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2 years ago

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A 2 000-kg car is slowed down uniformly from 20.0 m/s to 5.00 m/s in 4.00 s. (a) What average force acted on the car during that

time, and (b) how far did the car travel during that time

1 answer:

slava [35]2 years ago

3 0

Answer:

the answer is c

Explanation:

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What is the value of g on the surface of Saturn? Assume M-Saturn = 5.68×10^26 kg and R-Saturn = 5.82×10^7 m.Choose the appropria

Likurg_2 [28]

Answer:

Approximately $\rm 11.2 \; N \cdot kg^{-1}$ at that distance from the center of the planet.

Option A) The low value of $g$ near the cloud top of Saturn is possible because of the low density of the planet.

Explanation:

The value of $g$ on a planet measures the size of gravity on an object for each unit of its mass. The equation for gravity is:

$\displaystyle \frac{G \cdot M \cdot m}{R^2}$ ,

where

$G \approx 6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2$ .
$M$ is the mass of the planet, and
$m$ is the mass of the object.

To find an equation for $g$ , divide the equation for gravity by the mass of the object:

$\displaystyle g = \left.\frac{G \cdot M \cdot m}{R^2} \right/\frac{1}{m} = \frac{G \cdot M}{R^2}$ .

In this case,

$M = 5.68\times 10^{26}\; \rm kg$ , and
$R = 5.82 \times 10^7\; \rm m$ .

Calculate $g$ based on these values:

$\begin{aligned} g &= \frac{G \cdot M}{R^2}\cr &= \frac{6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2\times 5.68\times 10^{26}\; \rm kg}{\left(5.82\times 10^7\; \rm m\right)^2} \cr &\approx 11.2\; \rm N\cdot kg^{-1} \end{aligned}$ .

Saturn is a gas giant. Most of its volume was filled with gas. In comparison, the earth is a rocky planet. Most of its volume was filled with solid and molten rocks. As a result, the average density of the earth would be greater than the average density of Saturn.

Refer to the equation for $g$ :

$\displaystyle g = \frac{G \cdot M}{R^2}$ .

The mass of the planet is in the numerator. If two planets are of the same size, $g$ would be greater at the surface of the more massive planet.

On the other hand, if the mass of the planet is large while its density is small, its radius also needs to be very large. Since $R$ is in the denominator of $g$ , increasing the value of $R$ while keeping $M$ constant would reduce the value of $g$ . That explains why the value of $g$ near the "surface" (cloud tops) of Saturn is about the same as that on the surface of the earth (approximately $9.81\; \rm N \cdot kg^{-1}$ .

As a side note, $5.82\times 10^7\rm \; m$ likely refers to the distance from the center of Saturn to its cloud tops. Hence, it would be more appropriate to say that the value of $g$ near the cloud tops of Saturn is approximately $\rm 11.2 \; N \cdot kg^{-1}$ .

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2 years ago

Three wires are made of copper having circular cross sections. Wire 1 has a length l and radius r. Wire 2 has a length l and rad

Alex73 [517]

Explanation:

Below is an attachment containing the solution.

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2 years ago

A car travels at a constant rate for 25 miles, going due east for one hour. Then it travels at a constant rate another 60 miles

egoroff_w [7]

60 mph east...........

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2 years ago

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For a Physics course containing 10 students, the maximum point total for the quarter was 200. The point totals for the 10 studen

Talja [164]

Answer:

130.5

Explanation:

According to the stemplot attached (Which I think it is, and if not, then you only need to replace the procedure with your data and you should be fine), you need to calculate first the points of all ten students. In that plot, we can easily calculate the points.

The first number in the colum represents the centen of the point, while the numbers of the second column, represents the units of that centen, for example if you see:

16 | 8 5 6

This means that the point for the students are 168, 165 and 166. Three students, three points.

If you watch the stemplot, the points for the students are:

116, 118, 121, 124, 128, 133, 137, 142, 146 and 179.

The median can be calculated using the mean between the two values in the middle of the sequence.

In this case, half of ten is 5, so, the numbers from the middle in this sequence are 128 and 133, therefore:

Median = 128 + 133 / 2 = 130.5

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2 years ago

calculate the time rate of change in air density during expiration. Assume that the lung has a total volume of 6000mL, the diame

kipiarov [429]

Answer:

The time rate of change in air density during expiration is 0.01003kg/m³-s

Explanation:

Given that,

Lung total capacity V = 6000mL = 6 × 10⁻³m³

Air density p = 1.225kg/m³

diameter of the trachea is 18mm = 0.018m

Velocity v = 20cm/s = 0.20m/s

dv /dt = -100mL/s (volume rate decrease)

= 10⁻⁴m³/s

Area for trachea =

$\frac{\pi }{4} d^2\\= 0.785\times 0.018^2\\= 2.5434 \times10^-^4m^2$

0 - p × Area for trachea =

$\frac{d}{dt} (pv)=v\frac{ds}{dt} + p\frac{dv}{dt}$

$-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)$

$-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)$

⇒ $-0.623133\times10^-^4+1.225\times10^-^4=6\times10^-^3\frac{ds}{dt}$

$\frac{ds}{dt} = \frac{0.6018\times10^-^4}{6\times10^-^3} \\\\= 0.01003kg/m^3-s$

ds/dt = 0.01003kg/m³-s

Thus, the time rate of change in air density during expiration is 0.01003kg/m³-s

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2 years ago

Read 2 more answers