Answer:
A Fan Cart Initially Has An Acceleration Of 1.6m/s/s When It's Fan Is Directed Straight Backwards. If You Rotate The Fan By 45o, By What Percentage Do You Expect The Fan Cart's Thrust To Decrease? (Answer Should Be In Units Of 96)
a. 45%
b. 29%
c. 71%
d. 50%
The correct answer is d.
d. 50%
Explanation:
Fan cart acceleration = 1.6 m/s²
Thrust = 0.25×π×D²×ρ×v×Δv
where Δv = acceleration component and all factors remaining cconstant, when the fan is rotated by 45 ° the diameter changes to D₂ = sin 45 ×D
or 0.707×D. The thrust becomes 0.25×π×(0.707×D)²×ρ×v×Δv
=0.25×π×0.5×D²×ρ×v×Δv or 0.5(0.25×π×D²×ρ×v×Δv)
That is the thrust reduces by 50 %
The 48 and 47 are different atomic masses, this is caused by having a different number of neutrons.
The answer is 10,560 Joules or 1.1*10^4
Explanation:
Step 1: Calculate
The equation for Kinetic Energy is
Kinetic energy=.5 times Mass times Velocity²
KE=.5*m*v²
so we plug in our numbers
KE=.5*600*35.2²
This works out to be 10,560 Joules or 1.1*10^4
The strength of the electric field on the point charge at this distance will be 4000 V/m.
<h3>What is the strength of the electric field?</h3>
The strength of the electric field is the ratio of electric force per unit charge.
The given data in the problem is;
Qis the unit charge = 4.0 × 10⁻⁶ C
E is the strength of the electric field
R is the distance from point charge = 3 m
The strength of the electric field is;
Hence, the strength of the electric field on the point charge at this distance will be 4000 V/m.
To learn more about the strength of the electric field refer to the link;
brainly.com/question/15170044
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