How much is a city of a body when it covers 600m in 5 min?

Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the eq

gayaneshka [121]

Answer:

The volume is decreasing at 160 cm³/min

Explanation:

Given;

Boyle's law, PV = C

where;

P is pressure of the gas

V is volume of the gas

C is constant

Differentiate this equation using product rule:

$V\frac{dp}{dt} +P\frac{dv}{dt} = \frac{d(C)}{dt}$

Given;

$\frac{dP}{dt}$ (increasing pressure rate of the gas) = 40 kPa/min

V (volume of the gas) = 600 cm³

P (pressure of the gas) = 150 kPa

Substitute in these values in the differential equation above and calculate the rate at which the volume is decreasing ( $\frac{dv}{dt}$ );

(600 x 40) + (150 x $\frac{dv}{dt}$ ) = 0

$\frac{dv}{dt} = -\frac{(600*40)}{150} = -160 \ cm^3/min$

Therefore, the volume is decreasing at 160 cm³/min

3 0

3 years ago

If an object falling freely were somehow equipped with an odometer to measure the distance it travels, then the amount of distan

ioda

Answer:c

Explanation:

Given

object is falling Freely with an odometer

Suppose it falls with zero initial velocity

so distance fallen in time t is given by

$h=ut+\frac{1}{2}gt^2$

here u=0 and t=time taken

$h=\frac{1}{2}gt^2$

for $t=1 s$

$h_1=\frac{1}{2}g$

for $t=2 s$

$h_2=\frac{1}{2}g(2)^2=\frac{4}{2}g=2g$

distance traveled in 2 nd sec $=2g-\frac{1}{2}g=\frac{3}{2}g$

for $t=3 s$

$h_3=\frac{1}{2}g(3)^2=\frac{9}{2}g$

distance traveled in 3 rd sec $=\frac{9}{2}g-2g=\frac{5}{2}g$

so we can see that distance traveled in each successive second is increasing

5 0

3 years ago

Which conditions are necessary for clouds to form?

VikaD [51]

Answer: Some of the thing that are important for clouds to form are: Moisture - There must be sufficient water vapor in the air for a cloud to form. Cooling air - The air temperature must decrease enough for water vapor to condense

Explanation:

7 0

3 years ago

Read 2 more answers

How do you think overpumping groudwater is related to the formation of sinkholes?

vladimir2022 [97]

Ground water keept the ground at a stable level when it is gone the cavern it was in has no support and is at risk of callaps

3 0

3 years ago

A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t

Yuliya22 [10]

Answer:

a) $W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

b) $W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

c) $V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

d) $d_1 =0.183m$ or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

$W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

Part b

For this case first we can convert the spring constant to N/m like this:

$2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}$

And the work donde by the spring on this case is given by:

$W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

$W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i$

And if we solve for the initial velocity we got:

$V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

$-1/2mV^2_i = W_g + W_{spring}$

And replacing we got:

$-1/2mV^2_i =mg d_1 -1/2 k d^2_1$

And we can put the terms like this:

$\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0$

If we multiply all the equation by 2 we got:

$k d^2_1 -2 mg d_1 -m V^2_i =0$

Now we can replace the values and we got:

$200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0$

$200 d^2_1 -2.058 d_1 -6.3525=0$

And solving the quadratic equation we got that the solution for $d_1 =0.183m$ or 18.3 cm because the negative solution not make sense.

5 0

3 years ago