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DerKrebs [107]
3 years ago
5

Consider the roller coaster on the track. If you want to increase the kinetic energy of the roller coaster as it proceeds throug

h the course, what is the FIRST thing you should do?
A) Increase the height at point A.
B) Decrease the height at point A.
C) Increase the height at point B.
D) Decrease the height at point B.

Physics
1 answer:
a_sh-v [17]3 years ago
5 0

Answer:

The answer is A and D but if you want one anwer i think the anwer is A

Explanation:

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The speed of the ball when it returns to the same horizontal level
stellarik [79]

8.0 m/s if there is no air resistance. (B)

Less if there IS any air resistance.

4 0
3 years ago
A vertical cylindrical tank 10 ft in diameter, has an inflow line of 0.3 ft inside diameter and an outflow line of 0.4 ft inside
neonofarm [45]

Answer:

\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}, level is rising.

Explanation:

Since liquid water is a incompresible fluid, density can be eliminated of the equation of Mass Conservation, which is simplified as follows:

\dot V_{in} - \dot V_{out} = \frac{dV_{tank}}{dt}

\frac{\pi}{4}\cdot D_{in}^2 \cdot v_{in}-\frac{\pi}{4}\cdot D_{out}^2 \cdot v_{out}= \frac{\pi}{4}\cdot D_{tank}^{2} \cdot \frac{dh}{dt} \\D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out} = D_{tank}^{2} \cdot \frac{dh}{dt} \\\frac{dh}{dt}  = \frac{D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out}}{D_{tank}^{2}}

By replacing all known variables:

\frac{dh}{dt} = \frac{(0.3 ft)^{2}\cdot (5 \frac{ft}{s} ) - (0.4 ft)^{2} \cdot (2 \frac{ft}{s} )}{(10 ft)^{2}}\\\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}

The positive sign of the rate of change of the tank level indicates a rising behaviour.

6 0
3 years ago
A 2.0 kg block is pulled across a horizontal surface by a 15 N force at a constant velocity. What is the force of friction actin
lianna [129]

Answer:

<em>The force of friction acting on the block has a magnitude of 15 N and acts opposite to the applied force.</em>

Explanation:

<u>Net Force </u>

The Second Newton's law states that an object acquires acceleration when an unbalanced net force is applied to it.

The acceleration is proportional to the net force and inversely proportional to the mass of the object.

If the object has zero net force, it won't get accelerated and its velocity will remain constant.

The m=2 kg block is being pulled across a horizontal surface by a force of F=15 N and we are told the block moves at a constant velocity. This means the acceleration is zero and therefore the net force is also zero.

Since there is an external force applied to the box, it must have been balanced by the force of friction, thus the force of friction has the same magnitude acting opposite to the applied force.

The force of friction acting on the block has a magnitude of 15 N opposite to the applied force.

6 0
2 years ago
if vector u has lenght 70 and direction 40 degrees, and vector v has length 85 and direction 335 degrees what is the length and
Anastaziya [24]

Answer:

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

Explanation:

||u|| = 70

θ = 40°

\vec{u}_x=||u||cos\theta \\\Rightarrow \vec{u}_x=70cos40=53.62

\vec{u}_y=||u||sin\theta \\\Rightarrow \vec{u}_y=70sin40=44.99

||v|| = 85

θ = 335°

\vec{v}_x=||v||cos\theta \\\Rightarrow \vec{v}_x=85cos335=77.03

\vec{v}_y=||v||sin\theta \\\Rightarrow \vec{v}_y=85sin335=-35.92

Resultant

R=\sqrt{R_x^2+R_y^2}\\\Rightarrow R=\sqrt{(\vec{u}_x+\vec{v}_x)^2+(\vec{u}_y+\vec{v}_y)^2}\\\Rightarrow R =\sqrt{(70cos40+85cos335)^2+(70sin40+85sin335)^2}\\\Rightarrow R =131.15

\theta=tan^{-1}\frac{R_y}{R_x}\\\Rightarrow \theta=tan^{-1}\frac{70sin40+85sin335}{70cos40+85cos335}\\\Rightarrow \theta=tan^{-1}0.069=3.97^{\circ}

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

4 0
3 years ago
You toss a walnut at a speed of 15.0 m/s at an angle of 50.0∘ above the horizontal. The launch point is on the roof of a buildin
kaheart [24]

Answer:

a.3.51s

b.33.8m

c. 9.64,-22.9

Explanation:

8 0
2 years ago
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