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Elenna [48]
3 years ago
13

Compute your average velocity in the following two cases: (a) You walk 50.2 m at a speed of 2.21 m/s and then run 50.2 m at a sp

eed of 4.11 m/s along a straight track. (b) You walk for 1.00 min at a speed of 2.21 m/s and then run for 1.16 min at 4.11 m/s along a straight track.
Physics
1 answer:
Readme [11.4K]3 years ago
5 0

Answer:

a) 2.87 m/s

b) 3.23 m/s

Explanation:

The avergare velocity can be found dividing the length traveled d by the total time t.

a)

For the first part we easily know the total traveled length which is:

d = 50.2 m + 50.2 m = 100.4 m

The time can be found dividing the distance by the velocity:

t1 = 50.2 m / 2.21 m/s = 22.7149 s

t2 = 50.2 m / 4.11 m/s = 12.2141 s

t = t1 +t2 = 34.9290 s

Therefore, the average velocity is:

v = d/t =2.87 m/s

b)

Here we can easily know the total time:

t = 1 min + 1.16 min = 129.6 s

Now the distance wil be found multiplying each velocity by the time it has travelled:

d1 = 2.21 m/s * 60 s = 132.6 m

d2 = 4.11 m/s *(1.16 * 60 s) = 286.056 m

d = 418.656 m

Therefore, the average velocity is:

v = d/t =3.23 m/s

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HACTEHA [7]
They do not demonstrate Earth's tilt. In fact, they're not "used" to demonstrate anything. It works the other way:. When you observe the Coriolis effect and the behavior of the Foucault pendulum, and you try to explain why the behave the way they do, one possible simple explanation for both of them is the Earth's ROTATION. Then, when you also observe the rising and setting of the sun and moon, and you also notice how the NUMBERS all go together, the case for the rotating, spherical Earth gets stronger and stronger.
7 0
3 years ago
If the mercury barometer is 75.58 cm at the base of the mountain and 66.37 cm at the summit. What is the height of the mountain?
Licemer1 [7]

Answer:

5016.2446

Explanation:

3 0
2 years ago
if an average cloud has a density of 0.5 g/m3 and has a volume of 1,000,000,000 m3,what is the weight of an average cloud
sveta [45]

If you have (1 x 10⁹) cubic meters of volume, and 1/2 gram of mass
in each cubic meter, then you have

          (1/2 x 10⁹) grams = (5 x 10⁸) grams = (5 x 10⁵) kilograms of mass . 

On Earth, that mass weighs  4,900,000 newtons .

                                            (about  1,102,300 pounds)

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3 0
3 years ago
A horizontal spring with stiffness 0.4 N/m has a relaxed length of 11 cm (0.11 m). A mass of 21 grams (0.021 kg) is attached and
riadik2000 [5.3K]

Answer:

0.6983 m/s

Explanation:

k = spring constant of the spring = 0.4 N/m

L₀ = Initial length = 11 cm = 0.11 m

L = Final length = 27 cm = 0.27 m

x = stretch in the spring = L - L₀ = 0.27 - 0.11 = 0.16 m

m = mass of the mass attached = 0.021 kg

v = speed of the mass

Using conservation of energy

Kinetic energy of mass = Spring potential energy

(0.5) m v² = (0.5) k x²

m v² = k x²

(0.021) v² = (0.4) (0.16)²

v = 0.6983 m/s

5 0
3 years ago
A real object is 10.0 cm to the left of a thin, diverging lens having a focal length of magnitude 16.0 cm. What is the location
amm1812

Answer:

A)6.15 cm to the left of the lens

Explanation:

We can solve the problem by using the lens equation:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}

where

q is the distance of the image from the lens

f is the focal length

p is the distance of the object from the lens

In this problem, we have

f=-16.0 cm (the focal length is negative for a diverging lens)

p=10.0 cm is the distance of the object from the lens

Solvign the equation for q, we find

\frac{1}{q}=\frac{1}{-16.0 cm}-\frac{1}{10.0 cm}=-0.163 cm^{-1}

q=\frac{1}{-0.163 cm^{-1}}=-6.15 cm

And the sign (negative) means the image is on the left of the lens, because it is a virtual image, so the correct answer is

A)6.15 cm to the left of the lens

6 0
3 years ago
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