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musickatia [10]
3 years ago
12

3. An athlete makes a long jump and follows a projectile motion. Air resistance is negligible. Which one of the following statem

ents is true about the athlete? (a) "The athlete has a constant hanzontal und vertical velocities. (h) The athlete han a constant horizontal velocity anil constant downward acceleration ic) The athlete has a constant upward acceleration followed by a constant downward acceleration d) The athlete has a constant upward velocity followed by a constant downwani velo​
Physics
1 answer:
yulyashka [42]3 years ago
6 0

Answer:

Option (b) is correct.

Explanation:

The motion under the influence of gravity is called projectile motion.

The acceleration due to gravity is constant through out the motion and it is always acting downwards.

When an athlete jumps and follow the projectile path, it always have the same horizontal velocity as there is no acceleration in the horizontal direction.

Also he has the vertical acceleration constant which is equal to the acceleration due to gravity and acts towards the center of earth.  

Option (b) is correct.

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A gasoline tank has the shape of an inverted right circular cone with base radius 4 meters and height 5 meters. Gasoline is bein
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Answer:

h'=0.25m/s

Explanation:

In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).

So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of 8m^{3}/s. As you  may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

V_{cone}=\frac{1}{3} \pi r^{2}h

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.

If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

\frac {r}{h}=\frac{4}{5}

When solving for r, we get:

r=\frac{4}{5}h

so we can substitute this into our volume of a cone formula:

V_{cone}=\frac{1}{3} \pi (\frac{4}{5}h)^{2}h

which simplifies to:

V_{cone}=\frac{1}{3} \pi (\frac{16}{25}h^{2})h

V_{cone}=\frac{16}{75} \pi h^{3}

So now we can proceed and find the partial derivative over time of each of the sides of the equation, so we get:

\frac{dV}{dt}= \frac{16}{75} \pi (3)h^{2} \frac{dh}{dt}

Which simplifies to:

\frac{dV}{dt}= \frac{16}{25} \pi h^{2} \frac{dh}{dt}

So now I can solve the equation for dh/dt (the rate of height over time, the velocity at which height is increasing)

So we get:

\frac{dh}{dt}= \frac{(dV/dt)(25)}{16 \pi h^{2}}

Now we can substitute the provided values into our equation. So we get:

\frac{dh}{dt}= \frac{(8m^{3}/s)(25)}{16 \pi (4m)^{2}}

so:

\frac{dh}{dt}=0.25m/s

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