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lakkis [162]
3 years ago
11

A plane is flying horizontally with speed 292 m/s at a height 3880 m above the ground, when a package is dropped from the plane.

the acceleration of gravity is 9.8 m/s 2 . neglecting air resistance, when the package hits the ground, the plane will be 1. ahead of the package. 2. directly above the package. 3. behind the package. 008 (part 2 of 4) 10.0 points what is the horizontal distance from the release point to the impact point?
Physics
1 answer:
kap26 [50]3 years ago
4 0
The plane is straight above the package when it hits the ground. 
Find in what way long a free fall from a height of 4240 m takes: s = 1/2 gt^2 
t^2 = 2s/g 
t^2 = 2* 3880/9.8 = 791.8367
t = √791.8367
t = 28.1396 seconds 
Now find the horizontal displacement with v = 292 m/s in 28.1396 s: 
s = v*t 
s = 292 * 28.1396 = 8216.76 m
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a sample contains 100 g of radioactive isotope. How much radioactive isotope will remain in the sample after 1 half-life?
kap26 [50]

Answer:

\huge\boxed{50g}

Definition:

Half-life- The time taken for half of the radioactive isotopes to decay.

Explanation:

How does radioactive decay work? Radioactive decay is a process by which unstable nuclei become more stable through the emission of alpha or beta particles or gamma rays.

Since a half-life is the time taken for half of the isotopes to decay, we can simply divide the initial mass of 100 grams by 2; this gives us 50 grams.

1) Divide 100g by 2.

\frac{100g}{2}=50g

8 0
2 years ago
The relationship between the decay constant (λ) and the half-life (t1⁄2) is given by the equation t(1/2) = 0.693/λ. Use the equa
pshichka [43]

Answer: A: 70/2=35

B: 35/2=17

C: 17.5/2=8.75

D: 8.75 of C-14 will be left

E: 5,730 years

F: 5,631

Explanation:

that’s all I got, hope I helped kinda

6 0
2 years ago
You want the current amplitude through a 0.450-mH inductor (part of the circuitry for a radio receiver) to be 1.90 mA when a sin
faust18 [17]

Answer:

Frequency required will be 2421.127 kHz

Explanation:

We have given inductance L=0.450H=0.45\times 10^{-3}H

Current in the inductor i=1.90mA=1.90\times 10^{-3}A

Voltage v = 13 volt

Inductive reactance of the circuit X_l=\frac{v}{i}

X_l=\frac{13}{1.9\times 10^{-3}}=6842.10ohm

We know that

X_l=\omega L=2\pi fL

2\times 3.14\times  f\times 0.45\times 10^{-3}=6842.10

f = 2421.127 kHz

6 0
3 years ago
A large power plant generates electricity at 12.0 kV. Its old transformer once converted the voltage to 385 kV. The secondary of
enot [183]

Answer:

a) In the new transformer there are 42 turns in the secondary per turn in the primary, while in the old transformer there were 32 turns per turn in the primary.

b) The new output is 86% of the old output

c) The losses in the new line are 74% the losses in the old line.

Explanation:

a) To relate the turns of primary and secondary to the ratio of voltage we have this expression:

\frac{n_1}{n_2}=\frac{V_1}{V_2}

In the old transformer the ratio of voltages was:

\frac{n_1}{n_2}=\frac{V_1}{V_2}=\frac{12}{385} =0.03117\\\\n_2=n_1/0.03117=32.1n_1

In the new transformer the ratio of voltages is:

\frac{n_1}{n_2}=\frac{V_1}{V_2}=\frac{12}{500} =0.024\\\\n_2=n_1/0.24=41.7n_1

In the new transformer there are 42 turns in the secondary per turn in the primary, while in the old transformer there were 32 turns per turn in the primary.

b) The new current ratio is

\frac{V_1}{V_2}=\frac{I_2}{I_1}=\frac{12}{500}= 0.024\\\\I_2=0.024I_1

If the old current output was 425 kV, the ratio of current was:

\frac{V_1}{V_2}=\frac{I_2}{I_1}=\frac{12}{425}= 0.028\\\\I_2=0.028I_1

Then, the ratio of the new output over the old output is:

\frac{I_{2new}}{I_{2old}} =\frac{0.024\cdot I_1}{0.028\cdot I_1}= 0.86

The new output is 86% of the old output (smaller output currents lower the losses on the transmission line).

c) The power loss is expressed as:

P_L=I^2\cdot R

Then, the ratio of losses is (R is constant for both power losses):

\frac{P_n}{P_o} =\frac{I_n^2R}{I_o^2R} =(\frac{I_n}{I_o} )^2=0.86^2=0.74

The losses in the new line are 74% the losses in the old line.

7 0
3 years ago
If an automobile battery is rated at 20 ampere-hours, what is the maximum current that can be supplied for 19 minutes?
Elenna [48]
20Ah is the number of charge that can be supplied at 20A for 1 hour. If you wish to drain it in 19 minutes, then the current is:
20*60 = A_{2}*19
A_{2}=63.2A


5 0
2 years ago
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