B, then object B exerts a force equal and opposite on object B. In this case, objects A and B are the bat and the baseball, therefore the two forces are equal in magnitude and opposite in direction.

25. B. 2940 N

The mass of the boulder is equal to its weight divided by the acceleration of gravity (9.81 m/s^2):

m=frac{W}{g}=frac{2400 N}{9.81 m/s^2}=245 kg

So now we can calculate the force needed to accelerate the boulder to 12.0 m/s^2:

F=ma=(245 kg)(12.0 m/s^2)=2940 N

26. D. 32.2 N

The weight of the object on Mercury is given by:

W=mg=(8.69 kg)(3.71 m/s^2)=32.2 N

About eduhawks

Check Also

Are the phases of the moon the same everywhere on earth

Are the phases of the moon the same everywhere on earth