Answer:
<em>The increase of kinetic energy is 108,000 J</em>
Explanation:
<u>Kinetic Energy
</u>
Is the energy an object has due to its state of motion. It's proportional to the square of the speed and the mass.
The equation for the kinetic energy is:
![\displaystyle K=\frac{1}{2}mv^2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20K%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
Where:
m = mass of the object
v = speed at which the object moves
The kinetic energy is expressed in Joules (J)
A car has a total mass of m=1,200 kg and travels at v1=12 m/s. Then it increases its speed at v2=18 m/s.
It's required to compute the increase of kinetic energy. We'll calculate both energies K1 and K2 and then subtract them.
![\displaystyle K_1=\frac{1}{2}1,200*12^2=86,400\ J](https://tex.z-dn.net/?f=%5Cdisplaystyle%20K_1%3D%5Cfrac%7B1%7D%7B2%7D1%2C200%2A12%5E2%3D86%2C400%5C%20J)
![\displaystyle K_2=\frac{1}{2}1,200*18^2=194,400\ J](https://tex.z-dn.net/?f=%5Cdisplaystyle%20K_2%3D%5Cfrac%7B1%7D%7B2%7D1%2C200%2A18%5E2%3D194%2C400%5C%20J)
The increase of kinetic energy is:
![\Delta K=K_2-K_1 =194,400\ J-86,400\ J](https://tex.z-dn.net/?f=%5CDelta%20K%3DK_2-K_1%20%3D194%2C400%5C%20J-86%2C400%5C%20J)
![\Delta K=108,000\ J](https://tex.z-dn.net/?f=%5CDelta%20K%3D108%2C000%5C%20J)
The increase of kinetic energy is 108,000 J
The sine of an angle is (opposite side) / (hypotenuse) .
Multiply each side by (hypotenuse, and you have
Opposite side = (hypotenuse) x (sine of the angle) .
Sin(12°) = 0.2079...
Opposite side = (52 m) x (0.2079)
= 10.81 meters .
Well first of all, a planet doesn't have a semimajor axis, although it's orbit does.
In an orbit with a smaller semimajor axis, the planet moves faster, and its orbital period is shorter.
That's why the International Space Station circles the Earth in less time than the Moon does.
Answer:
The net force is zero, so the acceleration is zero
Explanation:
Newton's second law states that the acceleration of an object is proportional to the net force applied to it, according to the equation:
(1)
where
is the net force on the object
m is the mass of the object
a is its acceleration
In this problem, we have a sled acted upon two forces,
. So the net force on the sled is
(2)
however, we are told that the two forces are equal in magnitude but in opposite directions, so
![F_1 = F\\F_2 = -F](https://tex.z-dn.net/?f=F_1%20%3D%20F%5C%5CF_2%20%3D%20-F)
So, eq.(2) becomes
![\sum F = F+(-F) = 0](https://tex.z-dn.net/?f=%5Csum%20F%20%3D%20F%2B%28-F%29%20%3D%200)
and so eq.(1) becomes
![\sum F = ma = 0](https://tex.z-dn.net/?f=%5Csum%20F%20%3D%20ma%20%3D%200)
which means
![a=0](https://tex.z-dn.net/?f=a%3D0)
so the acceleration of the sled is zero, and if the sled was at rest, it will not move.
Answer:
The image height is 3.0 cm
Explanation:
Given;
object distance,
= 15.0 cm
image distance,
= 5.0 cm
height of the object,
= 9.0 cm
height of the image,
= ?
Apply lens equation;
![\frac{h_i}{h_o} = -\frac{d_i}{d_o}\\\\ h_i = h_o(-\frac{d_i}{d_o})\\\\h_i = -9(\frac{5}{15} )\\\\h_i = -3 \ cm](https://tex.z-dn.net/?f=%5Cfrac%7Bh_i%7D%7Bh_o%7D%20%3D%20-%5Cfrac%7Bd_i%7D%7Bd_o%7D%5C%5C%5C%5C%20h_i%20%3D%20h_o%28-%5Cfrac%7Bd_i%7D%7Bd_o%7D%29%5C%5C%5C%5Ch_i%20%3D%20-9%28%5Cfrac%7B5%7D%7B15%7D%20%29%5C%5C%5C%5Ch_i%20%3D%20-3%20%5C%20cm)
Therefore, the image height is 3.0 cm. The negative values for image height indicate that the image is an inverted image.