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andre [41]
4 years ago
11

A reconnaissance plane flies 556 km away from

Physics
1 answer:
9966 [12]4 years ago
5 0

We use the formula, to calculate the average speed of the round trip,

v_{av} =\frac{d_{total}}{t_{total}}

Here, d_{total}, is total distance covered by plane in total time, t_{total}.

For the round trip,

d_{total=556\ km+556\ km=1112 \times 10^3\ m

t_{total}=\frac{556 \times 10^3 m}{832 m/s} + \frac{556 \times 10^3 m}{1248 m/s}.

Thus,

v_{av} =\frac{1112 \times 10^3\ m}{\frac{556 \times 10^3 m}{832 m/s} + \frac{556 \times 10^3 m}{1248 m/s}} =\frac{1112 \times 10^3\ m}{668.26\ s+445.51\s} \\\\v_{av}=998.41\ m/s.


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Empty space; an atom consists of a dense, positively charged nucleus surrounded by a "cloud" of negatively charged electrons
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Edward runs with a speed of 3m/s. How long would it take him to run 2km? Give your answer in seconds to four significant figures
Whitepunk [10]

Answer:

666.6 seconds

Explanation:

if he runs at 3m/sec he will achieve the goal of 2000m in 666.6 seconds. just divide - 3/2000.

note we have changed 2km to 2000metres

5 0
3 years ago
How to find initial velocity without time?
77julia77 [94]
1. The problem statement, all variables and given/known data Knowing that snow is discharged at an angle of 40 degrees, determine the initial speed, v0 of the snow at A. Answer: 6.98 m/s 2. Relevant equations 3. The attempt at a solution I have found the x and y velocity and position formulas. Now since I don't know time, should I solve both position equations for time (t) and set them equal to each other to get my only unknown, vi? The quadratic equation for time in the y-dir seems a bit hectic. Is there an easier way to go about trying to find vi?

6 0
4 years ago
An engine absorbs 1.69 kJ from a hot reservoir at 277°C and expels 1.25 kJ to a cold reservoir at 27°C in each cycle.
Anna35 [415]

Answer

Given,

Energy absorbed, Q_H = 1.69\ kJ

Energy expels,Q_C =  1.25\ kJ

Temperature of cold reservoir, T = 27°C

a) Efficiency of engine

 \eta = \dfrac{Q_H - Q_C}{Q_H}\times 100

 \eta = \dfrac{1.69 - 1.25}{1.69}\times 100

\eta =26.03 %

b) Work done by the engine

 W = Q_H- Q_C

 W =1.69 - 1.25

 W = 0.44\ kJ

c) Power output

     t = 0.296 s

   P = \dfrac{W}{t}

   P = \dfrac{0.44}{0.296}

   P = 1.486\ kW

8 0
3 years ago
The largest and the smallest balls used in the experiment are with diameter 9.52 mm, and 2.38 mm respectively. For a glycerin wi
svlad2 [7]

Answer

given,

largest diameter of  balls = 9.52 mm = 0.00476 m

                                 radius = 0.00476

smallest diameter of ball = 2.38 mm = 0.00238 m

                                 radius = 0.00119

viscosity = 1.5 Pa.s

density of the ball = 1.42 g/cm

F = 6\pi \eta r V_t

F = \dfrac{mv}{t}

F = \dfrac{\dfrac{4}{3}\pi\ r^3\times (\rho-\sigma) \times 0.99 V_T}{t}

6\pi \eta r V_t = \dfrac{\dfrac{4}{3}\pi\ r^3\times rho \times 0.99 V_T}{t}

t = \dfrac{\dfrac{4}{3}\pi\ r^3\times rho \times 0.99 V_T}{6\pi \eta r V_t}

t= \dfrac{0.22 r^2 (\rho-\sigma) }{\eta}

for small balls

t= \dfrac{0.22\times 0.00119^2 (1460-1300)}{1.5}

t = 0.033 ms

for larger ball

t= \dfrac{0.22\times 0.00476^2 (1460-1300)}{1.5}

t = 0.531 ms

6 0
4 years ago
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