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
spayn [35]
3 years ago
7

A boulder rolls off a 100 meter cliff with a horizontal velocity of 2 m/s.

Physics
1 answer:
bonufazy [111]3 years ago
7 0

Explanation:

It is given that,

Height of the cliff is 100 m

Its horizontal velocity is 2 m/s

The attached figure shows the sketch of the cliff and the boulder. It will go in the form of parabola. Its trajectory is parabolic in nature. The horizontal distance is called the range of the parabola.

You might be interested in
How do u solve
Shalnov [3]

Time = distance/speed

         = 5/180

Time = 0.03 hrs

7 0
3 years ago
A high powered rifle can shoot a bullet at a speed of 1500 mi/hr.
kumpel [21]
Initial velocity of the billet is maximum, once out if the rifle it begins to slow down
7 0
3 years ago
An object-spring system undergoes simple harmonic motion with an amplitude A. Does the total energy change if the mass is double
son4ous [18]

Answer: No, The energy will remain the same

Explanation: Doubling the mass and leaving the amplitude unchanged won't have any effect on the total energy of the system.

At maximum displacement, E=0.5kA^2

Where E = total energy

K = spring constant

A = Amplitude

From the formula above : Total Energy is independent of mass,. Therefore, total energy won't be affected by Doubling the mass value of the object.

Also when the object is at a displacement 'x' from its equilibrium position.

E = Potential Energy(P.E) + Kinetic Energy(K.E)

P.E = 0.5kx^2

Where x = displacement from equilibrium position

E = Total Energy

K. E= E-0.5kx^2

From the relation above, total energy is independent of its mass and therefore has no effect on the total energy.

6 0
3 years ago
A small rock is launched straight upward from the surface of a planet with no atmosphere. The initial speed of the rock is twice
Scorpion4ik [409]

If gravitational effects from other objects are negligible, the speed of the rock at a very great distance from the planet will approach a value of \sqrt{3} v_{e}

<u>Explanation:</u>

To express velocity which is too far from the planet and escape velocity by using the energy conservation, we get

Rock’s initial velocity , v_{i}=2 v_{e}. Here the radius is R, so find the escape velocity as follows,

            \frac{1}{2} m v_{e}^{2}-\frac{G M m}{R}=0

            \frac{1}{2} m v_{e}^{2}=\frac{G M m}{R}

            v_{e}^{2}=\frac{2 G M}{R}

            v_{e}=\sqrt{\frac{2 G M}{R}}

Where, M = Planet’s mass and G = constant.

From given conditions,

Surface potential energy can be expressed as,  U_{i}=-\frac{G M m}{R}

R tend to infinity when far away from the planet, so v_{f}=0

Then, kinetic energy at initial would be,

                  k_{i}=\frac{1}{2} m v_{i}^{2}=\frac{1}{2} m\left(2 v_{e}\right)^{2}

Similarly, kinetic energy at final would be,

                k_{f}=\frac{1}{2} m v_{f}^{2}

Here, v_{f}=\text { final velocity }

Now, adding potential and kinetic energies of initial and final and equating as below, find the final velocity as

                 U_{i}+k_{i}=k_{f}+v_{f}

                 \frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}+0

                  \frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}

'm' and \frac{1}{2} as common on both sides, so gets cancelled, we get as

                   4\left(v_{e}\right)^{2}-\frac{2 G M}{R}=v_{f}^{2}

We know, v_{e}=\sqrt{\frac{2 G M}{R}}, it can be wriiten as \left(v_{e}\right)^{2}=\frac{2 G M}{R}, we get

                4\left(v_{e}\right)^{2}-\left(v_{e}\right)^{2}=v_{f}^{2}

                v_{f}^{2}=3\left(v_{e}\right)^{2}

Taking squares out, we get,

                v_{f}=\sqrt{3} v_{e}

4 0
3 years ago
A jet is circling an airport control tower at a distance of 20.6 km. An observer in the tower watches the jet cross in front of
lesya [120]

Answer:

197.76 m

Explanation:

r = Radius of the path = 20.6 km = 20.6\times 10^3\ m

\theta = The angle subtended by moon = 9.6\times 10^{-3}\ rad

Distance traveled is given by

s=r\times\theta

\Rightarrow s=20.6\times 10^3\times 9.6\times 10^{-3}

\Rightarrow s=197.76\ m

The distance traveled by the jet is 197.76 m

8 0
3 years ago
Other questions:
  • Explain relationship between thermal conductivity and thermal behaviour of glass , plastic and wood.
    9·2 answers
  • Which of the following is not a prediction of General Relativity?(A) gravitational waves(B) spacetime is distorted by mass and e
    5·1 answer
  • A 15.0 kg block is pulled by a horizontal force of 30.0 N along a rough horizontal surface at a constant velocity. What is the c
    11·1 answer
  • At the pizza party you and two friends decide to go to Mexico City from El Paso, TX where y'all live. You volunteer your car if
    15·1 answer
  • Need help on this please
    13·1 answer
  • An object is hanging by a string from the ceiling of an elevator. The elevator is moving upward with a constant speed. What is t
    12·1 answer
  • Riparian zones are attractive to livestock because _____. they are far from human presence they contain lush vegetation that liv
    9·2 answers
  • At what speed does the kg ball move ?
    13·1 answer
  • A certain element has a half-life of 200 years. How much of a 1600 gram sample would be left after 800 years?
    15·1 answer
  • HELP ME PLEASE BRAINIEST IF CORRECT!!!!!
    8·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!