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Sergio [31]
2 years ago
8

A small air plane's takeoff speed is 30 m/s. With constant acceleration it takes 150 m of runway to become airbourne. How long d

oes it take to become airbourne?
Physics
2 answers:
UkoKoshka [18]2 years ago
8 0
T=vf-vi/a
t=30-0/150m
t=0.2s
Butoxors [25]2 years ago
8 0
The plane moves at a constant rate of 30m/s

there are 150m in the runway

150/30 = 5

5 seconds is the time it takes for it to be airborne

hope this helps
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Two point charges are 10.0cm apart and have charges of 2.0uC and -2.0uC, respectively. What is the magnitude of the electric fie
elena-14-01-66 [18.8K]
The electric field generated by a point charge is given by:
E= k_e \frac{Q}{r^2}
where
k_e = 8.99 \cdot 10^9 Nm^2 C^{-2} is the Coulomb's constant
Q is the charge
r is the distance from the charge

We want to know the net electric field at the midpoint between the two charges, so at a distance of r=5.0 cm=0.05 m from each of them. 

Let's calculate first the electric field generated by the positive charge at that point:
E_1=k_e  \frac{Q_1}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(2.0 \cdot 10^{-6} C)}{(0.05 m)^2} =+7.19 \cdot 10^6 N/C
where the positive sign means its direction is away from the charge.

while the electric field generated by the negative charge is:
E_2=k_e \frac{Q_1}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(-2.0 \cdot 10^{-6} C)}{(0.05 m)^2} =-7.19 \cdot 10^6 N/C
where the negative sign means its direction is toward the charge.

If we assume that the positive charge is on the left and the negative charge is on the right, we see that E1 is directed to the right, and E2 is directed to the right as well. This means that the net electric field at the midpoint between the two charges is just the sum of the two fields:
E_{tot} =E_1 + E_2 = 7.19 \cdot 10^6 N/C+7.19 \cdot 10^6 N/C=1.44 \cdot 10^7 N/C
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3 years ago
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Explanation:

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Imagine that you are working as a roller coaster designer. You want to build a record breaking coaster that goes 70.0 m/s at the
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Wow !  This is not simple.  At first, it looks like there's not enough information, because we don't know the mass of the cars.  But I"m pretty sure it turns out that we don't need to know it.

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                                  PE = (mass) x (gravity) x (height) .

At the bottom, the car's kinetic energy is

                                 KE = (1/2) (mass) (speed²) .

You said that the car's speed is 70 m/s at the bottom of the hill,
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                                   KE = 0.9  PE

        (1/2) (mass) (70 m/s)² = (0.9) (mass) (gravity) (height)     

Divide each side by (mass): 

               (0.5) (4900 m²/s²) = (0.9) (9.8 m/s²) (height)

(There goes the mass.  As long as the whole thing is 90% efficient,
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Divide each side by (0.9):

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Divide each side by (9.8 m/s²):

               Height = (5/9)(4900 m²/s²) / (9.8 m/s²)

                          =  (5 x 4900 m²/s²) / (9 x 9.8 m/s²)

                          =  (24,500 / 88.2)  (m²/s²) / (m/s²)

                          =        277-7/9    meters
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nalin [4]
<span>The inner core is liquid and moving.</span>
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Answer:

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