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stealth61 [152]

3 years ago

12

If a thief jumps from the tried floor of a house while holding a box on

2 answers:

Ksju [112]3 years ago

7 0

What do you mean? it doesn't make sense

Firdavs [7]3 years ago

3 0

What do I mean it makes no sense

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A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What

AleksandrR [38]

Answer:

The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

Explanation:

We are given that

Angular acceleration, $\alpha=3.3 rad/s^2$

Diameter of the wheel, d=21 cm

Radius of wheel, $r=\frac{d}{2}=\frac{21}{2}$ cm

Radius of wheel, $r=\frac{21\times 10^{-2}}{2} m$

1m=100 cm

Magnitude of total linear acceleration, a= $1.7 m/s^2$

We have to find the linear speed of a at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.

Tangential acceleration, $a_t=\alpha r$

$a_t=3.3\times \frac{21\times 10^{-2}}{2}$

$a_t=34.65\times 10^{-2}m/s^2$

Radial acceleration, $a_r=\frac{v^2}{r}$

We know that

$a=\sqrt{a^2_t+a^2_r}$

Using the formula

$1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}$

Squaring on both sides

we get

$2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}$

$\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}$

$v^4=r^2\times 2.7699$

$v^4=(10.5\times 10^{-2})^2\times 2.7699$

$v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}$

$v=0.418 m/s$

Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

6 0

2 years ago

A container is filled with water and the pressure at the container bottom is P. If the container is instead filled with a liquid

den301095 [7]

Eat ahah is when s susuehs

7 0

3 years ago

PLEASE ANSWER BOTH QUICK!! 50 POINTS!!! PLEASE DON'T IGNORE!!

RUDIKE [14]

Answer:

A. B. Pls Give Branliest

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3 0

2 years ago

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you spray your sister with water from a garden hose. the water is supplied to the hose at a rate of 0.649x10-3 m^3/s and the dia

Ivanshal [37]

Answer:

27.82 m/s

Explanation:

The radius of the hose is half of its diameter

$r = d/2 = 5.45\times10^{-3}/2 = 0.002725 m$

So its area must be

$A = \pi r^2 = \pi 0.002725^2 = 2.33\times10^{-5} m^2$

The speed of water coming out of the hose is its flow rate divided by the cross-section area of the hose

$v = \dot{V}/A =0.000649 / 2.33\times10^{-5} = 27.82 m/s$

7 0

2 years ago

A right-circular cylindrical tank of height 8 ft and radius 4 ft is laying horizontally and is full of fuel weighing 52 lb/ft3

tresset_1 [31]

Given:

Height of tank = 8 ft

and we need to pump fuel weighing 52 lb/ $ft^{3}$ to a height of 13 ft above the tank top

Solution:

Total height = 8+13 =21 ft

pumping dist = 21 - y

Area of cross-section = $\pi r^{2}$ = $\pi 4^{2}$ =16 $\pi$ $ft^{2}$

Now,

Work done required = $\int_{0}^{8} 52\times 16\pi (21 - y)dy$

= $832\pi \int_{0}^{8} (21 - y)dy$

= 832 $\pi($ $[ 21y ]_{0}^{8} - [\frac{y^{2}}{2}]_{0}^{8}\\$ )

= 113152 $\pi$ = 355477 ft-lb

Therefore work required to pump the fuel is 355477 ft-lb

7 0

3 years ago