Kinetic Energy = (1/2) (mass) (speed)
First runner: KE = (1/2) (45kg) (49 m/s) = 1,102.5 Joules
Second runner: KE = (1/2) (93kg) (9 m/s) = 418.5 Joules
The <em>first runner </em><em>has 163</em>% more kinetic energy than the second runner has.
Answer:
Explanation:
Momentum is the product of mass and velocity.
The mass of the truck is 2,000 kilograms and the velocity is 35 meters per second.
Substitute the values into the formula and multiply.
The truck's momentum is <u>70,000 kilograms meters per second.</u>
Answer:
<em>A</em><em>.</em><em>increases</em><em> </em><em>friction</em>
Explanation:
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