It increases, f = ma both mass and acceleration are directly proportional to force so if mass is constant acceleration must increase to increase the force.

Answer:

3.32 atm

__________________________________________________________

We are given:

side of the cubical container = 28 cm

number of molecules in the container = 3 * Nₐ  

[where Nₐ is the Avogadro's number]

Temperature = 24°C  OR  297 K

We need to find the pressure exerted by the gas on the walls of the container

__________________________________________________________

Some Calculations:

Volume of the container

we are given the side of the cubical container = 28 cm

Volume of the cubical container = side³

Volume = 28³

Volume = 21952 cm³

We know that 1 cm³ = 1 mL

So,

Volume = 21952 mL

We also know that 1 L = 1000 mL

Volume = 21.952 L

Number of moles of Gas

We know that:

number of moles = number of molecules / Avogadro's number

number of moles = 3 * Nₐ / Nₐ                   [number of molecules = 3 * Nₐ]

number of moles = 3 moles

__________________________________________________________

Pressure Exerted by the Gas:

Using the ideal gas equation:

PV = nRT

Since the volume is in L, and Temperature is in K. R is equal to

0.082 L atm /mol K and the pressure will be in atm

P(21.952) = 3*(0.082)*(297)        

P = 3.32 atm

Hence, the gas will exert a pressure of 3.32 atm on the walls of the container

Answer:

Q = PV(In 4)

Explanation:

We are told that the volume expands from V to a state with volume 4V.

Thus, initial volume is V and Final volume is 4V.

We want to find How much heat is added to the expanding gas.

For an isothermal process, the work done is calculated from;

W = nRT(In(V_f/V_i))

Where;

V_f is final volume

V_i is initial volume

Thus;

W = nRT(In(4V/V))

W = nRT(In 4)

Now, from ideal gas equation, we know that;

PV = nRT

Thus;

W = PV(In 4)

Now from first law of thermodynamics, we know that internal energy is zero and thus; Q = W

Where Q is quantity of heat

Thus;

Q = PV(In 4)

Answer:

Explanation:

step one:

given data

velocity v=34.2m/s

time t= 8.7s

Step two

Required is the distance the cheetah has covered on the condition

we know that speed= distance/time

make distance subject of formula we have

distance= velocity *time

distance= 34.2*8.7

distance = 297.54m

Therefore the displacement the cheetah covered at that velocity

is 297.54m

Answer:

penis

Explanation:

Answer:

false ..........false

Answer:

FALSE                                            

Explanation:

Answer:

You will be very glad to know that my preparation for HSC examination is very good. I have prepared myself for the HSC Examination 2015. I am confident that I will obtain very good marks in the examination. I have revised them several times and I am hopeful of obtaining distinction marks in all the subjects.

Explanation:

This is from Brainly hope this person helped you

Answer:

Ek = 196.2 [J]

Explanation:

The question concerns the KE kinetic energy.

That is, we must find the kinetic energy at the moment the cannon is fired and the kinetic energy of when the ball hits the ground after having fallen 20 meters.

At the moment when the ball is fired it is 20 meters above ground level. If the ground level is taken as the reference level of potential energy, where it is equal to zero, in this way when the ball is at the highest (20 meters) you have the maximum potential energy.

In this way, the energy in the initial state is equal to the sum of the kinetic energy plus the potential energy. As the energy is conserved this same energy will be present when the ball hits the ground, where the potential energy is zero and will have only kinetic energy.

[tex]E_{1}=E_{2}\\E_{k1}+E_{p1}=E_{k2}\\\frac{1}{2} *m*v^{2} +m*g*h=E_{k2}\\E_{k2}=0.5*1*(5)^{2} +1*9.81*20\\E_{k2}=208.7[J][/tex]

The kinetic energy in the initial state can be easily calculated by means of the following equation.

[tex]E_{k1}=\frac{1}{2} *m*v^{2}\\E_{k1}=0.5*1*(5)^{2}\\E_{k1}=12.5 [J][/tex]

Therefore the change in KE

[tex]E_{k} = 208.7 - 12.5\\E_{k} = 196.2 [J][/tex]

Answer:

320 paper clips

Explanation:

mass = volume × density = 20cm³ × 8g/cm³ = 160g

mass of 1 paper clip = 0.50g

mass of x paper clips = 160g

x = 160/0.50 = 320

Answer:

The mass of the catfish is 2.13 kg

Explanation:

Period of oscillation, T = 0.19 s

spring constant, k = 2330 N/m

The period of oscillation of the spring is given by;

[tex]T = 2\pi \sqrt{\frac{m}{k} }\\\\\frac{T}{2\pi} = \sqrt{\frac{m}{k} }\\\\\frac{T^2}{4\pi^2} = \frac{m}{k}\\\\m = \frac{kT^2}{4\pi^2}[/tex]

where;

m is mass of the catfish

substitute the given values and solve for m;

[tex]m = \frac{kT^2}{4\pi^2} \\\\m = \frac{(2330)(0.19)^2}{4\pi^2} \\\\m = 2.13 \ kg[/tex]

Therefore, the mass of the catfish is 2.13 kg

Answer:

e. 55.0 m

Explanation:

Given;

diameter of the aluminum wire, d = 0.865 mm

radius of the wire, r = d/2 = 0.4325 mm = 0.4325 x 10⁻³ m

resistivity of the wire, ρ = 2.65 x 10⁻⁸ Ω-m

resistance of the wire, R = 2.48 Ω

The resistance of a wire is given by;

[tex]R = \frac{\rho \ L}{A} \\\\[/tex]

where;

L is length of the wire

A is area of the wire = πr² = π(0.4325 x 10⁻³ )² = 5.877 x 10⁻⁷ m²

Substitute the givens and solve for L,

[tex]L = \frac{RA}{\rho} \\\\L = \frac{(2.48)(5.877*10^{-7})}{2.65*10^{-8}}\\\\L = 55.0 \ m[/tex]

Therefore, the length of the wire is 55.0 m

Answer:

contains many young stars

Explanation:

Irregular galaxies have no definite shape, which means that the first option is incorrect. They are definitely not round.

However, they contain many young stars because the degree of star formation is fast. They also contain old stars. Thus, the second choice is correct.

The "spiral galaxy" is the type of galaxy that has arms that extend from the center. These arms look "spiral," which influenced its name. This makes the last choice incorrect.

They are actually smaller than the other types of galaxies. This makes them prone to collisions. This makes the last choice incorrect.

Answer:

Contains many young stars

Explanation:

Answer:

3.125 meters.

Explanation:

(3.0*10^8)/(96*10^6)

= 3.125 meters.

Hope this helped!

Answer:

451 milliliters equals 15.25 fluid ounces

Explanation:

The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them. That is, what is intended with it is to find the fourth term of a proportion knowing the other three.

To solve a direct rule of three, the following formula must be followed:

a ⇒ b

c ⇒ x

So: [tex]x=\frac{c*b}{a}[/tex]

The direct rule of three is the rule applied in this case where there is a change of units.

In this case, the rule of three can be applied in the following way: if there are 29.57 milliliters in 1 fluid ounce, in 451 milliliters how many fluid ounces are there?

[tex]fluid ounces=\frac{451 mL*1 fluid ounce}{29.57 mL}[/tex]

fluid ounces= 15.25

451 milliliters equals 15.25 fluid ounces

Answer:

5×10^4Pa

Explanation:

Given force of 108 lb is required to pull the lid off the box,

To convert "Ib"to Newton ,we use conversation rate below

1 pounds = 4.4482216282509 newtons

Then 108 lb=x Newton

Cross multiply we have

X= 480.41Newton

The force that is needed to open the lid is F and pressure P.

We know that Pressure= Force/Area

Area is given as 80.0 cm^2, we can convert to m^2 for unit consistency since 1cm^2= 0.001m^2 then

80.0 cm^2 = 80×10^-4m^2

Substitute to the equation of the pressure we have

P= 480.41Newton/(80×10^4m^2)

P=6×10^4 Pa

The pressure in the box will be difference between the initial pressure and final pressure

=( 1.01 ×10^5 Pa)-(6×10^4 Pa)

= 50100Pa

= 5×10^4Pa

Therefore, The pressure in the box was

5×10^4Pa

Answer:

v₁ = 0.375 m / s ,   x = 0.335 m

Explanation:

Let's analyze this interesting exercise, the block moves and has a friction force with the tile, we assume that the speed of the block is constant, so the friction force opposes the block movement. For the only force that acts (action and reaction) this friction force exerted by the block that is in the direction of movement of the tile.

We can also see that the isolated system formed by the block and the tile will reach a stable speed where friction cannot give the system more energy, this speed can be found by treating the system with the conservation of linear momentum.

initial moment. Right at the start of the movement

       p₀ = m v₀ + 0

final moment. Just when it comes to equilibrium

      [tex]p_{f}[/tex] = (m + M) v₁

how the forces are internal

       p₀ =p_{f}

       m v₀ = (m + M) v₁

       v₁ = m /m+M    v₀

let's calculate

       v₁ = 0.4 /(0.4 + 2.8)  3

       v₁ = 0.375 m / s

Let's apply Newton's second law to the Block, to find the friction force

Y axis

       N - W = 0

       N = W

       N = m g

where m is the mass of the block

the friction force has the formula

      fr = μ N

      fr = μ m g

We apply Newton's second law to slab    

X axis

       fr = M a

where M is the mass of the slab

       μ m g = M a

       a = μ g m / M

let's calculate

       a = 0.15  9.8  0.4 / 2.8

       a = 0.21 m / s²

With kinematics we can find the position

       v²= v₀²+2 a x

as the slab is initially at rest, its initial velocity is zero

       v² = 2 a x

       x = v2 / 2a

let's calculate

        x = 0.375²/2 0.21

        x = 0.335 m

Answer:

horizontal component=fcostita

=150cos60

use calculator to evaluate it

for vertical=fsintita

=150sin60

Answer:100m west

Explanation:

Answer:

ill get back to this question once i find the answer to it

Answer:

[tex]v=\sqrt{26}~m/s[/tex]

Explanation:

Parametric Equation of the Velocity

Given the position of the particle at any time t as

[tex]r(t) = (x(t),y(t))[/tex]

The instantaneous velocity is the first derivative of the position:

[tex]v(t)=(v_x(t),v_y(t))=(x'(t),y'(t))[/tex]

The speed can be calculated as the magnitude of the velocity:

[tex]v=\sqrt{v_x^2+v_y^2}[/tex]

We are given the coordinates of the position of a particle as:

[tex]x=5t-3t^2[/tex]

[tex]y=5t[/tex]

The coordinates of the velocity are:

[tex]v_x(t)=(5t-3t^2)'=5-6t[/tex]

[tex]v_y(t)=(5t)'=5[/tex]

Evaluating at t=1 s:

[tex]v_x(1)=5-6(1)=-1[/tex]

[tex]v_y(1)=5[/tex]

The velocity is:

[tex]v=\sqrt{(-1)^2+5^2}[/tex]

[tex]v=\sqrt{1+25}[/tex]

[tex]\mathbf{v=\sqrt{26}~m/s}[/tex]

Answer:

10 seconds

Explanation:

Explanation:

Perspective- the art of drawing solid objects on a two-dimensional surface so as to give the right impression of their height, width, depth, and position in relation to each other when viewed from a particular point.

Answer:

Specific Gravity = 0.378

Explanation:

First, we will find the force exerted by the can on the table. This force will be equal to the weight of the can:

Pressure = Force/Area = Weight/Area

Weight = Pressure*Area

where,

Area = πdiameter²/4 = π[(3.1 in)(0.0254 m/1 in)]²/4 = 4.8 x 10⁻³ m²

Weight = (510 N/m²)(4.8 x 10⁻³ m²)

Weight = 2.48 N

Now, the weight is given as:

Weight = mg

2.48 N = m(9.8 m/s²)

m = (2.48 N)/(9.8 m/s²)

m = 0.25 kg

Now, we calculate volume of can:

Volume = (Area)(Height) = (4.8 x 10⁻³ m²)(5.42 in)(0.0254 m/1 in)

Volume = 6.6 x 10⁻⁴ m³

Hence, the density of can will be:

Density of Can = m/Volume = 0.25 kg/6.6 x 10⁻⁴ m³

Density of Can = 378.32 kg/m³

So, the specific gravity of Can will be:

Specific Gravity = Density of Can/Density of Water

Specific Gravity = (378.32 kg/m³)/(1000 kg/m³)

Specific Gravity = 0.378

A spring balance measures the weight of an object by opposing the force of gravity acting with force of an extending spring. May be used to determine mass as well as weight by recalibrating the scale. Some spring balances are available in gram or kilogram markings and are used to measure the mass of an object. Spring balances consist of a cylindrical tube with a spring inside. One end (at the top) is fixed to an adjuster which can be used to calibrate the device. The other end is attached to a hook on which you can hang masses etc.

Answer:

[tex]\displaystyle x=10\sqrt{3}\ m/s[/tex]

[tex]y=10\ m/s[/tex]

Explanation:

Rectangular coordinates of vectors in 2D

Given a vector with a magnitude v and angle θ with respect to the positive horizontal direction, the x and y components of the vector are given by:

[tex]x=v\cos\theta[/tex]

[tex]y=v\sin\theta[/tex]

The soccer ball is kicked upward at an angle θ = 30° and at a speed v=20 m/s.

The rectangular components of the vector are:

[tex]x=20\cos 30^\circ[/tex]

[tex]\displaystyle x=20\cdot \frac{\sqrt{3}}{2}[/tex]

Operating:

[tex]\mathbf{\displaystyle x=10\sqrt{3}\ m/s}[/tex]

[tex]y=20\sin 30^\circ[/tex]

[tex]\displaystyle y=20\cdot \frac{1}{2}[/tex]

Operating:

[tex]\mathbf{y=10\ m/s}[/tex]

Answer:

For any given projector, the width of the image (W) relative to the throw distance (D) is know as the throw ratio D/W or distance over width. So for example, the most common projector throw ratio is 2.0. This means that for each foot of image width, the projector needs to be 2 feet away or D/W = 2/1 = 2.0.

Answer:

Potential energy is the energy that is stored in an object. ... When you park your car at the top of a hill, your car has potential energy because the gravity is pulling your car to move downward; if your car's parking brake fails, your vehicle may roll down the hill because of the force of gravity.

Answer:

v = 120 m/s

Explanation:

We are given;

earth's radius; r = 6.37 × 10^(6) m

Angular speed; ω = 2π/(24 × 3600) = 7.27 × 10^(-5) rad/s

Now, we want to find the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator.

The angle will be;

θ = ¾ × 90

θ = 67.5

¾ is multiplied by 90° because the angular distance from the pole is 90 degrees.

The speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator will be:

v = r(cos θ) × ω

v = 6.37 × 10^(6) × cos 67.5 × 7.27 × 10^(-5)

v = 117.22 m/s

Approximation to 2 sig. figures gives;

v = 120 m/s

Answer:

10778292789403987593790

Explanation:

I am a Cow'

Answer:

Explanation:

Given;

initial velocity of the ball, u = 23 m/s

time of motion, t = 2 s

The position of the ball after 2 s is given by;

h = ut - ¹/₂gt²

h = (23 x 2) - ¹/₂ x 9.8 x 2²

h = 46 - 19.6

h = 26.4 m

The velocity of the ball after 2 s is given by;

v² = u² + 2(-g)h

v² = u² - 2gh

v² = 23² - (2 x 9.8 x 26.4)

v² = 529 - 517.44

v² = 11.56

v = √11.56

v = 3.4 m/s

Answer:

describe, explain, predict, and change/control behavior.

Explanation:

describe: What are they doing? -Pavlov noticed that dogs were salivating when they would see his lab assistant before food was presented to them. This observation acted as a description of what was happening to them.

explain: Why are they doing that?- Pavlov started to look into why they were doing it. There was a stimulus, the assistant giving them food in the past to where they started to salivate at the sight of the lab assistant

predict: What would happen if I responded in this way?- Pavlov predicted that he could get the same reaction if he used a bell as a stimulus. Using this he was able to condition dogs to salivate at the ring of the bell.

change/control: What can I do to get them to stop doing that? Because of this discovery we can use conditioning today. For example, in the classroom teachers can use conditioning with their students to make it easier, parents to teach their children right from wrong and to have good behavior. (you do this bad thing you get time out, do a good thing and I will praise you, etc) It can be used when training employees and many other places.