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1 year ago

8

Consider a ball rolling around in a circular path on the inner surface of a cone. the weight of the ball is shown by the vector

W . There us no friction, so only one other force acts on the ball a normal force.
1.The normal force is greater than the weight and greater than the centripetal force
2.The normal force is greater than the weight and less that the centripetal force
3.The normal force is less than the weight and the greater than the centripetal force
4. The normal force less than the weight and less than the centripetal force.

1 answer:

zimovet [89]1 year ago

5 0

Answer:

option 2 is righttttt.................

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A particular planet has a moment of inertia of 9.74 × 1037 kg ⋅ m2 and a mass of 5.98 × 1024 kg. Based on these values, what is

malfutka [58]

Answer: A) $6.38(10)^{6} m$

Explanation:

The equation for the moment of inertia $I$ of a sphere is:

$I=\frac{2}{5}mr^{2}$ (1)

Where:

$I=9.74(10)^{37}kg m^{2}$ is the moment of inertia of the planet (assumed with the shape of a sphere)

$m=5.98(10)^{24}kg$ is the mass of the planet

$r$ is the radius of the planet

Isolating $r$ from (1):

$r=\sqrt{\frac{5I}{2m}}$ (2)

Solving:

$r=\sqrt{\frac{5(9.74(10)^{37}kg m^{2})}{2(5.98(10)^{24}kg)}}$ (3)

Finally:

$r=6381149.077m \approx 6.38(10)^{6} m$

Therefore, the correct option is A.

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1 year ago

(a) What is the sum of the following four vectors in unit-vector notation? For that sum, what are (b) the magnitude, (c) the ang

avanturin [10]

6m at 0.9 radian means 6m at $51.57^0$

Since positive angles are counter clockwise

6m at $51.57^0$ can be written as 6*cos $51.57^0$ i+6*sin $51.57^0$ j = 3.73i+4.70j

5m at $-75^0$ can be written as 5*cos $(-75)^0$ i+5*sin $(-75)^0$ j = 1.294i-4.83j

4m at 1.2 radian means 4m at $68.75^0$

4m at $68.75^0$ can be written as 4*cos $68.75^0$ i+4*sin $68.75^0$ j = 1.45i+3.73j

6m at $-210^0$ can be written as 6*cos $(-210)^0$ i+6*sin $(-210)^0$ j = -5.20i-3j

a) So sum of the vector = 1.274i+0.6j

b) Magnitude = $\sqrt{1.274^2+0.6^2} = 1.98 m$

c) Angle, $tan\theta = \frac{0.6}{1.274} \\ \\ \theta = 25.22^0$

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1 year ago

Michael Phelps needs to swim at an average speed of 2.00 m/s in order to set a new world record in the 200 m freestyle. If he sw

Natasha_Volkova [10]

Answer:

Explanation:

Given

average speed of Phelps $v_{avg}=2\ m/s$

total distance $d=200\ m$

he swims first 100 m at an average speed if $1.8 m/s$

so time taken is $t_1=\frac{100}{1.8}=55.55\ s$

suppose $t_2$ is the time taken to swim remaining half

average velocity is $v_{avg}=\frac{displacement}{total\ time}$

$v_{avg}=\frac{100+100}{55.55+t_2}$

$t_2+55.55=\frac{200}{2}=100$

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so velocity in the second half is

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2 years ago

Raphael refers to a wave by noting its wavelength. lucinda refers to a wave by noting its frequency. which student is correct an

blsea [12.9K]

They are both right because you can note both things, I mean Raphael and Lucinda, both has a right statement or explanation about the wave. Wave by nothing is both for its wavelength and for its frequency. So Raphael and Lucinda are both correct because you can note both wavelength and frequency.

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2 years ago

If the radius of the sun is 7.001×105 km, what is the average density of the sun in units of grams per cubic centimeter? The vol

xenn [34]

Answer:

Average density of Sun is 1.3927 $\frac{g}{cm}$ .

Given:

Radius of Sun = 7.001 × $10^{5}$ km = 7.001 × $10^{10}$ cm

Mass of Sun = 2 × $10^{30}$ kg = 2 × $10^{33}$ g

To find:

Average density of Sun = ?

Formula used:

Density of Sun = $\frac{Mass of Sun}{Volume of Sun}$

Solution:

Density of Sun is given by,

Density of Sun = $\frac{Mass of Sun}{Volume of Sun}$

Volume of Sun = $\frac{4}{3} \pi r^{3}$

Volume of Sun = $\frac{4}{3} \times 3.14 \times [7.001 \times 10^{10}]^{3}$

Volume of Sun = 1.436 × $10^{33}$ $cm^{3}$

Density of Sun = $\frac{ 2\times 10^{33} }{1.436 \times 10^{33} }$

Density of Sun = 1.3927 $\frac{g}{cm}$

Thus, Average density of Sun is 1.3927 $\frac{g}{cm}$ .

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2 years ago