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2 years ago

Suppose an object’s initial velocity is 10 m/s and its final velocity is 4 m/s. Mass is constant.

Physics

2 answers:

Anna [14]2 years ago

8 0

Its C I just did it on edge

Ostrovityanka [42]2 years ago

3 0

The correct answer is:

Work is negative, the environment did work on the object, and the energy of the system decreases.

In fact, the work-energy theorem states that the work done by the system is equal to its variation of kinetic energy:

$W=\Delta K=K_f -K_i$

In this problem, the variation of kinetic energy $\Delta K$ is negative (because the final velocity is less than the initial velocity), so the work is negative, and this means that the environment did work on the object, and its energy decreased.

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Objects 1 and 2 attract each other with a gravitational force of 12 units. If the mass of Object 2 is tripled, then the new grav

olasank [31]

Explanation:

Fgravity = G*(mass1*mass2)/D².

G is the gravitational constant, which has the same value throughout our universe.

D is the distance between the objects.

so, if you triple one of the masses, what does that do to our equation ?

Fgravitynew = G*(3*mass1*mass2)/D²

due to the commutative property of multiplication

Fgravitynew = 3* G*(mass1*mass2)/D² = 3* Fgravity

so, the right answer is 3×12 = 36 units.

5 0

2 years ago

While running at a constant velocity, how should you throw a ball with respect to you so that you can catch it yourself?

timurjin [86]

You are running at constant velocity in the x direction, and based on the 2D definition of projectile motion, Vx=Vxo. In other words, your velocity in the x direction is equal to the starting velocity in the x direction. Let's say the total distance in the x direction that you run to catch your own ball is D (assuming you have actual values for Vx and D). You can then use the range equation, D= (2VoxVoy)/g, to find the initial y velocity, Voy. g is gravitational acceleration, -9.8m/s^2. Now you know how far to run (D), where you will catch the ball (xo+D), and the initial x and y velocities you should be throwing the ball at, but to find the initial velocity vector itself (x and y are only the components), you use the pythagorean theorem to solve for the hypotenuse. Because you know all three sides of the triangle, you can also solve for the angle you should throw the ball at, as that is simply arctan(y/x).

5 0

2 years ago

True or false an experiment in investigating the effects and development variable on the independent variable

Sergeu [11.5K]

true if you are refering to the desing of the experimnt as it does identify the variable

8 0

2 years ago

A 0.600 kg block is attached to a spring with spring constant 15 N/m. While the block is sitting at rest, a student hits it with

gulaghasi [49]

Answer:

8.8 cm

31.422 cm/s

Explanation:

m = Mass of block = 0.6 kg

k = Spring constant = 15 N/m

x = Compression of spring

v = Velocity of block

A = Amplitude

As the energy of the system is conserved we have

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{0.6\times 0.44^2}{15}}\\\Rightarrow A=0.088\ m\\\Rightarrow A=8.8\ cm$

Amplitude of the oscillations is 8.8 cm

At x = 0.7 A

Again, as the energy of the system is conserved we have

$\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\dfrac{15(0.088^2-(0.7\times 0.088)^2)}{0.6}}\\\Rightarrow v=0.31422\ m/s$

The block's speed is 31.422 cm/s

4 0

3 years ago

Figure 10.20 in your textbook shows an energy diagram for a system with total energy E1. Suppose the system's total energy is E2

wolverine [178]

The particles can undergo small oscillations around x₂.

The given parameters;

<em>initial energy of the particles = E₁</em>
<em>final energy of the particles, E₂ = 0.33E₁</em>

The movement of the particles depends on the kinetic energy of the particles.

When kinetic energy of the particles is 100%, the particles can oscillate from x₁ to x₅.

However, when the total energy of this particles is reduced to one-third (¹/₃) or 33% of the initial energy of the particle, the oscillation of the particles will be reduced.

The maximum position the particle can oscillate is x₅

The half position the particles can oscillate is x₃

Since 33% is less than the half of the energy of the particle, the particle will oscillate between x₁ and x₂.

Thus, we can conclude that the particles can undergo small oscillations around x₂.

Learn more here:brainly.com/question/23910777

3 0

2 years ago