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This cannot be answered. We don't have weight, height, etc.
Answer:
<em>500Joules</em>
Explanation:
Kinetic energy = 1/2mv²
m is the mass of the wood
v is the velocity
Given
Mass = 10kg
Velocity v = 10m/s
Substitute into the formula and get KE
KE = 1/2 * 10 * 10²
KE = 1/2 * 1000
KE = 500Joules
<em>Hence the kinetic energy of the wood during delivery is 500Joules</em>
The direction of a vector multiplied by a scalar is only affected if the scalar is negative, in which case the vector will now be in the opposite direction. If the scalar is positive, the vector will only change in magnitude
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:
Magnitude of force the ground exerts on the plow = 263.234 N
Explanation:
Magnitude of force the ground exerts on the plow = Fground - Fplow
We are given that:
Fgound = 275 N
We will now calculate Fplow as follows:
Fplow = mass of horse * acceleration of plow
Fplow = 53 * 0.222
Fplow = 11.766 N
Now, substitute in the above equation to get magnitude of force the ground exerts on the plow as follows:
Magnitude of force the ground exerts on the plow = Fground - Fplow
Magnitude of force the ground exerts on the plow = 275 - 11.766
Magnitude of force the ground exerts on the plow = 263.234 N
Hope this helps :)
