Answer:
Explanation:
Given that,
The mass of a bag of acorns, m = 8.5 kg
The force acting on the bag = 80 N
The force of friction = 32 N
We need to find the acceleration of the bag of acorns.
Net force acting on it is = 80 N - 32 N
= 48 N
Let a be the acceleration of the bag of acorns. So,
F = ma
So, the acceleration of the bag of acorns is .
<span>The difficulty is to find the acceleration with which the bucket will fall. It will not fall with "g" because of the tension in the rope.
</span><span>On pole torque ,τ = Tr = Iα where I is moment of inertia of pole =Mr^2/2 and r is the radius of the pole
α = Tr/I ..........................................(1)
Applying newtons law on bucket,
ma = mg-T a=g-T/m .................................................(2)
If the rope does not slip
a=αr
g-T/m = Tr^2/I
Putting I = Mr^2/2 ,we get
g-T/m = 2T/M T = g/(1/m+2/M)
Putting this value of T in eqn (2), we will get the value of "a" the downward accelration of bucket.
(Please find a by putting the values of m and M yourself).
Now that we know the value of "a", find time taken to travel 20m is very easy.
Apply the equation of motion s=ut+1/2at^2
u=0
s=20m
</span>
I hope my answer has come to your help. Have a nice day ahead and may God bless you always!
Answer:
the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m
Explanation:
Given the data in the question;
Kinetic energy of each proton that makes up the beam = 3.25 × 10⁻¹⁵ J
Mass of proton = 1.673 × 10⁻²⁷ kg
Charge of proton = 1.602 × 10⁻¹⁹ C
distance d = 2 m
we know that
Kinetic Energy = Charge of proton × Potential difference ΔV
so
Potential difference ΔV = Kinetic Energy / Charge of proton
we substitute
Potential difference ΔV = ( 3.25 × 10⁻¹⁵ ) / ( 1.602 × 10⁻¹⁹ )
Potential difference ΔV = 20287.14 V
Now, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m will be;
E = Potential difference ΔV / distance d
we substitute
E = 20287.14 V / 2 m
E = 10143.57 V/m or 1.01 × 10⁴ V/m
Therefore, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m
That would be t<span>he law of conservation of energy.</span>