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Burka [1]
3 years ago
5

Gordon throws a baseball into the air. It rises, stops when it reaches its greatest height, and then falls back down to the grou

nd. At what point does kinetic energy convert to potential energy?a. when the baseball is rising
b. when the baseball is falling
c. just after the baseball hits the ground
d. while the baseball is stopped in the air
Physics
1 answer:
Arte-miy333 [17]3 years ago
7 0

Answer:

A) When the baseball is rising

Explanation:

When an object like the baseball is thrown into the air, the kinetic energy  is gradually converted into potential energy. When the baseball rises to its maximum height, the kinetic energy becomes zero and the kinetic energy is fully converted into potential energy.

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Studying this brochure from nasa, which explains more detail the instruments carried by the Juno spacecraft which scientific act
adoni [48]

Answer:

Juno scientific payload includes:

  • A gravity/radio science system (Gravity Science)
  • A six-wavelength microwave radiometer for atmospheric sounding and composition (MWR)
  • A vector magnetometer (MAG)
  • Plasma and energetic particle detectors (JADE and JEDI)
  • A radio/plasma wave experiment (Waves)
  • An ultraviolet imager/spectrometer (UVS)
  • An infrared imager/spectrometer (JIRAM)

Explanation:

Each mission of NASA has a specific set of instruments that it uses to perform scientific experiments on the desired heavenly body. In case of Juno, the mission for Jupiter has a series of instruments that would study domains of gravitational forces, magnetic effect, particle detection, radiation detection, UV/IR imaging, and plasma experiments.

3 0
2 years ago
Help with either I don’t understand this I know W=f*d but don’t get how it applies here
rosijanka [135]

Answer:

5) Displacement = +3.125 m

Displacement is in the same direction as the force vector.

6) Force = -53.89 N

Force is in an opposite direction relative to the displacement.

Explanation:

5) We are given;

Force; F = 160 N.

Workdone; W = +500 J

Now, formula for workdone is;

W = Force × displacement

Thus, displacement = Work/force

Displacement = 500/160

Displacement = +3.125 m

Thus, displacement is in the same direction as the force vector.

6) We are given;

Displacement; d = 18 m.

Workdone; W = -970 J

Like in the first answer above,

Workdone = Force × Displacement

Thus;

Force = Workdone/Displacement

Force = -970/18

Force = -53.89 N

Since force is negative and displacement is positive, it means force is in an opposite direction relative to the displacement.

3 0
2 years ago
The weight of a rock is 8.5 pounds. Does this statement include
Kaylis [27]

Answer:

quantitative

Explanation:

4 0
2 years ago
A 5.00 kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the en
Brut [27]

Answer:

75 N

Explanation:

In this problem, the position of the crate at time t is given by

y(t)=2.80t+0.61t^3

The velocity of the crate vs time is given by the derivative of the position, so it is:

v(t)=y'(t)=\frac{d}{dt}(2.80t+0.61t^3)=2.80+1.83t^2

Similarly, the acceleration of the crate vs time is given by the derivative of the velocity, so it is:

a(t)=v'(t)=\frac{d}{dt}(2.80+1.83t^2)=3.66t [m/s^2]

According to Newton's second law of motion, the force acting on the crate is equal to the product between mass and acceleration, so:

F(t)=ma(t)

where

m = 5.00 kg is the mass of the crate

At t = 4.10 s, the acceleration of the crate is

a(4.10)=3.66\cdot 4.10 =15.0 m/s^2

And therefore, the force on the crate is:

F=ma=(5.00)(15.0)=75 N

7 0
3 years ago
Water is leaking out of an inverted conical tank at a rate of 10,500 cm3/min at the same time that water is being pumped into th
satela [25.4K]

The tank has a volume of \dfrac\pi3R^2H, where H=6\,\rm m is its height and R=\dfrac d2=2\,\rm m is its radius.

At any point, the water filling the tank and the tank itself form a pair of similar triangles (see the attached picture) from which we obtain the following relationship:

\dfrac26=\dfrac rh\implies r=\dfrac h3

The volume of water in the tank at any given time is

V=\dfrac\pi3r^2h

and can be expressed as a function of the water level alone:

V=\dfrac\pi3\left(\frac h3\right)^2h=\dfrac\pi{27}h^3

Implicity differentiating both sides with respect to time t gives

\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9h^2\,\dfrac{\mathrm dh}{\mathrm dt}

We're told the water level rises at a rate of \dfrac{\mathrm dh}{\mathrm dt}=20\,\frac{\rm cm}{\rm min} at the time when the water level is h=2\,\mathrm m=200\,\mathrm{cm}, so the net change in the volume of water \dfrac{\mathrm dV}{\mathrm dt} can be computed:

\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9(200\,\mathrm{cm})^2\left(20\,\dfrac{\rm cm}{\rm min}\right)=\dfrac{800,000\pi}9\,\dfrac{\mathrm{cm}^3}{\rm min}

The net rate of change in volume is the difference between the rate at which water is pumped into the tank and the rate at which it is leaking out:

\dfrac{\mathrm dV}{\mathrm dt}=(\text{rate in})-(\text{rate out})

We're told the water is leaking out at a rate of 10,500\,\frac{\mathrm{cm}^3}{\rm min}, so we find the rate at which it's being pumped in to be

\dfrac{800,000\pi}9\,\dfrac{\mathrm{cm}^3}{\rm min}=(\text{rate in})-10,500\,\dfrac{\mathrm{cm}^3}{\rm min}

\implies\text{rate in}\approx289,753\,\dfrac{\mathrm{cm}^3}{\rm min}

4 0
3 years ago
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