1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Lesechka [4]
3 years ago
9

Three balls are thrown off the top of a building, all with the same speed but with different launch angles (position is given in

meters and time is given in seconds). The components of the initial velocities are given.
The blue ball has an initial velocity of (6 m/s, 8 m/s).
The green ball has an initial velocity of (10 m/s, 0 m/s).
The red ball has an initial velocity of (8 m/s, -6 m/s).

a. Rank the three balls according to which one hits the ground first.
b. Rank the three balls according to which one has the greatest speed the instant before impact with the ground.
c. Calculate the speed of each of the balls the instant before impact with the ground.
Physics
1 answer:
jek_recluse [69]3 years ago
4 0

Answer:

A. Red, Green, Blue

B. None. They all hit the ground with the same speed.

C. Final Kinetic Energy = Initial Gravitational Potential Energy

Explanation:

A. This one is actually just common sense. If you think about it, the object that is being thrown downward will reach the ground first. The object thrown upwards will hit the ground the last. If you don't believe me, use just the vertical component of velocity and use acceleration = 9.8 m/s^2. Find the time it takes for an arbitrary displacement each ball will travel. You will find that the one pointed downward will reach the ground in the shortest amount of time. Just do the same math over for the rest of the balls and you can find the time it takes for each ball to hit the ground.

B. This one is actually really simple if you think about it. Using the Work-Energy theorem, using the system consisting a ball and the earth there will be no external forces doing work on the system. Since each ball is at some height "h" and the same initial speed, they will all have the same amount of energy in the beginning. If we set the ground as our reference point for gravitational potential energy, then there will be no gravitational energy as the height at that point will be 0. They all must have the same kinetic energy at the ground.

C.

Mechanical Energy Final = Mechanical Energy Initial + Net Work from non-conservative forces

Kinetic Energy Final + Gravitational Potential Energy Final = Kinetic Energy Initial + Gravitational Potential Energy Initial

I am using the symbol " ' " to indicate final speeds and heights

0.5*m*v'^2 + mgh' = 0.5*m*v^2 + mgh

0.5*m*v'^2 + 0 = 0.5*m*v^2 + mgh

(cancel out the mass)

0.5*v'^2 = 0.5*v^2 +gh

Since the "v" is the same initially for all three balls and "g" and "h" are the same for all three balls. The final speed v' must be the same for all three situations regardless of mass.

If you have any questions feel free to comment again. I'll try to clarify.

You might be interested in
Compute the density in g/cm? of a piece of metal that has a mass of 0.450 kg and a volume of 52 cm3
Y_Kistochka [10]

Answer:

Ro = 8.65 [g/cm³]

Explanation:

We must remember that density is defined as the ratio of mass to volume.

Ro=m/V

where:

m = mass = 0.450 [kg] = 450 [g]

V = volumen = 52 [cm³]

Ro = density [g/cm³]

Now replacing:

Ro = 450/52\\Ro = 8.65 [g/cm^{3} ]

8 0
3 years ago
Describe two reasons why an alpha particle is less penetrating than a beta or gamma particle.
uysha [10]
Alpha particles travel through the air they collide with oxygen and nitrogen molecules. While they collide with these molecules, they lose some energy until all energy are used up and they are absorbed. These particles can be absorbed by a sheet of paper or by the air. On the other hand, beta particles and gamma particles move faster than the alpha particles and are poor at ionizing atoms or molecules thus it takes more of the material to be able to absorb these particles.
8 0
3 years ago
Read 2 more answers
A proposed communication satellite would revolve around the earth in a circular orbit in the equatorial plane at a height of 358
aleksandr82 [10.1K]

Answer:

Period is 86811.5 seconds.

Explanation:

{ \boxed{ \bf{T {}^{2} =  (\frac{4 {\pi}^{2} }{GM}) {r}^{3}   }}}

{ \tt{T {}^{2}  =  \frac{4 {(3.14)}^{2} }{(6.6 \times  {10}^{ - 11} ) \times (5.98 \times  {10}^{24} )} \times  {((35880\times  {10}^{3}) } + (6370 \times  {10}^{3} )) {}^{3}   }} \\  \\ { \tt{T {}^{2}  =  7.54 \times {10}^{9} }} \\ { \tt{T =  \sqrt{7.54 \times  {10}^{9} } }} \\ { \tt{T = 86811.5 \: seconds}}

6 0
3 years ago
Suppose an electron is trapped within a small region and the uncertainty in its position is 24.0 x 10-15 m. What is the minimum
Alina [70]

Answer:

  • Uncertainty in position (∆x) = 24 × 10⁻¹⁵ m
  • Uncertainty in momentum (∆P) = ?
  • Planck's constant (h) = 6.26 × 10⁻³⁴ Js

\longrightarrow \:  \:  \sf\Delta x .\Delta p =  \dfrac{h}{4\pi}

\longrightarrow \:  \:  \sf24 \times  {10}^{ - 15}  .\Delta p =  \dfrac{6.26 \times  {10}^{ - 34}} {4 \times  \frac{22}{7} }

\longrightarrow \:  \:  \sf24 \times  {10}^{ - 15}  .\Delta p =  \dfrac{6.26 \times  {10}^{ - 34}} { \frac{88}{7} }

\longrightarrow \:  \:  \sf24 \times  {10}^{ - 15}  .\Delta p =  \dfrac{6.26 \times  {10}^{ - 34} \times 7} { 8 }

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ - 34} } { 8  \times 24 \times  {10}^{ - 15} }

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ - 34} } { 192 \times  {10}^{ - 15} }

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ - 34}  \times  {10}^{15} } { 192}

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ -19}   } { 192}

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{4382 \times  {10}^{ - 2}  \times  {10}^{ -19}   } { 192}

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{4382 \times  {10}^{ - 21}   } { 192}

\longrightarrow \:  \:  \sf\Delta p = 22.822\times  {10}^{ - 21}

\longrightarrow \:  \:  \sf\Delta p = 2.2822 \times  {10}^{1} \times  {10}^{ - 21}

\longrightarrow \:  \: \underline{ \boxed{ \red{  \bf\Delta p = 2.2822 \times  {10}^{ - 20}  \:  kg/ms}}}

4 0
3 years ago
A stationary 15 kg object is located in a table near the surface of the earth. The coefficient of static friction between the su
Marta_Voda [28]

The net force acting on the object perpendicular to the table is

∑ F[perp] = F[normal] - mg = 0

where mg is the weight of the object. Then

F[normal] = mg = (15 kg) (9.8 m/s²) = 147 N

The maximum magnitude of static friction is then

0.40 F[normal] = 58.8 N

which means the applied 40 N force is not enough to make the object start to move. So the object has zero acceleration and does not move.

8 0
2 years ago
Other questions:
  • An object is placed 12.5 cm from a lens of focal length 22.0 cm. What is the image distance?
    14·1 answer
  • How many electrons must be removed from each of two 4.85-kg copper spheres to make the electric force of repulsion between them
    5·1 answer
  • The energy efficiency of an incandescent light bulb (= the percentage of consumed power that is actually converted into radiated
    8·1 answer
  • A parallel-plate capacitor is constructed of two horizontal 12.0-cm-diameter circular plates. A 1.0 g plastic bead, with a charg
    13·1 answer
  • What was world war 2 about?
    14·1 answer
  • In which image below is the angle of refraction the greatest?
    8·1 answer
  • Calculate the heat energy needed to change the temperature of 2 kg of copper from 10°C to 110°C.
    15·2 answers
  • is it possible that the philippines can solely rely from geothermal energy? state your opinion why or why not?​
    11·1 answer
  • What is the amplitude of a wave related to?<br> speed<br><br> time<br><br> mass<br><br> height
    12·1 answer
  • Hummingbirds may seem fragile, but their wings are capable of sustaining very large forces and accelerations. (Figure 1) shows d
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!