A 2kg water balloon is flying at a rate of 4m/s^2. With what force will it hit its target?

If a bird lands on a tree, is that an example of speed, velocity, or acceleration?

vazorg [7]

I think the answer is:
Velocity

4 0

3 years ago

What is the effect on the force of gravity between two objects if the mass of one object remains unchanged while the distance to

drek231 [11]

Answer:

The force of gravity after you double the mass and the distance is half of the initial force: $F_{2}=\frac{1}{2}F_{1}$

Explanation:

The initial force of gravity is:

$F_{1}=\frac{Gm_{1}m_{2}}{r^2}$

where $G$ is the universal gravitational constant, $m_{1}$ is the mass of the first object, $m_{2}$ is the mass of the second object, and $r$ is the distance between the objects.

If the mass of the second object is doubled, now we have $2m_{2}$ , and if the distance between the objects is also doubled instead of $r$ now we have $2r$ .

So the force of gravity now is:

$F_{2}=\frac{Gm_{1}(2m_{2})}{(2r)^2}\\ F_{2}=\frac{2Gm_{1}m_{2}}{4r^2} \\F_{2}=\frac{1}{2} \frac{Gm_{1}m_{2}}{r^2}$

and we know that $F_{1}=\frac{Gm_{1}m_{2}}{r^2}$

so the new force of gravity is:

$F_{2}=\frac{1}{2}F_{1}$

The force of gravity after you double the mass and the distance is half of the initial force.

3 0

3 years ago

A 0.40 kg ball is suspended from a spring with spring constant 12 n/m . part a if the ball is pulled down 0.20 m from the equili

PolarNik [594]

The total mechanical energy of the system at any time t is the sum of the kinetic energy of motion of the ball and the elastic potential energy stored in the spring:
$E=K+U= \frac{1}{2}mv^2+ \frac{1}{2}kx^2$
where m is the mass of the ball, v its speed, k the spring constant and x the displacement of the spring with respect its rest position.

Since it is a harmonic motion, kinetic energy is continuously converted into elastic potential energy and vice-versa.

When the spring is at its maximum displacement, the elastic potential energy is maximum (because the displacement x is maximum) while the kinetic energy is zero (because the velocity of the ball is zero), so in this situation we have:
$E=U_{max}= \frac{1}{2}k(x_{max})^2$

Instead, when the spring crosses its rest position, the elastic potential energy is zero (because x=0) and therefore the kinetic energy is at maximum (and so, the ball is at its maximum speed):
$E=K_{max}= \frac{1}{2}m(v_{max})^2$

Since the total energy E is always conserved, the maximum elastic potential energy should be equal to the maximum kinetic energy, and so we can find the value of the maximum speed of the ball:
$U_{max}=K_{max}$
$\frac{1}{2}k(x_{max})^2 = \frac{1}{2}m(v_{max})^2$
$v_{max}= \sqrt{ \frac{k x_{max}^2}{m} }= \sqrt{ \frac{(12 N/m)(0.20 m)^2}{0.4 kg} }=1.1 m/s$