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3 years ago

If you were to walk with a speed if 2m/s for 2 hours how far would you travel

Physics

1 answer:

olga nikolaevna [1]3 years ago

7 0

So simply it to 120m/m for 120 minutes. So then you multiply 120x120 and that equals 14,400

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A truck using a rope to tow a 2230-kg car accelerates from rest to 13.0 m/s in a time of 15.0s. How strong must the rope be? μk

Leokris [45]

Answer:

The rope must have a force of 10084,21 N

Explanation

Acceleration calculation

The car acceleration is equal to the acceleration of the truck

ac: car acceleration $\frac{m}{s^{2} }$

at: truck acceleration $\frac{m}{s^{2} }$ )

$ac = at= \frac{vf-vi}{t-ti}$ equation(1)

Known information:

vi = Initial speed = 0, ti = initial time = 0

vf = Final speed = 13 $\frac{m}{s}$ , t = final time =5 s

We replaced the known information in the equation(1):

$ac = at = \frac{13-0}{15-0}$

$ac=ac=\frac{13}{15} \frac{m}{s}$

Dynamic analysis

The forces acting on the car are the following:

Wc: Car weight

N: normal force, road force on the car

Ff: Friction force

T: Force of tension

Car weight calculation:

$Wc=mc*g$

mc = Car mass = 2230kg

g = Gravity acceleration=9.8 $\frac{m}{s^{2} }$

$Wc= 2230*9.8$

$Wc=21854 N$

Normal force calculation:

Newton's first law

$sum Fy= 0$

$N-W=0$

$N=W$

$N=21854 N$

Friction force calculation (Ff):

We have the formula to calculate the friction force:

Ff = μk * N Equation (3)

μk kinetic coefficient of friction

We know that μk = 0.373and N= 21854N ,then:

$Ff=0.373*21854$

$Ff=8151.54 N$

Calculation of the tension force in the rope (T):

Newton's Second law

$sum Fx= mc*ac$

$T-Ff=mc*ac$

$T=2230(\frac{13}{15}) + 8151.54$

T=10084,21 N

Answer: The rope must have a force of 10084,21 N

8 0

3 years ago

A 3.5-kg object placed on an inclined plane (angle 30? above the horizontal) is connected by a string going over a pulley to a 1

Blababa [14]

Answer:

a= 0.22 m/s²

Explanation:

Given that

M = 3.5 kg

θ = 30°

m = 1 kg

μ= 0.3

The force due to gravity

F₁= M g sinθ

F₁=3.5 x 10 x sin 30

F₁= 17.5 N

F₂ = m g

F₂ = 1 x 10 = 10 N

The maximum value of the friction force on the incline plane

Fr = μ M g cosθ

Fr = 0.3 x 2.5 x 10 cos30°

Fr= 6.49 N

Lets take acceleration of the system is a m/s²

F₁ - F₂ - Fr = (M+m) a

17.5 - 10 - 6.49 = (3.5+1)a

a= 0.22 m/s²

7 0

3 years ago

An engineering team has come to the stage in the engineering design

SIZIF [17.4K]

Answer: D

Explanation: read the answers above and compare to your test

6 0

3 years ago

Which of these is a result of gravitational lensing?

valentina_108 [34]

Answer:

Answer is the formation of multiple images of the same object .

Explanation:

I hope it's helpful!

7 0

3 years ago

ASTM A229 oil-tempered carbon steel is used for a helical coil spring. The spring is wound with D = 50 mm d = 10.0 mm, and a pit

horrorfan [7]

Answer

given,

D = 50 mm = 0.05 m

d = 10 mm = 0.01 m

Force to compress the spring

$F = \dfrac{d^4G\delta}{8D^3N}$

$\dfrac{\delta}{N} = p - d = 14 - 10 = 4 mm$

$F = \dfrac{d^4G}{8D^3}\times 0.004$

$F = \dfrac{0.1^4\times 79\times 10^9}{8\times 0.05^3}\times 0.004$

F = 3160 N

stress correction factor from stress correction curve is equal to 1.1

now, calculation of corrected stress

$\tau = \dfrac{8FDk_s}{\pi d^3}$

$\tau = \dfrac{8\times 3160 \times 0.05 \times 1.1}{\pi \times 0.01^3}$

= 442.6 Mpa

The tensile strength of the steel material of ASTM A229 is equal to 1300 Mpa

now,

$\tau_s \leq 0.45 S_u$

$\tau_s \leq 0.45 \times 1300$

$\tau_s \leq 585\ Mpa$

since corrected stress is less than the $\tau_s$

hence, spring will return to its original shape.

6 0

3 years ago

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