What is the gradual process through which humans change from birth to<br> adulthood?

How long would it take a drag racer to increase her speed from 10.m/s to 20 m/s if her car accelerates at a uniform rate of 15 m

garik1379 [7]

It would take about 2 thirds of a second or .66666666 repeating of a second. please give brainliest?

5 0

3 years ago

Value of g is independent of

Margaret [11]

The lord of the greeks answer d

5 0

3 years ago

Read 2 more answers

One hazard of space travel is the debris left by previous missions. There are several thousand objects orbiting Earth that are l

MariettaO [177]

Answer:

F = 6666.7 N

Explanation:

Given that,

Mass of a chip, m = 0.1 mg

Initial speed, u = 0

Final speed, $v=4\times 10^{3}\ m/s$

Time of collision, $t=6\times 10^{-8}\ s$

We know that,

Force, F = ma

Put all the values,

$F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.1\times 10^{-6}\times (4\times 10^3-0)}{6\times 10^{-8}}\\\\F=6666.7\ N$

So, the required force is 6666.7 N.

3 0

3 years ago

A piston-cylinder device contains Helium gas initially at 150 kPa, 20 o C, and 0.5m 3 . The helium is now compressed in a polytr

Molodets [167]

Answer:

Explanation:

Given

$P_1=150 kPa$

$T_1=20^{\circ}C$

$V_1=0.5 m^3$

$T_2=140^{\circ}C$

$P_2=400 kPa$

R for Helium $R=2.076$

$c_v=3.115 kJ/kg-K$

mass of gas $m=\frac{P_1V_1}{RT_1}$

$m=\frac{150\times 0.5}{2.076\times 293}$

$m=0.123 kg$

Similarly $V_2$ can be found

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

$V_2=0.264 m^3$

Work done $W=\int_{V_1}^{V_2}PdV$

$W=\frac{P_2V_2-P_1V_1}{n-1}$

$W=\frac{mR(T_2_T_1)}{n-1}$

Since it is a polytropic Process

therefore $PV^n=c$

$P_1V_1^n=P_2V_2^n$

$(\frac{V_1}{V_2})^n=\frac{P_2}{P_1}$

$(\frac{0.5}{0.264})^n=\frac{400}{150}$

$n=\frac{\ln 2.66}{\ln 1.893}$

$n=1.533$

$W=\frac{0.123\times 2.076(140-20)}{1.533-1}$

$W=57.48 kJ$

From Energy balance

$E_{in}-E_{out}=\Delta E_{system}$

Neglecting kinetic and Potential Energy change

$Q_{in}+W_{in}=change\ in\ Internal\ Energy$

Change in Internal Energy $\Delta U=u_2-u_1$

$\Delta U=mc_v(T_2-T_1)$

$\Delta U=0.123\times 3.115(140-20)$

$\Delta U=45.977 kJ$

$Q_{in}+57.48=45.977$

$Q_{in}=-11.50 kJ$

i.e. Heat is being removed

3 0

3 years ago

A 0.02kg dart is loaded into a toy spring gun by compressing the spring 6cm. If the spring has a constant of 20 N/m, find… a. Th

katrin2010 [14]

Answer:

a) 0.036 J b) 0.036J c) 0.036 d) 1.9m/s e) 0.18 m

Explanation:

Mass of the dart = 0.02kg, the spring was compressed to 6cm

Work needed to compress the spring = 1/2*k*x ^2 where k is the force constant of the spring in N/m, x is the distance it was compressed in m

Work needed to compress the spring = 0.5 * 20* 0.06^2 since 6cm = 6 / 100 = 0.06 m

Work needed to compress the spring = 0.036J

b) the total energy stored in the spring = the work done to compress the spring = 0.036J

c) kinetic energy of the dart as it leaves the the spring = elastic potential energy stored in the spring = the work done in compressing the = 0.036J using the law of conservation of energy; energy is neither created nor destroyed but transformed from one form to another.

d) 1/2mv^2 = 0.036

mv^2 = 0.036*2

v^2 = 0.036*2 / 0.02 = 3.6

v = √3.6 = 1.897 approx 1.9m/s

e) kinetic energy of the dart = work done against gravity to get the body to height h

Work done against gravity = potential energy conserved at height = -mgh g is negative because the motion is upward while gravity acts downward

0.036 = 0.02 * 9.81 * h

0.036 / ( 0.02*9.81) = h

h = 0.18 m

6 0

3 years ago

What is the gradual process through which humans change from birth to adulthood?