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kotegsom [21]
2 years ago
15

A 0.023 kg beetle is sitting on a record player 0.15 m from the center of the record. if it takes 0.070 n of force to keep the b

eetle moving in a circle on the record, what is the tangential speed of the beetle? 0.68 m/s 0.46 m/s 0.33 m/s 0.11 m/s
Physics
2 answers:
Margaret [11]2 years ago
6 0
The answer is 0.68 meters per second, swagteam
vova2212 [387]2 years ago
6 0

Answer:

Tangential speed is 0.68 m/s

Explanation:

It is given that,

Mass of the beetle, m = 0.023 kg

It is placed at a distance of 0.15 m from the center of record i.e. r = 0.15 m.

If it takes 0.070 n of force to keep the beetle moving in a circle on the record i.e. centripetal force acting on it is, F = 0.070 N

We have to find the tangential speed of the beetle. The formula for centripetal force is given by :

F=\dfrac{mv^2}{r}

v is tangential speed

v=\sqrt{\dfrac{Fr}{m}}

v=\sqrt{\dfrac{0.070\times 0.15}{0.023}}

v = 0.675 m/s

or

v = 0.68 m/s

Hence, the correct option for tangential speed is (A).

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A typical jet airliner has a cruise airspeed of 900 km/h , which is its speed relative to the air through which it is flying. If
Luba_88 [7]

Explanation:

It is given that,

Speed of the jet airplane with respect to air, v_{PA} = 900\ km/h          

If the wind at the airliner’s cruise altitude is blowing at 100 km/h from west to east, v_{AG}= 100\ km/h

(A) Let v_{PG} is the speed of the airliner relative to the ground if the airplane is flying from west to east,

v_{PG}=900-100=800\ km/h

(B) Let v'_{PG} is the speed of the airliner relative to the ground if the airplane is flying from east to west,

v'_{PG}=900+100=1000\ km/h

Hence, this is the required solution.                                            

8 0
2 years ago
a 4357 kg roller coaster car starts from rest at the top of a 36.5 m high track. determine the speed of the car at the top of a
andrey2020 [161]
The correct answer is 17.24 m/s. You get the answer by subtracting the two heights of the tracks which are 36.5 and 10.8 m, and the answer is 25.7. Since you already know the height at which the kinetic energy will be coming from, you then divide the amount of weight the roller coaster has to the distance it needs to travel in order for you to determine the speed of the car. So that is, 4,357 kg and 25.7 m and the answer is 169 kg/m. Dividing it to the earth's gravity of 9.8 m/s you'll get 17.24 m/s.
4 0
2 years ago
Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller
kobusy [5.1K]

Answer:

σ₁ = 3.167 * 10^{-6} C/m²

σ₂ = 7.6 * 10 ^{-6}  C/m²

Explanation:

The given data :-

i) The radius of smaller sphere ( r ) = 5 cm.

ii) The radius of larger sphere ( R ) = 12 cm.

iii) The electric field at of larger sphere  ( E₁ ) = 358 kV/m. = 358 * 1000 v/m

E_{1} = (\frac{1}{4\pi\epsilon  }) (\frac{Q_{1} }{R^{2} } )

358000 = 9 * 10^{9 } *\frac{Q_{1} }{0.12^{2} }

Q₁ = 572.8 * 10^{-9} C

Since the field inside a conductor is zero, therefore electric potential ( V ) is constant.

V = constant

∴\frac{Q_{1} }{R} = \frac{Q_{2} }{r}

Q_{2}  = \frac{r}{R} *Q_{1}

Q_{2} = \frac{5}{12} *572.8*10^{-9}   = 238.666 *10^{-9} C

Surface charge density ( σ₁ ) for large sphere.

Area ( A₁ )  = 4 * π * R²  = 4 * 3.14 * 0.12 = 0.180864 m².

σ₁  = \frac{Q_{1} }{A_{1} } = \frac{572.8 *10^{-9} }{0.180864} = 3.167 * 10^{-6}  C/m².

Surface charge density ( σ₂ ) for smaller sphere.

Area ( A₂ )  = 4 * π * r²  = 4 * 3.14 * 0.05²  =0.0314 m².

σ₂ =\frac{Q_{2} }{A_{2} } = \frac{238.66 *10^{-9} }{0.0314} = 7.6 * 10 ^{-6} C/m²

8 0
2 years ago
Although blood cells are contained within a special liquid called plasma, the cells themselves are___________.
valentina_108 [34]

Answer:

Solid

Explanation:

The plasma is the liquid part of blood, it is 90% and accounts for 55% of blood volume. It is what red blood cells, white blood cells, and platelets move around in. These cells remain solid within the plasma. I hoped this helped!

8 0
2 years ago
1. In a single atom, no more than 2 electrons can occupy a single orbital? A. True B. False
Studentka2010 [4]
1. In a single atom, no more than 2 electrons can occupy a single orbital? A. True

2. The maximum number of electrons allowed in a p sublevel of the 3rd principal level is? 
B.6

3. A neutral atom has a ground state electronic configuration of 1s^2 2s^2. Which of the following statements concerning this atom is/are correct?
B. All of the above.
4 0
2 years ago
Read 2 more answers
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