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Elodia [21]
3 years ago
7

If a weight hanging on a string of length 5 feet swings through 5^\circ on either side of the vertical, how long is the arc thro

ugh which the weight moves from one high point to the next high point?
Physics
1 answer:
Zielflug [23.3K]3 years ago
7 0
<span>First we can find the circumference of the whole circle with a radius of 5 feet. circumference = 2 pi radius circumference = (2 pi) (5 feet) circumference = (10 pi) feet From one high point to the other high point, the string moves through an angle of 10 degrees. Since a full circle is 360 degrees, this angle is 1/36 of a full circle. Therefore, the arc length is 1/36 of the whole circumference. arc length = (1/36) (circumference) arc length = (1/36) (10 pi) feet arc length = 0.873 feet</span>
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A large piece of jewelry has a mass of 130.8 g. A graduated cylinder initially contains 47.7 mL water. When the jewelry is subme
Shkiper50 [21]

Answer: The density of this piece of jewelry is 8.90g/cm^3

Explanation:

To calculate the density, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Mass of piece of jewellery = 130.8 g

Density of piece of jewellery = ?

Volume of piece of jewellery =( 62.4-47.7 ) ml = 14.7 ml = 14.7cm^3   1cm^3=1ml

Putting values in above equation, we get:

\text{Density of piece of jewellery}=\frac{130.8g}{14.7cm^3}=8.90g/cm^3

Thus density of this piece of jewelry is 8.90g/cm^3

8 0
3 years ago
Which of these best defines weather? (3 points) a Atmospheric conditions over 30-year period Ob Day-to-day condition of the atmo
prisoha [69]

Answer:

b Day-to-day condition of the atmosphere

Explanation:

Weather is short term, Climate is long term

3 0
3 years ago
Read 2 more answers
A car is traveling 35 mph on a smooth surface. If a balanced force is applied to the car, what happens?
zepelin [54]

Answer:

Here is the answer.

Explanation:

Balanced forces- they are those forces that produce 0 resultant forces.

therefore, on applying a balanced force on the object, it wouldn't result in any change, as resultant force would be 0.

7 0
3 years ago
Help with these please?
Anuta_ua [19.1K]

E=kq/r^2

q=(E*r^2)/k

q=(.086N/C)(1.7m^2)/(8.99*10^9N*m^2/C^2)

q=2.76*10^-11 C

q=2.8*10^-11 C

5 0
3 years ago
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Ali is whirling a 2.0 kg bunch of bananas in a circular path having a radius of 0.50 m. The bananas complete 2 revolutions every
skad [1K]

Answer:

1) 2.467 N

2) a) 0.248m

   b) 2.3π rad/sec

Explanation:

Given data:

mass of Banana bunch ( m ) = 2.0 kg

radius of circular path ( R ) = 0.5 m

number of revolutions completed = 2

Time to complete 2 revolutions = 6 seconds

1) Determine the force to keep the motion constant for one complete revolution in every 4 seconds

F = mv^2 / r ----- ( 1 )

where V = 2πR/T

where : R = 0.5 , m = 2, T = 4 seconds

Insert values into equation 1

F = 2 * 4π^2 * 0.5/4^2

 = 2.467 N

2a) Calculate the maximum distance of coin from center

angular velocity ( w ) = v/r

coefficient of static friction  ( μ ) = 0.25

F_{c} = u mg  ---- ( 1 )

mv^2/r = μmg --- ( 2 )        cancelling the mass on both sides eqn 2 becomes

v^2 = μ*g*r

dividing both sides of equation by r^2

w^2 = μ*g/r

hence determine distance ( r ) of coin from center

r = 0.25 * 9.81 / π^2 =  0.248 m

2b ) determine the maximum speed of rotation of the turntable for the coin to move relative to the turntable without slipping

distance coin is placed ( r ) = 4.7 cm = 0.047 m

find speed of rotation ( w )

w^2 =  μ*g/r

w = √ 0.25 * 9.81/ 0.047

   = 7.2236 rad/secs ≈  2.3π rad/sec

       

3 0
3 years ago
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