Wavelength = c/f.
Wavelength =0.5km
The initial kinetic energy of the car is
Then, the velocity of the car is decreased by half:
so, the new kinetic energy is
So, the new kinetic energy is 1/4 of the initial kinetic energy of the car. Numerically:
Answer:
35 mph
Explanation:
The key of this problem lies in understanding the way that projectile motion works as we are told to neglect the height of the javelin thrower and wind resistance.
When the javelin is thown, its velocity will have two components: a x component and a y component. The only acceleration that will interact with the javelin after it was thown will be the gravety, which has a -y direction. This means that the x component of the velocity will remain constant, and only the y component will be affected, and can be described with the constant acceleration motion properties.
When an object that moves in constant acceleration motion, the time neccesary for it to desaccelerate from a velocity v to 0, will be the same to accelerate the object from 0 to v. And the distance that the object will travel in both desaceleration and acceleration will be exactly the same.
So, when the javelin its thrown, it willgo up until its velocity in the y component reaches 0. Then it will go down, and it will reach reach the ground in the same amount of time it took to go up and, therefore, with the same velocity.
Answer:
The total resistance in a parallel circuit is always less than any of the branch resistances. Adding more parallel resistances to the paths causes the total resistance in the circuit to decrease. As you add more and more branches to the circuit the total current will increase because Ohm's Law states that the lower the resistance, the higher the current.
Explanation:
The <em>mass</em> of an object is a measure of how much stuff it's made of. So it's
a property of the object, like the object's length or color. It doesn't depend on
where the object is located right now, or on what else is around it..
The <em>weight</em> of an object is the measure of the gravitational force between it
and something else. The strength of the force depends on the size of both
masses, and also on how far apart they are. So the weight does depend on
where the object is located right now, and on what else is around it.
I think that sums it up fairly well.
