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Gwar [14]
3 years ago
13

15. A body moving with a velocity of 20 m/s begins to accelerate at 3 m/s2. How far does the body move in 5 seconds? A. 137.5 m

B. 196 m C. 260 m D. 162.5 m
Physics
2 answers:
Rudik [331]3 years ago
7 0

Answer is B. According to the equation of motion s = vt + 1/2 at2 Where s is distance covered, v is velocity, a is acceleration and t is time taken. So, by putting all the values, we get s = (20)(5) + 1/2 (3)(5)2 s = 100 + 1/2 (3)(25) s = 100 + 1/2 75 s = 100 + 37.5 s = 137.5 meters



adelina 88 [10]3 years ago
5 0

The Correct answer to this question for Penn Foster Students is: 137.5 m

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Carbon is allowed to diffuse through a steel plate 11 mm thick. The concentrations of carbon at the two faces are 0.88 and 0.41
Serhud [2]

Answer:

The temperature is 2584.5 K

Explanation:

Given:

Activation energy Q = 80000 \frac{J}{mol}

Preexponential D= 6.2 \times 10^{-7} \frac{m^{2} }{s}

Diffusion flux J = 6.4 \times 10^{-10} \frac{kg}{m^{2} s}

Thickness of plate \Delta x =  11 \times 10^{-3} m

Concentration of carbon at two faces \Delta C = (0.88 - 0.41 ) = 0.47 \frac{kg}{m^{3} }

From the formula of temperature in terms of diffusion flux,

  T = (\frac{Q}{R} ) \frac{1}{\ln (\frac{D\Delta C}{J\Delta x} )}

Where R = 8.314 \frac{J}{mol.K} ( gas constant )

Put the values and find the temperature,

  T = (\frac{80000}{8.314} ) \frac{1}{\ln (\frac{6.2 \times 10^{-7} \times 0.47 }{6.4 \times 10^{-10}\times 11 \times 10^{-3} } )}

  T = 2584.5 K

Therefore, the temperature is 2584.5 K

8 0
3 years ago
How is a controlled variable different from a responding variable?
Arisa [49]

Answer:

The answer is a controlled variable stays the same throughout an experiment, but a responding variable changes

Explanation:

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5 0
3 years ago
True or false, only a radioactive isotope will have a half-life
maks197457 [2]
False not radioactive isotope will have a half-life
5 0
3 years ago
A stone is dropped from the upper observation deck of a tower, "500" m above the ground. (Assume g = 9.8 m/s2.) (a) Find the dis
Yuri [45]

Answer:

h₍₁₎ = 495,1 meters

h₍₂₎ = 480,4 m

h₍₃₎ = 455,9 m

...

..

Explanation:

The exercise is "free fall". t = \sqrt{\frac{2h}{g} }

Solving with this formula you find the time it takes for the stone to reach the ground (T) = 102,04 s

The heights (h) according to his time (t) are found according to the formula:

h(t) = 500 - 1/2 * g * t²

Remplacing "t" with the desired time.

4 0
3 years ago
An electron passes location < 0.02, 0.04, -0.06 > m and 5 us later is detected at location < 0.02, 1.62,-0.79 > m (1
hram777 [196]

Answer:

(a)

Average velocity = < 0, 316000, 146000> m/s

(b) < 0, 2.844, 1.314 > m

Explanation:

r1 = < 0.02, 0.04, - 0.06 > m

r2 = < 0.02, 1.62, - 0.79 > m

time, t = 5 micro second = 5 x 10^-6 s

(a) Average velocity is defined as the ratio of total displacement to the total time taken.

Displacement = r = r2 - r1

r = < 0.02 - 0.02, 1.62 - 0.04, - 0.79 + 0.06 > m

r = < 0, 1.58, - 0.73 > m

So. Average velocity =  \frac{< 0, 1.58, - 0.73 >}{5 \times 10^{-6}}

Average velocity = < 0, 316000, 146000> m/s

Average velocity = < 0, 316000, 146000> m/s

(b)

Distance = velocity x time

Here time, t = 9 micro second

d =  < 0, 316000, 146000>  x 9 x 10^-6 m

d = < 0, 2.844, 1.314 > m

5 0
3 years ago
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