A student was walking with a mass of 40 kilograms and an acceleration of 1 m/s/s. What is the force exerted on her?

I a liter jar contains 0.5 moles of gas at a pressure of 2 ATM what is the temperature of the gas

sweet [91]

Answer:

<h2>The temperature of the gas is 48.75 Kelvin.</h2>

Explanation:

Using the ideal gas equation as shown

PV = nRT where;

P is the pressure of the gas in ATM

V is the volume of the gas

n is the number of moles

R is the ideal gas constant

T is the temperature in Kelvin

From the formula, $T = \frac{PV}{nR}$

Given the following parameters V = 1litre, n = 0.5moles. pressure = 2ATM

R = 0.08206 atm L/molK

On substituting to get the temperature we have:

$T = \frac{2*1}{0.08206*0.5} \\T = \frac{2}{0.04103}\\T = 48.75Kelvin$

5 0

2 years ago

If a species experiences a helpful mutation, like camouflage, explain how that mutation would help the species to better survive

ExtremeBDS [4]

Answer:

A helpful mutation like camoflage would immensely help a species survive and flourish. The reason being the appliances of such a mutation. Camoflage is one of the most apllicable of abilities. It can be used as a defense mechanism, to hide in plain sight, it can be used offensively to avoid being seen while hunting and it can even be used as a form of communication between the species. All of these appliances would greatly benefit the species and would most definitely help it survive and flourish.

4 0

2 years ago

At what distance does a 100 Watt lightbulb deliver the same power per unit surface area as a 75 Watt lightbulb produces 10 m awa

ch4aika [34]

Answer:

At 11.5 m

Explanation:

The power per unit area corresponds to the intensity, which is given by

$I=\frac{P}{4\pi r^2}$

where

P is the power

$4\pi r^2$ is the area irradiated at a distance r from the source (it corresponds to the surface area of a sphere of radius r)

Here we want the intensity of the two light bulbs to be the same, so

$I_1 = I_2\\\frac{P_1}{4 \pi r_1^2}=\frac{P_2}{4\pi r_2^2}$

where we have

P1 = 100 W is the power of the first light bulb

P2 = 75 W is the power of the second light bulb

r2 = 10 m is the distance from the second light bulb

Solving for r1, we find

$r_1 = r_2 \sqrt{\frac{P_1}{P_2}}= (10 m) \sqrt{\frac{100 W}{75 W}} = 11.5 m$

4 0

2 years ago

g In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period

NeTakaya

The question is incomplete. The complete question is :

In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period of the oscillations is 1.2 microseconds (1.2*10^-6). What is Q?

Solution :

The underdamped RLC circuit

$v_{t} = ve^{-\frac{R}{2L}t} \cos \omega t$

$\omega = \sqrt{\frac{1}{LC}-\frac{R^2}{4L^2}}= \frac{2 \pi}{T}$

We know in one time period, v = 2v, at t = T, $v_t = 3.8 v$

so, $9.8 = 5 e^{-\frac{R}{2L}T} \cos \frac{2 \pi}{T}T$

$e^{-\frac{R}{2L}T} = \frac{3.8}{5} \times 1$

$\frac{R}{2L}T= \ln \frac{5}{3.8}$

$\frac{R}{L}= \frac{2}{1.2 \times 10^{-6}} \ln \frac{5}{3.8}$

$\frac{R}{L} = 457.3 \times 10^3$

Now, Q value $= \frac{1}{R}\sqrt{\frac{L}{C}}$

$=\sqrt{\frac{L}{R^2C}\times \frac{L}{L}}$

$=\sqrt{(\frac{L}{R})^2 \times \frac{1}{LC}}$

$\frac{1}{LC}=27.43 \times 10^{12}$

∴ $Q=\sqrt{\left(\frac{1}{457.3 \times 10^3}\right)^2 \times 27.43 \times 10^{12}}$

$Q=\sqrt{131.166}$

= 11.45

4 0

2 years ago

A snowball is launched horizontally from the top of a building at v = 16.9 m/s. If it lands d = 44 meters from the bottom, how h

Aloiza [94]

$v_x=16,9\frac{m}{s}\\
s=44m\\
g=10\frac{m}{s^2}\\\\
v_x=\frac{s}{t}\Rightarrow t=\frac{s}{v_x}\\\\
t=\frac{44}{16,9}\\\\
t=2,6s\\\\
h=\frac{gt^2}{2}\\\\
h=\frac{10*2,6^2}{2}\\\\
h=5*6,76\\\\
h=33,8m$

6 0

3 years ago