Sally wins the race by 0.03 min
Explanation:
To find time = distance / velocity
Sally;
1 mile / 12 mph
= ¹/12 hrs * 60
= 5 min
Ramona;
³/4 mile ÷ 11 mph + ¹/4 mile ÷ 16 mph
= ³/44 hrs + ¹/64 hrs
= 4.09 min + 0.94 min
= 5.03 min
The power is 
Explanation:
First of all, we need to find the acceleration of the car, which is given by
where
v = 60 mph = 26.8 m/s is the final velocity
u = 0 is the initial velocity
t = 10.0 s is the time
Substituting,
Now we can find the mass of the car by using Newton's second law:
where
F = 5300 N is the force applied
m is the mass
is the acceleration
Solving for m,
Now we can use the work-energy theorem, which states that the work done is equal to the change in kinetic energy of the car, to find the work:
And substituting,
Finally, we can find the power output of the car:
where
W is the work
t = 10.0 s is the time elapsed
Substituting,
Learn more about power:
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Aerospace engineering is responsible for a plane design
The biggest thing you're doing wrong is ignoring the units
when you're working with the quantities.
Now let's look at the rest of the problem:
The formula you used is correct:
Net flux through the surface = (net charge inside) / ε₀
and ε₀ = 8.85 x 10⁻¹² farad/meter.
What's the net charge inside the surface in this problem ?
It's (5.85 x 10⁷ electrons) x (the charge on each electron)
= (5.85 x 10⁷ electrons) x (-1.6 x 10⁻¹⁹ coulomb/electron)
= -9.36 x 10⁻¹² coulomb .
Finally, (net charge inside) / ε₀
= (-9.36 x 10⁻¹² coulomb) / (8.85 x 10⁻¹² farad/meter)
= -1.058 newton-m²/coulomb .
The sign and the significant figures in your answer are correct, so
we can see that you know what you're doing. The only thing left is
the order of magnitude. You most likely took one of the negative
exponents and made it positive. You got an answer that's 10²² too
small. Big deal. You could claim "that's close", and see whether you
can convince a teacher.
Answer:
Part (a) The magnitude of the deflection of electron beam on the screen due to the Earth's gravitational field is 5.97*
m.
Part (b) The magnitude of the deflection of electron beam on the screen due to the vertical component of the Earth's magnetic field is 
m
