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HACTEHA [7]
2 years ago
4

A 0.800-kg ball is tied to the end of a string 1.60 m long and swung in a vertical circle. (a)During one complete circle, starti

ng anywhere, calculate the total work done on the ballby (i) the tension in the string and (ii) gravity. (b) Repeat part (a) for motion along thesemicircle from the lowest to the highest point on the path.
Physics
1 answer:
kipiarov [429]2 years ago
5 0

Answer: a) i) 0 ii) 0  b) i) 0 ii) -25.1 J

Explanation:

a) For the first part, as work is defined as the process through an applied force produces a displacement in the direction of the movement, if the start and the end points are the same (as for a complete circle), this means that there is no displacement, i.e. total work (for any applied force) is 0.

b) Taking only a semicircle from the lowest to the highest point on the path, as tension force is always perpendicular to the displacement, we can conclude that the tension force does no work.

Now, for gravity, the displacement, is just the difference in height between the highest and lowest point, which is equal to twice the length of the string, and is a vector with only a vertical component, as follows:

Δy = y₂ - y₁ (by definition of displacement)

Now, the work done by gravity is just the product of the weight of  the ball, times the vertical displacement (as the work done by gravity is independent of the trajectory) , as follows :

W = m. g. (y₂-y₁) = 0.8 Kg. 9.8 m/s². (-2. 1.6 m) = -25.1 J

(The minus sign is due to gravity and the displacement point in opposite directions, in this case we take as positive the downward direction).

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If you double the mass of an object while leaving the net force unchanged what is the result
valentinak56 [21]

Answer: If the net force acting on an object doubles, its acceleration is doubled. If the mass is doubled, then acceleration will be halved. If both the net force and the mass are doubled, the acceleration will be unchanged.

Explanation:

5 0
2 years ago
A star orbiting a black hole in a clockwise direction at a radial distance of 1.0 × 106 km is acted upon by a counterclockwise f
snow_tiger [21]

Answer:

2.5 \times 10^{11} N-m upwards        

Explanation:

Torque is the vector cross product of the force and radial distance.

\tau = rF sin \theta

\tau = (1.0\times 10^6 \times 10^3) m \times 250 N\times sin 90^o \\\Rightarrow \tau= 2.5 \times 10^{11} N-m

The direction of the torque would be perpendicular to the direction of the force and radial distance. The direction of the force is counter-clockwise. The direction of the torque would be upwards.

4 0
2 years ago
A sample of an unknown volatile liquid was injected into a Dumas flask (mflask = 27.0928 g, Vflask = 0.1040 L) and heated until
NNADVOKAT [17]

Answer:

The gas was Hexane

Explanation:

taking the diference between the mass of the flask and the final mass qe can calculate the mass of liquid injected (assuming none escaped the flask):

m_{l}  = 27.4593g - 27.0928g = 0.3665g

with the volume of the flask we can get the density of the gas at the indicated pressure and temperature:

d_{g}  = \frac{0.3665 g}{0.1040L} = 3.524 g/L

From the ideal gases law we have that the density can be calculated as:

d_{g}  = \frac{P*M}{R*T}

Where R is the ideal gases constant = , and M the molecular weight of the fluid. Solving for M:

M=\frac{d_{g}*R*T}{P}=\frac{3.524g/L*0,082atmL/molK*291K}{0.976atm}

M=86.16 g/mol

Note that the temperature is computed in Kelvin T= 18+273=291K

The gas with the closer molar mass is Hexane

4 0
2 years ago
A gas station owner suspects that he is being overcharged for gasoline deliveries by a gasoline supplier. The overcharge seems p
Tems11 [23]

Answer:

Explanation:

delta V = v * alpha * delta T

= V * 0.00053 * (92.2 - 55.0)

= 0.019716 V

percentage that the owner

= [delta V / V] * 100

= [0.019716 V / V] * 100

= 1.9716 %

4 0
2 years ago
What is the gravitational force of attraction between a planet and a 17-kilogram mass that is falling freely toward the surface
PolarNik [594]

Answer:

a. 150 N

Explanation:

Gravitational Force: This is the force that act on a body under gravity.

The gravitational force always attract every object on or near the earth's surface. The earth therefore, exerts an attractive force on every object on or near it.

The S.I unit of gravitational force is Newton(N).

Mathematically, gravitational force of attraction is expressed as

(i) F = GmM/r² ........................ Equation 1 ( when it involves two object of different masses on the earth)

(ii) F = mg ............................... Equation 2 ( when it involves one mass and the gravitational field).

Given: m = 17 kg, g = 8.8 m/s²

Substituting into equation 2,

F = 17(8.8)

F = 149.6 N

F ≈ 150 N.

Thus the gravitational force = 150 N

The correct option is a. 150 N

5 0
2 years ago
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