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4 years ago

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A boy whirls a stone in a horizontal circle of radius 1.1 m and at height 2.1 m above ground level. The string breaks, and the s

tone flies horizontally and strikes the ground after traveling a horizontal distance of 10 m. What is the magnitude of the centripetal acceleration of the stone while in circular motion? m/s2

1 answer:

DedPeter [7]4 years ago

8 0

Answer:

$212.8 m/s^{2}$

Explanation:

Time taken by stone to cover horizontal distance

$t=\sqrt{\frac {2h}{g}}$ where t is time, h is height of whirling the stone in horizontal circle, g is gravitational constant, Substituting h for 2.1 m and g for 9.81

$t=\sqrt{\frac {2*2.1}{9.81}}$ = 0.654654 seconds

t=0.65 s

Velocity, v= distance/time

v=10/0.65= 15.27525 m/s

v=15.3 m/s

$a=\frac {v^{2}}{r}$ where r is radius of circle, substituting r with 1.1m

$a=\frac {15.3^{2}}{1.1}$

$a=212.8 m/s^{2}$

Therefore, centripetal acceleration is $212.8 m/s^{2}$

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Maksim231197 [3]

Answer:

7.00 m

Explanation:

Given:

v₀ = 2.00 m/s

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3 years ago

The speed of the center of the earth as it orbits the sun is 107257 kmph and the absolute angular velocity of the earth about it

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Answer with Explanation:

We are given that

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Linear velocity, $v=r\omega=6371000\times 7.292\times 10^{-5}=464.57 m/s$

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Velocity at point B, $v_B=v_0j-vj=107257j-1672.46j=105584.54j kmph$

Velocity at point C, $v_C=v_0j+vi=1672.46 i+107257j kmph$

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3 years ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

<u>Distance of the center from each corners</u> $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

<u>For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:</u>

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

<u>Force by each of the charges at the remaining corners:</u>

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

<u> Now, net electric field in the vertical direction:</u>

$E_y=E_1-E_4$

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$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

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marysya [2.9K]

Answer:576

Explanation:

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3 years ago

The momentum of an object is determined to be 7.2 x 10-3 cm kg x m/s. Express this as provided or use any equivalent unit. How i

Leno4ka [110]

Complete Question:

The momentum of an object is determined to be 7.2 × 10-3 kg⋅m/s. Express this quantity as provided or use any equivalent unit. (Note: 1 kg = 1000 g).

Answer:

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Explanation:

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

$Momentum = mass * velocity$

Given the following data;

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<em>Momentum = 7.2 gm/s. </em>

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3 years ago