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Romashka [77]
3 years ago
7

A suspicious-looking man runs as fast as he can along a moving sidewalk from one end to the other, taking 2.20 s. Then security

agents appear and the man runs as fast as he can back along the sidewalk to his starting point, taking 10.3 s. What is the ratio of the man's running speed to the sidewalk's speed (running speed / sidewalk speed)?
Physics
1 answer:
Brilliant_brown [7]3 years ago
7 0

Answer: The ratio is 1.54

Explanation:

Firs, we have to find the relative speed of the man moving forward and backward.

Forward:

vf = vs + vm

vm = the man's speed

vs = the sidewalk's speed

vf = relative velocity moving forward

Because we don't know how much the man moved,

vf = distance (meters) / time (seconds)

vf = x / 2.20s

Backward:

vb = -vm + vs

vb = relative velocity moving backward

vf = distance (meters) / time (seconds)

vf = -x / 10.30s

We now divide the relative speeds

vf / vb = (x / 2.20) / (-x / 10.30)

We cancel the x

vf / vb = -10.3s / 2.2s = -4.68

vf = -4.68 . vb

We now substitute this in the equation we used for the forward travel

-4.68vb = vm + vs

Subtracting this from the backward travel equation

vb - (-4.68vb) = -vm - vm + vs -vs

5.68vb = -2vm

vb = -2vm / 5.68

Now, adding to the backward travel equation

vb + (-4.68vb) = -vm + vm + vs + vs

-3.68vb = 2vs

Using the two resulting equations

-3.68 . (-2 / 5.68) vm = 2vs

7.36 / 5.68 vm = 2vs

vm / vs = 1.54

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Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.
padilas [110]

Explanation:

We have,

Semimajor axis is 4\times 10^{12}\ m

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

T^2=\dfrac{4\pi ^2}{GM}a^3

G is universal gravitational constant

M is solar mass

Plugging all the values,

T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s

Since,

1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}

So, the orbital period of a dwarf planet is 138.52 years.

3 0
3 years ago
The density of a block of wood is 694 kg/m3. Its mass is 689 g. We tie the block to the bottom of a swimming pool using a single
Serhud [2]

Answer:

<em>The tension in the string = 2.065 N</em>

Explanation:

From Archimedes principle,

R.d = density of the wood block/density of water = weight of the wood block/Upthrust of the wood block in water.

R.d = D₁/D₂ = W/U

W/U =D₁/D₂.................................. Equation 1

Where W = weight of the wood block, U = upthrust of the wood block in water, D₁ = Density of the wood block, D₂ = Density of water.

Making U the subject of the equation,

U = WD₂/D₁........................... Equation 2

Given: W =  mg = (689/1000)9.8 = 6.75 N,  D₁ = 694 kg/m³, D₂ = 1000 kg/m³.

Substituting these values into equation 2,

U = 6.75(694)/1000

U = 4684.5/1000

U = 4.685 N.

Note: Three forces act on the wood block in the pool. and they are

(i) The weight(W) acting downs

(ii) The upthrust (U) acting upwards,

(iii) The Tension (T) in the string, acting upwards.

Thus,

W = U+T

T = W - U ................................. Equation 3

Where W = 6.75 N, U = 4.685 N

T = 6.75 - 4.685

T = 2.065 N.

T = 2.065 N

<em>Thus the tension in the string = 2.065 N</em>

7 0
3 years ago
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Gala2k [10]

Answer:

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7 0
3 years ago
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Anna11 [10]

Answer:

4.8 mph

Explanation:

From the question,

Average speed = total distance/total time

V = d/t....................... Equation 1

Where d = distance, t = time

Given: d = 26.2 miles, t = 5.5 hours.

Substitute these values into equation 1

V = 26.2/5.5

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miv72 [106K]

Answer:c

Explanation:

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