Answer:
Light consists of photons, which are produced when an object's atoms heat up. Light travels in waves and is the only form of energy visible to the human eye.
Explanation:
Answer:
O The environment did work on an object
Explanation:
Answer:
Explanation:
When a projectile is launched at some angle to the horizontal with some speed vi , and air resistance is negligible , it is definitely a freely falling body .
It is so because it is free to accelerate towards the earth with acceleration of g . Air has no resistance , hence no force is acting on it except the gravitational force . Hence it is a freely falling body .
b )
The acceleration in the vertical direction is due to force exerted by the earth that is gravitational force on it . Hence its acceleration is equal to g in vertically downward direction .
c )
It has zero acceleration in horizontal direction . It is so because no force is acting on it in horizontal direction . So no acceleration will be present in horizontal direction . It will move in horizontal direction with constant speed of vi cos θ where θ is the angle vi make with the horizontal .
Answer:
The power of the bulb is reduced to a quarter
Explanation:
The power in an electrical circuit is
P = V I = I² R
If the dimmer cuts the current in half
I = I₀ / 2
The resistance of the light bulb after hot is little affected by small changes in the current
We substitute in the equation
P = I₂ R
P = (I₀ /2)² R
P = ¼ I₀² R
P = ¼ P₀
The power of the bulb is reduced to a quarter
Answer:
Tension= 21,900N
Components of Normal force
Fnx= 17900N
Fny= 22700N
FN= 28900N
Explanation:
Tension in the cable is calculated by:
Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium
FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)
Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)
Ftorque= 2/3FBcostheta+ 4/3FWcostheta
Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°
Ftorque= 21900N
b) components of Normal force
Efx=FNx-FTcos(90-theta)=0 static equilibrium
Fnx=21900cos(90-55)=17900N
Fy=FNy+ FTsin(90-theta)-FB-FW=0
FNy= -FTsin(90-55)+FB+FW
FNy= -21900sin(35)+(1350+2250)×9.81=22700N
The Normal force
FN=sqrt(17900^2+22700^2)
FN= 28.900N