Answer: The coefficient of static friction is 3.85 and The coefficient of kinetic friction is 2.8
Explanation:
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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
A) No, the equations presented above are the product of the derivation of position and velocity when the acceleration is constant.
The equations change to polynomial function of the second degree for the description of the acceleration when described as a function of time.
B) Yes, when the acceleration is zero it is concluded that the velocity is constant, therefore they could be used to describe the position as a function of the change in velocity.
Answer:
Explanation:
According to “Newton's second law”
“Force” is “mass” times “acceleration”, or F = m× a. This means an object with a larger mass needs a stronger force to be moved along at the same acceleration as an object with a small mass
Force = mass × acceleration
Given that,
Mass = 5.32 kg
F = 12.7N
Normal force = mg + F sinx,
“m” being the object's "mass",
“g” being the "acceleration of gravity",
“x” being the "angle of the cart"
To find normal force substitute the values in the formula,
Normal force = 5.32 × 9.8 + 12.7 × sin(-28.7)
Normal force = 52.136 + 12.7 × 0.480
Normal force = 52.136 + 6.096
Normal force = 58.232 N
<u>Acceleration of the cart</u>:
