Ask question

Ask question

All categories

2 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)?

1 answer:

Brilliant_brown [7]2 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

You might be interested in

A rod 150 cm long and of diameter 2.0 cm is subjected to an axial pull of 20 kn. if the modulus of elasticity of the material of

Scilla [17]

Given:
rod of circular cross section is subjected to uniaxial tension.
Length, L=1500 mm
radius, r = 10 mm
E=2*10^5 N/mm^2
Force, F=20 kN = 20,000 N
[note: newton (unit) in abbreviation is written in upper case, as in N ]

From given above, area of cross section = π r^2 = 100 π =314 mm^2
(i) Stress,
σ
=force/area
= 20000 N / 314 mm^2
= 6366.2 N/mm^2
= 6370 N/mm^2 (to 3 significant figures)

(ii) Strain
ε
= ratio of extension / original length
= σ / E
= 6366.2 /(2*10^5)
= 0.03183
= 0.0318 (to three significant figures)

(iii) elongation
= ε * L
= 0.03183*1500 mm
= 47.746 mm
= 47.7 mm (to three significant figures)

5 0

2 years ago

How many significant figures are in 0.0069

posledela

Two significant figures, the 6 and the 9

7 0

2 years ago

An astronaut weighs 8.00 × 102 newtons on the sur- face of Earth. What is the weight of the astronaut 6.37 × 106 meters above th

kolbaska11 [484]

Answer:

$mg=200.4 N$ .

Explanation:

This problem can be solved using Newton's law of universal gravitation: $F=G\frac{m_{1}m_{2}}{r^{2}}$ ,

where F is the gravitational force between two masses $m_{1}$ and $m_{2}$ , $r$ is the distance between the masses (their center of mass), and $G=6.674*10^{-11}(m^{3}kg^{-1}s^{-2})$ is the gravitational constant.

We know the weight of the astronout on the surface, with this we can find his mass. Letting $w_{s}$ be the weight on the surface:

$w_{s}=mg$ ,

$mg=8*10^{2}$ ,

$m=(8*10^{2})/g$ ,

since we now that $g=9.8m/s^{2}$ we get that the mass is

$m=81.6kg$ .

Now we can use Newton's law of universal gravitation

$F=G\frac{Mm}{r^{2}}$ ,

where $m$ is the mass of the astronaut and $M$ is the mass of the earth. From Newton's second law we know that

$F=ma$ ,

in this case the acceleration is the gravity so

$F=mg$ , (<u>becarefull, gravity at this point is no longer</u> $9.8m/s^{2}$ <u>because we are not in the surface anymore</u>)

and this get us to

$mg=G\frac{Mm}{r^{2}}$ , where $mg$ is his new weight.

We need to remember that the mass of the earth is $M=5.972*10^{24}kg$ and its radius is $6.37*10^{6}m$ .

The total distance between the astronaut and the earth is

$r=(6.37*10^{6}+6.37*10^{6})=2(6.37*10^{6})=12.74*10^{6}$ meters.

Now we can compute his weigh:

$mg=G\frac{Mm}{r^{2}}$ ,

$mg=(6.674*10^{-11})\frac{(5.972*10^{24})(81.6)}{(12.74*10^{6})^{2}}$ ,

$mg=200.4 N$ .

5 0

2 years ago

A spring has a spring constant of 48 N/m. The end of the spring hangs 8 m above the ground. How much weight can be placed on the

Setler79 [48]

The answer is 96 N .....................................

7 0

2 years ago

Read 2 more answers

Which changes in an electric motor will make the motor stronger? Check all that apply.

ollegr [7]

To make the motor turn faster we can:
(a) increase the current
(b) use stronger magnets
(c) push the magnets closer to the coil
(d) put an iron centre piece into the coil
(e) adding more sets of coils

8 0

2 years ago