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3 years ago

How far can a person run in 15 minutes if he runs at an average speed of 16 km/hr?

Physics

1 answer:

anygoal [31]3 years ago

3 0

Answer:

4km

Explanation:

15 minutes is 1/4 of an hour.

1/4 of 16 is 4.

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A student hangs a weight on a newtonmeter. The energy currently stored in the spring in the newton meter is 0.045N. The student

castortr0y [4]

Answer:

5x10^-3

Explanation:

Hooke's Law states that the force needed to compress or extend a spring is directly proportional to the distance you stretch it.

Hooke's Law can be represented as

<em>where F is the force </em>

<em> k is the spring constant</em>

<em> x is the extension of the material </em>

<em />

Plug values in the equation

Step 1 find the original extension

0.045 = (400)x

x = 1.125x 10^-4 m d

Step 2 find the new extension

0.045+2 = 400(x)

2.045 = 400x

x = 5.1125x10^-3

Step 3 subtract the new extension with original

Total extension of the spring = 5.1125x10^-3 - 1.125x 10^-4 m = 5x10^-3

8 0

3 years ago

What do scientist use to determine the temperature of a star

REY [17]

Color is what scientist use to determine the temperature of a star!

hope this helps!

4 0

3 years ago

What is the equation for the enthalpy of vaporization?

Natalka [10]

Answer: Use the formula q = m·ΔHv in which q = heat energy, m = mass, and ΔHv = heat of vaporization.

4 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

3 years ago

When a metal bar is heated the increase in length is greater if

lorasvet [3.4K]

Answer:

Explanation:

i think the answer is A

6 0

3 years ago