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3 years ago

As its speed increases, what happens to its energy of motion?

Physics

1 answer:

irina [24]3 years ago

5 0

Answer:

The energy of motion increases

Explanation:

The reason the energy of motion increases is because as the speed increases there is more motion and more energy

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An object of mass 2.0 kg is attached to the top of a vertical spring that is anchored to the floor. The unstressed length of the

poizon [28]

Answer:

The value is $A = 0.014 \ m$

Explanation:

From the question we are told that

The mass of the object is $m = 2.0 \ kg$

The unstressed length of the string is $l = 0.08 \ m$

The length of the spring when it is at equilibrium is $l_e = 5.9 \ cm = 0.059 \ m$

The initial speed (maximum speed)of the spring when given a downward blow $v = 0.30 \ m/s$

Generally the maximum speed of the spring is mathematically represented as

$u = A * w$

Here A is maximum height above the floor (i.e the maximum amplitude)

and $w$ is the angular frequency which is mathematically represented as

$w = \sqrt{\frac{k}{m} }$

$u = A * \sqrt{\frac{k}{m} }$

=> $A = u * \sqrt{\frac{m}{k} }$

Gnerally the length of the compression(Here an assumption that the spring was compressed to the ground by the hammer is made) by the hammer is mathematically represented as

$b = l -l_e$

=> $b = 0.08 - 0.05 9$

=> $b = 0.021 \ m$

Generally at equilibrium position the net force acting on the spring is

$k * b - mg = 0$

=> $k * 0.021 - 2 * 9.8 = 0$

=> $k = 933 \ N/m$

$A = 0.30 * \sqrt{\frac{2}{933} }$

=> $A = 0.014 \ m$

8 0

3 years ago

George walks to a friend's house. He walks 850 meters north before he realizes he walked too far.

anyanavicka [17]

Answer:

Explanation:

In this problem we are expected to calculate the velocity of Georges movements.

Given data

Total distance covered by George= 850+250= 1100 meters

Time taken by George to cover the total distance= 25 seconds

We know that velocity is, v= distance/ time

Therefore substituting our data into the expression for velocity we have

v= 1100/ 25= 44 m/s

Hence the velocity in m/s is 44

7 0

3 years ago

A simple pendulum has length of 820mm. Calculate the frequency (g = 9.8 ms -2)<br>

Vadim26 [7]

Answer:

$\huge\boxed{\sf f=0.55 \ Hz}$

Explanation:

<u>Given Data:</u>

Length = l = 820 mm = 0.82 m

Acceleration due to gravity = g = 9.8 ms⁻²

<u>Required:</u>

Frequency = f = ?

<u>Formula:</u>

$\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} }$

<u>Solution:</u>

$\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} } \\\\Put\ the\ givens\\\\f=\frac{1}{2 \pi} \sqrt{\frac{9.8}{0.82} }\\\\ f = 0.159 \times \sqrt{11.95} \\\\f=0.159 \times 3.457\\\\f=0.55 \ Hz\\\\\rule[225]{225}{2}$

7 0

2 years ago

How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe

Rus_ich [418]

Explanation:

(a) For an isothermal process, work done is represented as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

Putting the given values into the above formula as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

= $- 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})$

= $-12280.82 \times ln (0.09)$

= $-12280.82 \times -2.41$

= 29596.78 J

or, = 29.596 kJ (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

W = $\frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}$

= $\frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}$

= $\frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}$

= 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c) We know that for an isothermal process,

$P_{1}V_{1} = P_{2}V_{2}$

or, $P_{2} = \frac{P_{1}V_{1}}{V_{2}}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})$

= 11 atm

Hence, the required pressure is 11 atm.

(d) For adiabatic process,

$P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}$

or, $P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}$

= 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0

4 years ago

Two factors that affect the speed of sound in media are how stiff a material is and blank??

alexgriva [62]

The speed of sound is influenced by several factors, including medium, density and temperature. The rate at which sound waves moves varies widely from one situation to the next and can change dramatically in a short period of time.

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3 years ago