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3 years ago

Can a runner's average speed be different from her average velocity?

Physics

1 answer:

Tomtit [17]3 years ago

5 0

Answer:

Yes

Explanation:

Since velocity is a vector, meaning it also relies on direction, the average speed can be different from her average velocity. An example would be if a runner turned around and ran backward after running 10 meters and returned to her starting point. If you took her average velocity of the entire trip it would actually be 0 but her average speed obviously would not be. This is why velocity can be negative but speed cannot.

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A 50 g sample of an unknown metal is heated to 90.0C. It is placed in a perfectly insulated container along with 100 g of water

Scrat [10]

Answer:

0.64 J/g°C

Explanation:

Using the formula;

Q = m × c × ∆T

Where;

Q = amount of heat

m = mass (g)

c = specific heat capacity

∆T = change in temperature (°C)

In this case:

Q (water) = - Q (metal)

mc∆T (water) = - mc∆T (metal)

According to the information in this question,

For water; m = 100g, c = 4.18J/g°C, ∆T = (25°C - 20°C)

For metal; m = 50g, c =?, ∆T = (25°C - 90°C)

mc∆T (water) = - mc∆T (metal)

100 × 4.18 × (25°C - 20°C) = - {50 × c × (25°C - 90°C)}

100 × 4.18 × 5 = - {50 × c × -65}

2090 = -{-3250c}

2090 = 3250c

c = 2090/3250

c = 0.643

c = 0.64J/g°C

7 0

3 years ago

Read 2 more answers

Anders suffered a shock when his electric radio dropped into the bathtub--while anders was taking a bath. Anders argued that he

blsea [12.9K]

The claim Anders is most likely to make is the failure of the manufacturer to warn about such risk.

This is defined as the possibility of something bad happening and in this case it is electric shock when dropped into the bathtub.

Anders can decide to sue for not warning against risk of electric shock when in contact with water.

Read more about Risk here brainly.com/question/1224221

3 0

2 years ago

Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.370 mm wide. The diffraction pattern is observed

erica [24]

Answer:

Δx = 6.33 x 10⁻³ m = 6.33 mm

Explanation:

We can use the Young's Double Slit Experiment Formula here:

$\Delta x = \frac{\lambda L}{d}\\\\$

where,

Δx = distance between consecutive dark fringes = width of central bright fringe = ?

λ = wavelength of light = 633 nm = 6.33 x 10⁻⁷ m

L = distance between screen and slit = 3.7 m

d = slit width = 0.37 mm = 3.7 x 10⁻⁴ m

Therefore,

$\Delta x = \frac{(6.33\ x\ 10^{-7}\ m)(3.7\ m)}{3.7\ x \ 10^{-4}\ m}$

8 0

3 years ago

Complete the electron-pushing mechanism for the e1 reaction when 2-methylbutan-2-ol is treated with 20% sulfuric acid.

Sav [38]

E1 reaction works in the mechanism that the removal of an HX substituent results in the formation of a double bond. The E1 reaction for 2-methylbutan-2-ol is shown in the figure. This reaction is called acid-catalyzed dehydration of a tertiary alcohol.

The mechanism works in three major steps:
1. The OH group of the main reactant is hydrated by H2SO4 so it becomes H2O.
2. The H2O leaves taking electrons with it. This results to a carbocation intermediate on the carbon atom where it was attached.
3. Another H2O protonates the beta carbon. This is the carbon atom next to the carbocation. It will donate its electrons to the neighboring C-C bond, as indicated by the arrow. The carbons are rehybridized from sp3 to sp2, which is a pi bond. As a result, a double bond forms.

The product is 2-methyl-2-butene.

6 0

4 years ago

Read 2 more answers

Two musicians are comparing their trombones. The first produces a tone that is known to be 438 Hz. When the two trombones play t

galben [10]

Answer:

It is producing either a 435-Hz sound or a 441-Hz sound.

Explanation:

When two sound of slightly different frequencies interfere constructively with each other, the resultant wave has a frequency (called beat frequency) which is equal to the absolute value of the difference between the individual frequencies:

$f_B = |f_1 -f_2|$ (1)

In this problem, we know that:

- The frequency of the first trombone is $f_1 = 438 Hz$

- 6 beats are heard every 2 seconds, so the beat frequency is

$f_B=\frac{6}{2 s}=3 Hz$

If we insert this data into eq.(1), we have two possible solutions for the frequency of the second trombone:

$f_2 = f_1 - f_B = 438 Hz - 3 Hz = 435 Hz\\f_2 = f_1+f_B = 438 Hz+3 Hz=441 Hz$

7 0

3 years ago