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3 years ago

7

An object that is dropped straight down from a height of 100 m has a vertical change in position that is less than that of an id

entical object thrown horizontally from the same height. true or false?

1 answer:

sergey [27]3 years ago

5 0

Objects dropped straight or thrown horizontally from the same height
change their vertical velocity at the same rate, and fall through equal
vertical distances in equal time intervals.

The statement is false.

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A permeability test was run on a compacted sample of dirty sandy gravel. The sample was 175 mm long and the diameter of the mold

LUCKY_DIMON [66]

Answer:

(a). The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

Explanation:

Given that,

Length = 175 mm

Diameter = 175 mm

Time = 90 sec

Volume= 405 cm³

We need to calculate the discharge

Using formula of discharge

$Q=\dfrac{V}{t}$

Put the value into the formula

$Q=\dfrac{405}{90}$

$Q=4.5\ cm^3/s$

(a). We need to calculate the coefficient of permeability

Using formula of coefficient of permeability

$Q=kiA$

$k=\dfrac{Q}{iA}$

$k=\dfrac{Ql}{Ah}$

Where, Q=discharge

l = length

A = cross section area

h=constant head causing flow

Put the value into the formula

$k=\dfrac{4.5\times175\times10^{-1}}{\dfrac{\pi(175\times10^{-1})^2}{4}\times38}$

$k=8.6\times10^{-3}\ cm/s$

The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(c). We need to calculate the discharge velocity during the test

Using formula of discharge velocity

$v=ki$

$v=\dfrac{kh}{l}$

Put the value into the formula

$v=\dfrac{8.6\times10^{-3}\times38}{17.5}$

$v=0.0187\ cm/s$

The discharge velocity during the test is 0.0187 cm/s.

(b). We need to calculate the volume of solid in the ample

Using formula of volume

$V_{s}=\dfrac{M_{s}}{V_{s}}$

Put the value into the formula

$V_{s}=\dfrac{4950\times10^{-3}}{2710}$

$V_{s}=1826.56\ cm^3$

We need to calculate the volume of the soil specimen

Using formula of volume

$V=A\times L$

Put the value into the formula

$V=\dfrac{\pi(17.5)^2}{4}\times17.5$

$V=4209.24\ cm^3$

We need to calculate the volume of the voids

$V_{v}=V-V_{s}$

Put the value into the formula

$V_{v}=4209.24-1826.56$

$V_{v}=2382.68\ cm^3$

We need to calculate the seepage velocity

Using formula of velocity

$Av=A_{v}v_{s}$

$v_{s}=\dfrac{Av}{A_{v}}$

$v_{s}=\dfrac{V}{V_{v}}\times v$

Put the value into the formula

$v_{s}=\dfrac{4209.24}{2382.68}\times0.0187$

$v_{s}=0.0330\ cm/s$

The seepage velocity is 0.0330 cm/s.

Hence, (a). The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

8 0

3 years ago

A sample of Bismuth-212 has a mass of 2.64 grams (g) after 121 seconds (s). What was the initial mass of the sample if Bismuth-2

uysha [10]

The mass of a radioactive element at time t is given by
$m(t) = m_0 ( \frac{1}{2} )^{ \frac{t}{t_{1/2}} }$
where $m_0$ is the mass at time zero, while $t_{1/2}$ is the half-life of the element.

In our problem, $m(t)=2.64 g$ , t=121.0 s and $t_{1/2}=60.5 s$ , so we can find the initial mass $m_0$ :
$m_0= \frac{m(t)}{ (\frac{1}{2})^{t/t_{1/2}} } = \frac{2.64 g}{( \frac{1}{2} )^{121/60.5}} =4 \cdot 2.64 g=10.56 g$

4 0

3 years ago

Suppose the maximum power delivered by a car's engine results in a force of 16000 N on the car by the road. In the absence of an

joja [24]

Answer:

Approximately $9.7\; \rm m \cdot s^{-2}$ .

Explanation:

Assuming that there is no other force on this vehicle, the $16000\; \rm N$ force from the road would be the only force on this vehicle. The net force would then be equal to this $16000\; \rm N\!$ force. The size of the net force would be $16000\; \rm N\!\!$ .

Let $m$ denote the mass of this vehicle and let $\Sigma F$ denote the net force on this vehicle.

By Newton's Second Law of motion, the acceleration of this vehicle would be proportional to the net force on this vehicle. In other words, the acceleration of this vehicle, $a$ , would be:

$\begin{aligned}a &= \frac{\Sigma F}{m}\end{aligned}$ .

For this vehicle, $\Sigma F = 16000\; \rm N$ whereas $m = 1650\; \rm kg$ . The acceleration of this vehicle would be:

$\begin{aligned}a &= \frac{16000\; \rm N}{1650\; \rm kg} \\ &= \frac{16000\; \rm kg \cdot m\cdot s^{-2}}{1650\; \rm kg}\\ &\approx 9.7 \; \rm m \cdot s^{-2}\end{aligned}$ .

8 0

3 years ago

If a 8.1g ring is heated using 10.0 ), its temperature rises 21.7°C. Calculate the specific

lakkis [162]

the water specific heat will remain at 4.184.

7 0

3 years ago

Explain the following defects of a simple electric cell:

eimsori [14]

Answer:

Explanation:

The two major defects of simple electric cells causes current supplied to be for short time. These defects are: polarization and local action.

a. Polarization: This is a defect caused by an accumulation of hydrogen bubbles at the positive electrode of the cell. It can be prevented by the use of vent, using a hydrogen absorbing material or the use of a depolarizer.

b. Local Action: This is the gradual wearing away of the electrode due to impurities in the zinc plate. It can be controlled by the amalgamation of the zinc plate before it is used.

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3 years ago