Answer:
When the object is placed between centre of curvature and principal focus of a concave mirror the image formed is beyond C as shown in the figure and it is real, inverted and magnified.
Answer:
The gravitational force on the moon is less than on Earth because the strength of gravity is determined by an object's mass. The bigger the object, the bigger the gravitational force. Gravity is pretty much everywhere. We just feel it in different ways depending on our state of motion.
Explanation:
Hope this helped!!
For Mass
K.E = (1/2*mv^2)
Explanation:
Kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. For example, if a an object with a mass of 10 kg (m = 10 kg) is moving at a velocity of 5 meters per second (v = 5 m/s), the kinetic energy is equal to 125 Joules, or (1/2 * 10 kg) * 5 m/s2.
Answer:
![t=12.25\ seconds](https://tex.z-dn.net/?f=t%3D12.25%5C%20seconds)
Explanation:
<u>Diagonal Launch
</u>
It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.
The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is
![x=v_ocos\theta t](https://tex.z-dn.net/?f=x%3Dv_ocos%5Ctheta%20t)
![\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3Dy_o%2Bv_osin%5Ctheta%20%5C%20t-%5Cfrac%7Bgt%5E2%7D%7B2%7D)
Where vo is the magnitude of the initial velocity,
is the angle, t is the time and g is the acceleration of gravity
The maximum height the object can reach can be computed as
![\displaystyle t=\frac{v_osin\theta}{g}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20t%3D%5Cfrac%7Bv_osin%5Ctheta%7D%7Bg%7D)
There are two times where the value of y is
when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making ![y=y_o](https://tex.z-dn.net/?f=y%3Dy_o)
![\displaystyle y_o=y_o+v_osin\theta\ t-\frac{gt^2}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y_o%3Dy_o%2Bv_osin%5Ctheta%5C%20t-%5Cfrac%7Bgt%5E2%7D%7B2%7D)
Removing
and dividing by t (t different of zero)
![\displaystyle 0=v_osin\theta-\frac{gt}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%200%3Dv_osin%5Ctheta-%5Cfrac%7Bgt%7D%7B2%7D)
Then we find the total flight as
![\displaystyle t=\frac{2v_osin\theta}{g}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20t%3D%5Cfrac%7B2v_osin%5Ctheta%7D%7Bg%7D)
We can easily note the total time (hang time) is twice the maximum (apex) time, so the required time is
![\boxed{t=24.5/2=12.25\ seconds}](https://tex.z-dn.net/?f=%5Cboxed%7Bt%3D24.5%2F2%3D12.25%5C%20seconds%7D)
Answer:
The bulb B glows brighter.
Explanation:
Given that,
A glows brightly and B glows dimly.
According to ohm's law,
Two light bulbs A and B are connected in series to a battery then the current will be same in both bulbs and the resistance is high of bulb A and low in bulb B.
If bulb A connect to a battery and bulb B connect to a same battery separately.
Then bulb B glows brighter because the resistance is high in bulb A so the current will be low.
The resistance is low in bulb B so the current will be high.
Hence, The bulb B glows brighter.