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3 years ago

10

A 900-N lawn roller is to be pulled over a 5-cm (high) curb. Radius of roller is 25 cm. What minimum pulling force is needed if

angle made by handle is (a) 0 deg (b) 30 degrees?

1 answer:

Feliz [49]3 years ago

8 0

I believe b 30 degrees

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A student connects an object with mass m to a rope with a length r and then rotates the rope around her head parallel to the gro

Alexandra [31]

The object takes 0.5 seconds to complete one rotation, so its rotational speed is 1/0.5 rot/s = 2 rot/s.

Convert this to linear speed; for each rotation, the object travels a distance equal to the circumference of its path, or 2<em>π</em> (1.2 m) = 2.4<em>π</em> m ≈ 7.5 m, so that

2 rot/s = (2 rot/s) • (2.4<em>π</em> m/rot) = 4.8<em>π</em> m/s ≈ 15 m/s

thus giving it a centripetal acceleration of

<em>a</em> = (4.8<em>π</em> m/s)² / (1.2 m) ≈ 190 m/s².

Then the tension in the rope is

<em>T</em> = (50 kg) <em>a</em> ≈ 9500 N.

7 0

2 years ago

Please answer help me

Andrews [41]

I believe the answer is x

7 0

2 years ago

A girl weighing 45kg is standing on the floor, exerting a downward force of 200N on the floor. The force exerted on her by the f

sukhopar [10]

Answer:

c.

Equal to 200 N..........

7 0

2 years ago

An ancient club is found that contains 100 g of pure carbon and has an activity of 6.5 decays per second. Determine its age assu

never [62]

Answer:

The age of living tree is 11104 years.

Explanation:

Given that,

Mass of pure carbon = 100 g

Activity of this carbon is = 6.5 decays per second = 6.5 x60 decays/min =390 decays/m

We need to calculate the decay rate

$R=\dfrac{-dN}{dt}=\lambda N=\dfrac{0.693}{t_{\frac{1}{2}}}N$ ....(I)

Where, N = number of radio active atoms

$t_{\frac{1}{2}}$ =half life

We need to calculate the number of radio active atoms

For $N_{12_{c}}$

$N_{12_{c}}=\dfrac{N_{A}}{M}$

Where, $N_{A}$ =Avogadro number

$N_{12_{c}}=\dfrac{6.02\times10^{23}}{12}$

$N_{12_{c}}=5.02\times10^{22}\ nuclie/g$

For $N_{c_{14}}$

$N_{c_{14}}=1.30\times10^{-12}N_{12_{c}}$

$N_{c_{14}}=1.30\times10^{-12}\times5.02\times10^{22}$

$N_{c_{14}}=6.526\times10^{10}\ nuclei/g$

Put the value in the equation (I)

$R=\dfrac{0.693\times6.526\times10^{10}\times60}{5700\times3.16\times10^{7}}$

$R=15.0650\ decay/min g$

100 g carbon will decay with rate

$R=100\times15.0650=1507\ decay/min$

We need to calculate the total half lives

$(\dfrac{1}{2})^{n}=\dfrac{390}{1507}$

$2^n=\dfrac{1507}{390}$

$2^n=3.86$

$n ln 2=ln 3.86$

$n=\dfrac{ln 3.86}{ln 2}$

$n =1.948$

We need to calculate the age of living tree

Using formula of age

$t=n\times t_{\frac{1}{2}}$

$t=1.948\times5700$

$t=11103.6 =11104\ years$

Hence, The age of living tree is 11104 years.

5 0

3 years ago

The pilot of an airplane traveling 45 m/s wants to drop supplies to flood victims isolated on a patch of land 160 m below. The s

Sidana [21]

Answer:

<h2>254.56 m</h2>

Explanation:

A object dropped from a plane from a certain height will follow a parabolic trajectory because it has a horizontal velocity equal to plane's velocity.

So, if supplies are to be dropped from a plane from a height of 160 m, let us calculate the time it takes to reach the ground.

$H=\frac{1}{2}gt^{2}\\160=\frac{1}{2}\times10\times t^{2}\\t=\sqrt{32}=4\sqrt{2}\text{ }sec$

So, in this time, the supply moves a horizontal distance of $(4\sqrt{2}\text{ }sec)\times(45\text{ }\frac{m}{sec})=180\sqrt{2}\text{ }m=254.56\text{ }m$ .

∴ The supply must be dropped when the plane is 255 m away.

5 0

2 years ago