Answer:
W=315 x 10⁵ J
Explanation:
Given that
F= 2.5 x 10⁵ N
d= 90 m
K.E.=5.4 x 10⁷ J
We know that work done by all force is equal to the change in kinetic energy
Lets take work done by catapult is W
W + F.d= K.E.
W= 5.4 x 10⁷ - 2.5 x 10⁵ x 90 J
W= (540 - 25 x 9) 10⁵ J
W=315 x 10⁵ J
Ionic compounds is your answer. What happens is one atom donates electron(s) to the other atom, making one positive and the other negative. The opposite atoms attract, forming an ionic bond.
Hope this helps! :)
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
Answer:
wrong question
Explanation:
unit of acceleration is written in m/s which is wrong
