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3 years ago

10

A spring of spring constant 25 N/m is hung vertically and a 0.300 kg mass is attached to one end, causing a displacement of the

end of the spring of _____ m.

1 answer:

Mamont248 [21]3 years ago

3 0

Answer:k=1175

Explanation:thank you for asking

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A 2kg mass is initially at rest on a frictionless surface. A 45N force acts at an angle of 300

raketka [301]

Answer:

a) The work done by the force is 136.400 joules.

b) The speed of the mass at the end of the 3.5 meters is approximately 11.679 meters per second.

Explanation:

The correct statement is shown below:

<em>A 2-kg mass is initially at reat on a frictionless surface. A 45 N force acts at an angle of 30º to the horizontal for a distance of 3.5 meters:</em>

<em>a)</em><em> Find the work done by the force.</em>

<em>b)</em><em> Find the speed of the mass at the end of the 3.5 meters.</em>

a) Given that a external constant force is acting on the mass on a frictionless surface, the work done by the force ( $W_{F}$ ), measured in joules, is:

$W_{F} = F\cdot \Delta s \cdot \cos \theta$ (1)

Where:

$F$ - External constant force exerted on the mass, measured in newtons.

$\Delta s$ - Horizontal travelled distance, measured in meters.

$\theta$ - Direction of the external force regarding the horizontal, measured in sexagesimal degrees.

If we know that $F = 45\,N$ , $\Delta s = 3.5\,m$ and $\theta = 30^{\circ}$ , then the work done by the force is:

$W_{F} = (45\,N)\cdot (3.5\,m)\cdot \cos 30^{\circ}$

$W_{F} = 136.400\,J$

The work done by the force is 136.400 joules.

b) The final speed of the mass at the end ot the 3.5 meter is calculated by means of the Work-Energy Theorem, which means that the work done on the mass is transformed into a translational kinetic energy. That is:

$W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2})$ (2)

Where:

$m$ - Mass, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speeds of the mass, measured in meters per second.

If we know that $W_{F} = 136.400\,J$ , $m = 2\,kg$ and $v_{1} = 0\,\frac{m}{s}$ , then the final speed of the mass is:

$\frac{2\cdot W_{F}}{m} = v_{2}^{2}-v_{1}^{2}$

$v_{2}^{2} =v_{1}^{2}+\frac{2\cdot W_{F}}{m}$

$v_{2} =\sqrt{v_{1}^{2}+\frac{2\cdot W_{F}}{m} }$

$v_{2} =\sqrt{\left(0\,\frac{m}{s} \right)^{2}+\frac{2\cdot (136.400\,J)}{2\,kg} }$

$v_{2} \approx 11.679\,\frac{m}{s}$

The speed of the mass at the end of the 3.5 meters is approximately 11.679 meters per second.

3 0

3 years ago

How do the earth's physical features affect people's activities?

masya89 [10]

Answer:

d.all of the above

Explanation:

The physical feature of the planet Earth affect people's activities in the following manner -

Choice of Occupation - the process of occupation is completely a land depended process , hence any changes on the physical features of Earth , like the climate , temperature , rain etc , will hinder the occupations process .
Method of transportation - it is also a Climate dependent process , so any change in climate may obstruct the transportation facility .

Recreational activities - also depends on the physical feature of the planet Earth .

4 0

3 years ago

Problem 10.64 the maximum energy a bone can absorb without breaking is surprisingly small. for a healthy human of mass 67 kg , e

kvasek [131]

Answer: h = 0.30 m

Explanation:

A person jumping from height h would possess potential energy = m g h

which will convert completely into kinetic energy as person hits the ground. Now, the maximum energy absorbed by the person can be = 200 J

m = 67 kg

g = 9.8 m/s²

⇒ m g h = 200 J

⇒ h = 200 J / (67 kg × 9.8 m/s²) = 0.30 m

Hence, a person can land safely on both legs without breaking them from a height of 0.30 m only.

8 0

3 years ago

Which two of the following reduce the background noise in magneto- medicine?

horrorfan [7]

Answer:

The answer is b.) and d.)

Explanation:

The options to reduce the background noise in magneto-medicine are given as follows:

a.) Orienting the heart parallel to the Earth's field

- This will have no significant effect on the measurements.

b.) Taking the difference of two nearby sensor measurements (gradiometer).

-This answer is TRUE.

c.) Placing the heart in a perpendicular fashion to the Earth's magnetic field.

_ This answer also will not have any significant effect on measurements.

d.) Using physical means to shield environmental fields.

- This answer is TRUE.

4 0

3 years ago

72) What is the freezing point (°C) of a solution prepared by dissolving 11.3 g of Ca(NO3)2 (formula weight = 164 g/mol) in 115

Rina8888 [55]

<u>Answer:</u> The freezing point of solution is -3.34°C

<u>Explanation:</u>

Vant hoff factor for ionic solute is the number of ions that are present in a solution. The equation for the ionization of calcium nitrate follows:

$Ca(NO_3)_2(aq.)\rightarrow Ca^{2+}(aq.)+2NO_3^-(aq.)$

The total number of ions present in the solution are 3.

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(Ca(NO_3)_2)$ = 11.3 g

$M_{solute}$ = Molar mass of solute $(Ca(NO_3)_2)$ = 164 g/mol

$W_{solvent}$ = Mass of solvent (water) = 115 g

Putting values in above equation, we get:

$\text{Molality of }Ca(NO_3)_2=\frac{11.3\times 1000}{164\times 115}\\\\\text{Molality of }Ca(NO_3)_2=0.599m$

To calculate the depression in freezing point, we use the equation:

$\Delta T=iK_fm$

where,

i = Vant hoff factor = 3

$K_f$ = molal freezing point depression constant = 1.86°C/m.g

m = molality of solution = 0.599 m

Putting values in above equation, we get:

$\Delta T=3\times 1.86^oC/m.g\times 0.599m\\\\\Delta T=3.34^oC$

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

$\Delta T=\text{freezing point of water}-\text{freezing point of solution}$

$\Delta T$ = 3.34 °C

Freezing point of water = 0°C

Freezing point of solution = ?

Putting values in above equation, we get:

$3.34^oC=0^oC-\text{Freezing point of solution}\\\\\text{Freezing point of solution}=-3.34^oC$

Hence, the freezing point of solution is -3.34°C

4 0

4 years ago