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3 years ago

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Một ô tô đi với vận tốc 60 km/h trên nửa phần đầu của đoạn đường AB. Trong nửa đoạnđường còn lại ô tô đi nửa thời gian đầu với v

ận tốc 40 km/h và nửa thời gian sau với vận tốc 20 km/h. Tính tốc độ trung bình của ô tô ?

1 answer:

melamori03 [73]3 years ago

4 0

Answer:

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Suppose that a balloon is being filled with air at a rate of 10 cm3/s. (Assume that theballoon is a perfect sphere.) At what rat

Basile [38]

Answer:

Therefore the surface area of the balloon is increased at 4 cm³/s.

Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

It means the volume of the balloon is increased at a rate 10 cm³/s.

i.e $\frac{dv}{dt} =10 cm^3/s$

Consider r be the radius of the balloon.

The volume of of a sphere is

$v=\frac{4}{3} \pi r^3$

Differentiate with respect to t

$\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}$

$\Rightarrow 10 =4\pi r^2\frac{dr}{dt}$

$\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}$

The surface of area of the balloon is(S) = $4\pi r^2$

$S=4\pi r^2$

Differentiate with respect to t

$\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}$

Putting the value of $\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}$

$\Rightarrow \frac{dS}{dt} =\frac{20}{ r}$

Given that r = 5 cm

$[\frac{dS}{dt}]_{r=5} =\frac{20}{ 5}$ =4 cm³/s

Therefore the surface area of the balloon is increased at 4 cm³/s.

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3 years ago

The pressure of a gas is reduced from 1200.0 mm hg to 850.0 mm hg as the volume of its container is increased by moving a piston

Kamila [148]

From gas laws:
$\frac{PV}{T} = Constant$

Therefore,
$\frac{ P_{1} V_{1} }{ T_{1} } = \frac{ P_{2} V_{2} }{ T_{2} }$

P1 = 1200 mm
V1 = 85 ml
T1 = 90°C = 363.15 K
P2 = 850 mm
V2 = 350 ml
T2 = ?

Substituting;
$T_{2} = \frac{ P_{2} V_{2} T_{1} }{ P^{1} V^{1} } = \frac{850*350*363.15}{1200*85} = 1059.19 K$ = 786.04 °C

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3 years ago

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A meter stick moves parallel to its axis with speed of 0.96 c relative to you. What would you measure for the length of the stic

Fed [463]

Answer:

The length of the stick is 0.28 m.

The time the stick take to move is 0.97 ns.

Explanation:

Given that,

Relative speed of stick v= 0.96 c

Speed of light $c= 2.99793\times10^{8}\ m/s$

Proper length of stick = 1 m

We need to calculate the length of the stick

Using formula of length

$\Delta l=\Delta l_{0}\sqrt{(1-\dfrac{v^2}{c^2})}$

Put the value into the formula

$\Delta l=1\sqrt{1-\dfrac{(0.96)^2c^2}{c^2}}$

$\Delta l=1\sqrt{1-(0.96)^2}$

$\Delta l=0.28\ m$

We need to calculate the time the stick take to move

Using formula of time

$t=\dfrac{\Delta l}{v}$

Put the value into the formula

$t=\dfrac{0.28}{0.96\times(2.99793\times10^{8})}$

$t=9.72\times10^{-10}\ sec$

$t=0.97\ ns$

Hence, The length of the stick is 0.28 m.

The time the stick take to move is 0.97 ns.

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3 years ago

It is recommended that children have no more than two sets of dental X-rays per year. Although X-rays have many benefits, they c

guapka [62]

Although X-rays have many benefits, they can be dangerous to human health because they are short wavelength, high frequency, with photons sufficiently energetic to damage DNA molecules.

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3 years ago

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Which jovian planet should have the most extreme seasonal changes? a. Saturn b. Neptune c. Jupiter d. Uranus

Maksim231197 [3]

Answer:

D). Uranus.

Explanation:

Jovian planets are described as the planets which are giant balls of gases and located farthest from the sun which primarily include Jupiter, Saturn, Uranus, and Neptune.

As per the question, 'Uranus' is the jovian planet that would have the most extreme seasonal changes as its tilted axis leads each season to last for about 1/4 part of its 84 years orbit. The strong tilted axis encourages extreme changes in the season on Uranus. Thus, <u>option D</u> is the correct answer.

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3 years ago