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4 years ago

Action reaction forces are in which one of newtons laws

Physics

1 answer:

Klio2033 [76]4 years ago

4 0

Answer:

the third law

Explanation:

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A car with good tires on a dry road can decelerate (slow down) at a steady rate of about 5.0 m/s2 when braking. if a car is init

Sladkaya [172]

The time required by the car to stop is 4.916 sec.

Since the car is moving with the constant deceleration we can apply the first equation of motion to calculate the time required by the car to stop.

The first equation of motion is given as

V=u+at

Here, V=final speed of the car=0 mi/h as the car stops

u =initial speed of the car=55 mi/hr=24.58 m/s

a= acceleartion =-5 m/s^2 (here negative sign indicates for deceleration)

Now applying the values in the first equation

V=u+at

0=24.58-5*t

t=4.916 sec

Therefore the car will stops in 4.916 sec.

8 0

3 years ago

A diverging mirror is .................

lyudmila [28]

Is there a multiple choice?

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4 years ago

Ryan is a nine-year-old boy whose father wants him to be a soccer player. Ryan does not have an aptitude for soccer, and keeps f

mihalych1998 [28]

A bad life when Ryan is older and will criticize his kids

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3 years ago

Read 2 more answers

4-12. The morning inspection of the tank farm finds a leak in the turpentine tank. The leak is repaired. An investigation finds

bonufazy [111]

Answer:

a) $V=759.8727\ ft^3$

b) $\dot V=1.403\times 10^{-3}\ ft^3.s^{-1}$

c) $t\approx29.541\ s$

Explanation:

Given:

diameter of hole in the tank, $d'=0.1\ in=\frac{1}{120}\ ft$

position of the hole form the tank bottom, $h' =7\ ft$

initial level of turpentine in the tank before the leakage, $h_i=17.3\ ft$

level of turpentine in the tank after the repair of leakage, $h_f=13\ ft$

diameter of the tank, $d=15\ ft$

density of turpentine oil, $\rho=55\ lbm.ft^3$

Now, volume of turpentine spilled:

$V=A.(h_i-h_f)$

where:

$A=$ area of the cross section of the tank's volume

$V=\pi .\frac{d^2}{4} \times(h_i-h_f)$

$V=\pi\times\frac{15^2}{4} \times(17.3-13)$

$V=759.8727\ ft^3$

When the tank was full the liquid level was highest:

so velocity form the height of the hole will be given as:

$v=\sqrt{2g.(h_i-h')}$

$v=\sqrt{2\times 32.12\times (17.3-7) }$

$v=25.722\ ft.s^{-1}$

<u>Now we have the flow rate of the spillage given by:</u>

$\dot V=(\pi.\frac{d'^2}{4}) \times v$

$\dot V=\pi\times \frac{(\frac{1}{120})^2 }{4} \times 25.722$

$\dot V=1.403\times 10^{-3}\ ft^3.s^{-1}$

Total time the leak was active can be calculated as:

$t=\frac{V}{\dot V}$

$t=\frac{759.8727}{25.722}$

$t\approx29.541\ s$

7 0

4 years ago

A satellite has a mass of 6463 kg and is in a circular orbit 4.82 × 105 m above the surface of a planet. The period of the orbit

Tems11 [23]

Answer:

The weight of the planet is 29083.5 N .

Explanation:

mass of satellite, m = 6463 kg

height of orbit, h = 4.82 x 10^5 m

period, T = 2 h

radius of planet, R = 4.29 x 10^6 m

Let the acceleration due to gravity at the planet is g.

$T = 2\pi\sqrt\frac{(R+h)^3}{gR^2}\\\\2\times 3600 = 2\times3.14\sqrt\frac{(4.29+0.482)^3\times10^{18}}{g\times 4.29\times 4.29\times 10^{12} }\\\\24.2 g =108.67\\\\g = 4.5 m/s^2$

The weight of the satellite at the surface of the planet is

W = m g = 6463 x 4.5 = 29083.5 N

6 0

3 years ago