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2 years ago

A rural mail carrier is driving slowly, putting mail in mailboxes near the road

Physics

1 answer:

kow [346]2 years ago

4 0

Incomplete question. I inferred you want access to the full question.

Explanation:

A rural mail carrier is driving slowly, putting mail in mailboxes near the road. He overshoots one mailbox, stops, shifts into reverse, and then backs up until he is at the right spot. The velocity graph given below represents his motion.

a. draw the mail carrier's position vs time graph. Assume that x = 0 m at t = 0 s.

b. what is the position of the mailbox?

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A glass jug can be used to play different pitched sounds by blowing air over the opening of the jug and vibrating the air molecu

Galina-37 [17]

Add more of the liquid because the length is more redemption.

6 0

2 years ago

Question 5 At 12:00 pm, a spaceship is at position ⎡⎣324⎤⎦ km ⎣ ⎢ ⎡ 3 2 4 ⎦ ⎥ ⎤ km away from the origin with respect to so

Anettt [7]

Answer:

[1, 6, -2]

Explanation:

Given the following :

Initial Position of spaceship : [3 2 4] km

Velocity of spaceship : [-1 2 - 3] km/hr

Location of ship after two hours have passed :

Distance moved by spaceship :

Velocity × time

[-1 2 -3] × 2 = [-2 4 -6]

Location of ship after two hours :

Initial position + distance moved

[3 2 4] + [-2 4 -6] = [3 + (-2)], [2 + 4], [4 + (-6)]

= [3-2, 2+4, 4-6] = [1, 6, -2]

4 0

2 years ago

The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371

arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

$q=m\times c\times (T_2-T_1)$

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = $4.18J/g.K$

$T_1$ = initial temperature of water = 293.0 K

$T_2$ = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

$q=250g\times 4.18J/g.K\times (371.2-293.0)K$

$q=81719J$

Now we have to calculate the enthalpy of combustion of octane.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

$\Delta H=\frac{-81719J}{0.015mole}$

$\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol$

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

5 0

2 years ago

A ball is dropped from the top of a building.After 2 seconds, it’s velocity is measured to be 19.6 m/s. Calculate the accelerati

zlopas [31]

Answer:

acceleration, a = 9.8 m/s²

Explanation:

'A ball is dropped from the top of a building' indicates that the initial velocity of the ball is zero.

u = 0 m/s

After 2 seconds, velocity of the ball is 19.6 m/s.

t = 2s, v = 19.6 m/s

Using

v = u + at

19.6 = 0 + 2a

a = 9.8 m/s²

6 0

2 years ago

Read 2 more answers

In a house the temperature at the surface of a window is 28.9 °C. The temperature outside at the window surface is 7.89 °C. Heat

Alenkasestr [34]

Answer:

-13.18°C

Explanation:

To develop the problem it is necessary to consider the concepts related to the thermal conduction rate.

Its definition is given by the function

$\frac{Q}{t} = \frac{kA\Delta T}{d}$

Where,

Q = The amount of heat transferred

t = time

k = Thermal conductivity constant

A = Cross-sectional area

$\Delta T =$ The difference in temperature between one side of the material and the other

d= thickness of the material

The problem says that there is a loss of heat twice that of the initial state, that is

$Q_2 = 2*Q_1$

Replacing,

$kA\frac{\Delta T_m}{x} = 2*kA\frac{\Delta T}{x}$

$\frac{\Delta T}{x}=2*\frac{\Delta T}{x}$

$\frac{T_i-T_o}{x} = 2\frac{T_1-T_2}{x}$

$\frac{28.9-T_o}{x} = 2\frac{28.9-7.86}{x}$

Solvinf for $T_o$ ,

$T_o = -13.18$

Therefore the temprature at the outside windows furface when the heat lost per second doubles is -13.18°C

3 0

2 years ago