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2 years ago

11

A security guard walks at a steady pace traveling 170 m in one trip around the perimeter of a building.

1 answer:

WARRIOR [948]2 years ago

6 0

Speed= distance/time
Speed= 170/230
Speed=0.74 m/s

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A car is travelling at a constant speed of 26.5 m/s. Its tires have a radius of 72 cm. If the car slows down at a constant rate

maksim [4K]

Answer:

Magnitude of angular acceleration = -3.95 rad/s²

Explanation:

Angular acceleration is the ratio of linear acceleration and radius.

That is

$\texttt{Angular acceleration}=\frac{\texttt{Linear acceleration}}{\texttt{Radius}}\\\\\alpha =\frac{a}{r}$

Radius = 72 cm = 0.72 m

Linear acceleration is rate of change of velocity.

$a=\frac{11.7-26.5}{5.2}=-2.85m/s^2$

Angular acceleration

$\alpha =\frac{a}{r}=\frac{-2.85}{0.72}=-3.95rad/s^2$

Angular acceleration = -3.95 rad/s²

Magnitude = 3.95 rad/s²

4 0

2 years ago

What is the term for an area of a country located away from capital?

8090 [49]

Province. Is what is located away from the capital.

5 0

3 years ago

A pipe is horizontal and carries oil that has a viscosity of 0.14 Pa s. The volume flow rate of the oil is 5.3 × 10-5 m3/s. The

Liula [17]

Answer:

Explanation:

Given

$\mu =0.14 Pa.s$

Volume Flow rate $Q=5.3\times 10^{-5} m^3/s$

Length $L=37 m$

radius $r=0.58 cm$

$D=1.16 cm$

According to Hagen-Poiseuille Equation

difference in Pressure is given

$\Delta P=\frac{128\mu L\cdot Q}{\pi D^4}$

$P_{in}-P_{out}=\frac{128\mu L\cdot Q}{\pi D^4}$

$P_{in}=P_{out}+\frac{128\mu L\cdot Q}{\pi D^4}$

$P_{in}=1.01325\times 10^5+\frac{128\times 0.14\times 37\times 5.3\times 10^{-5}}{\pi\cdot 1.16^4\times 10^{-8}}$

$P_{in}=1.01325\times 10^5+6.17\times 10^5$

$P_{in}=7.19\times 10^5$

3 0

2 years ago

A U-shaped tube open to the air at both ends contains some mercury. A quantity of water is carefully poured into the left arm of

Gekata [30.6K]

Answer:

a) $P=2450\ Pa$

b) $\delta h=23.162\ cm$

Explanation:

Given:

height of water in one arm of the u-tube, $h_w=25\ cm=0.25\ m$

a)

Gauge pressure at the water-mercury interface,:

$P=\rho_w.g.h_w$

we've the density of the water $=1000\ kg.m^{-3}$

$P=1000\times 9.8\times 0.25$

$P=2450\ Pa$

b)

Now the same pressure is balanced by the mercury column in the other arm of the tube:

$\rho_w.g.h_w=\rho_m.g.h_m$

$1000\times 9.8\times 0.25=13600\times 9.8\times h_m$

$h_m=0.01838\ m=1.838\ cm$

<u>Now the difference in the column is :</u>

$\delta h=h_w-h_m$

$\delta h=25-1.838$

$\delta h=23.162\ cm$

7 0

2 years ago

A wheel with moment of inertia 25 kg. m2 and angular velocity 10 rad/s begins to speed up, with angular acceleration 15 rad/sec2

Pani-rosa [81]

Answer:

(A) Angular speed 40 rad/sec

Rotation = 50 rad

(b) 37812.5 J

Explanation:

We have given moment of inertia of the wheel $I=25kgm^2$

Initial angular velocity of the wheel $\omega _0=10rad/sec$

Angular acceleration $\alpha =15rad/sec^2$

(a) We know that $\omega =\omega _0+\alpha t$

We have given t = 2 sec

So $\omega =10+15\times 2=40rad/sec$

Now $\Theta =\omega _0t+\frac{1}{2}\alpha t^2=10\times 2+\frac{1}{2}\times 15\times 2^2=50rad$

(b) After 3 sec $\omega =10+15\times 3=55rad/sec$

We know that kinetic energy is given by $Ke=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 25\times 55^2=37812.5J$

7 0

2 years ago