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salantis [7]
3 years ago
14

A jet with mass m = 5 × 104 kg jet accelerates down the runway for takeoff at 1.9 m/s2. What is the net vertical force on the ai

rplane as it levels off?
Physics
1 answer:
OLEGan [10]3 years ago
3 0
The first sentence got me all psyched up to answer the question "What
horizontal force do the engines generate in order to accelerate it ?". 

But the actual question, in the second sentence, turned out to be
a completely different one.

When the plane levels off and continues on at a constant altitude, it's
not accelerating up or down, so the net vertical force on it is zero. 
The lift generated by the wings is exactly balancing the downward
force of gravity on the airplane.
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A 16 g piece of Styrofoam carries a net charge of -8.6 µC and floats above the center of a large horizontal sheet of plastic tha
PilotLPTM [1.2K]

Answer:

the charge per unit area on the plastic sheet is - 3.23 x 10⁻⁷ C/m²

Explanation:

given information:

styrofoam mass, m = 16 g = 0.016 kg

net charge, q = - 8.6 μC

to calculate the charge per unit area on the plastic sheet, we can use the following equation:

F_{e} = mg

where

F_{e} = the force between the electric field

m = mass

g = gravitational force

F_{e} =qE

where

q = charge

E = electric field

and

E = σ/2ε₀

where

ε₀ = permitivity

thus

F_{e} =qE

mg = qσ/2ε₀

σ = (2mg ε₀)/q

  = 2 (0.016) (9.8)  (8.85 x 10⁻¹²)/( - 8.6 x 10⁻⁶)

  = - 3.23 x 10⁻⁷ C/m²

8 0
3 years ago
When connected to a battery, a lightbulb glows brightly. If the battery is reversed and reconnected to the bulb, the bulb will g
BartSMP [9]
If the battery is reversed and reconnected to the bulb, the bulb will glow <span>with the same brightness. The correct option among all the options that are given in the question is the third option or the penultimate option. I hope that this is the answer that you were looking for and it has actually come to your help.</span>
5 0
3 years ago
The first game in volleyball is played until a team reaches 25 points.
Zinaida [17]
I don't understand this question
7 1
3 years ago
Read 2 more answers
Plants grow over time by balancing two chemical reactions: photosynthesis and cellular respiration. Plants use the products of p
Andreyy89
The answer to that I think would be D because:

A is wrong because it doesn't make sense the way the order it is in,

B doesn't make sense because it is in the wrong order,  and

C says the same exact thing that photosynthesis says which would mean that it is representing the photosynthesis equation not the chemical equation for cellular respiration.

Also it is D because D shows the correct way that the chemical equation representing cellular respiration is supposed to be in.

Hope this helped!

8 0
3 years ago
1. Calculate the total binding energy of 12
ahrayia [7]

Answer:

1. B = 79.12 MeV

2. B = -4.39 MeV/nucleon

3. B = 2.40 MeV/nucleon

4. B = 7.48 MeV/nucleon

5. B = -18.72 MeV

6. B = 225.23 MeV            

Explanation:

The binding energy can be calculated using the followng equation:

B = (Zm_{p} + Nm_{n} - M)*931 MeV/C^{2}

<u>Where</u>:

Z: is the number of protons

m_{p}: is the proton's mass = 1.00730 u

N: is the number of neutrons

m_{n}: is the neutron's mass = 1.00869 u

M: is the mass of the nucleus

1. The total binding energy of ^{12}_{6}C is:

B = (Zm_{p} + Nm_{n} - M)*931.49 MeV/u

B = (6*1.00730 + 6*1.00869 - 12.011)*931.49 MeV/u = 79.12 MeV

 

2. The average binding energy per nucleon of ^{24}_{12}Mg is:

B = \frac{(Zm_{p} + Nm_{n} - M)}{A}*931.49 MeV/u

<u>Where</u>: A = Z + N

B = \frac{(12*1.00730 + 12*1.00869 - 24.305)}{(12 + 12)}*931.49 MeV/u = -4.39 MeV/nucleon                                  

   

3. The average binding energy per nucleon of ^{85}_{37}Rb is:

B = \frac{(Zm_{p} + Nm_{n} - M)}{A}*931.49 MeV/u

B = \frac{(37*1.00730 + 48*1.00869 - 85.468)}{85}*931.49 MeV/u = 2.40 MeV/nucleon

     

4. The binding energy per nucleon of ^{238}_{92}U is:

B = \frac{(92*1.00730 + 146*1.00869 - 238.03)}{238}*931.49 MeV/u = 7.48 MeV/nucleon

 

5. The total binding energy of ^{20}_{10}Ne is:

B = (Zm_{p} + Nm_{n} - M)*931.49 MeV/u

B = (10*1.00730 + 10*1.00869 - 20.180)*931.49 MeV/u = -18.72 MeV

6. The total binding energy of ^{40}_{20}Ca is:

B = (Zm_{p} + Nm_{n} - M)*931.49 MeV/u

B = (20*1.00730 + 20*1.00869 - 40.078)*931.49 MeV/u = 225.23 MeV

I hope it helps you!

8 0
3 years ago
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