Calculate the work WC done by the gas during the isothermal expansion. Express WC in terms of p0, V0, and Rv.

Answer:

108 metres

Explanation:

Given

[tex]Force = 900N[/tex]

[tex]Mass = 600kg[/tex]

[tex]Time = 12s[/tex]

Required

Determine the distance moved

First, we need to determine the acceleration.

[tex]Force = Mass * Acceleration[/tex]

[tex]900N = 600kg * a[/tex]

Solve for a

[tex]a = 900/600[/tex]

[tex]a = 1.5m/s^2[/tex]

Next, we determine the distance using:

[tex]S = ut + \frac{1}{2}at^2[/tex]

Since it starts from rest,

[tex]u = 0[/tex]

[tex]t = 12[/tex]

[tex]a = 1.5[/tex]

So:

[tex]S = 0 * 12 + \frac{1}{2} * 1.5 * 12^2[/tex]

[tex]S = \frac{1}{2} * 1.5 * 144[/tex]

[tex]S = \frac{1}{2} * 216[/tex]

[tex]S = 108m[/tex]

Answer:

B. iron ore

Explanation:

Hope this helps

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Answer:

20000

Explanation:

Newtons Second law states that the force acting on an object is equal to its mass times its acceleration, f=ma. To solve for force, plug in your values for m and a, and then solve. f = (1000)*(20) = 20000

Answer:

881.84905 LBS

Explanation:

ThErE :p

3922.66 newtons.

This is an exact amount, to get newtons form kg, multiply by 9.8, or in this case, 10.

This gives you 4000 newtons

Answer:

A window

Explanation:

Answer:

a window.. duh

Explanation:

Answer:

milliliters (ml)

Explanation:

millileters is the correct measurement for liquids

Answer:

We are given:

initial velocity (u) = 12 m/s

final velocity (v) = 0 m/s

time taken (t) = 15 seconds

acceleration (a) = a m/s²

Solving for acceleration:

from the first equation of motion

v = u + at

replacing the variables

0 = 12 + (a)(15)

0 = 15a + 12

a = -12 / 15

a = -4 / 5 m/s²

Answer:

Explanation:

To find the force given the mass , velocity and time can be found by using the formula

[tex]f = \frac{m \times v}{t} \\ [/tex]

where

m is the mass

v is the velocity

t is the time

From the question

m = 1500 kg

v = 30 m/s

t = 3 s

We have

[tex]f = \frac{1500 \times 30}{3} = \frac{45000}{3} \\ [/tex]

We have the final answer as

Hope this helps you

Answer:

(a) [tex]2\vec A=12\hat i-4\hat j+6\hat k[/tex]

(b) [tex]\displaystyle \vec{U_A}=12/7\hat i-4/7\hat j+6/7\hat k[/tex]

(c) [tex]-4\vec{U_A}=-48/7\hat i+16/7\hat j-24/7\hat k[/tex]

Explanation:

Vectors

Given a vector

[tex]\vec A=6\hat i-2\hat j+3\hat k[/tex]

We must determine the following:

a) A vector in the same direction as A with double magnitude 2A.

If the vector goes in the same direction but has a different magnitude, we only need to multiply each component by a common factor, in this case, by 2. Thus, the required vector is:

[tex]2\vec A=12\hat i-4\hat j+6\hat k[/tex]

b) A unit vector in the same direction of A.

The unit vector needs to compute the magnitude of the vector:

[tex]\mid A\mid=\sqrt{6^2+2^2+3^2}[/tex]

[tex]\mid A\mid=\sqrt{36+4+9}=\sqrt{49}=7[/tex]

[tex]\mid A\mid=7[/tex]

The unit vector is:

[tex]\displaystyle \vec{U_A}=\frac{\vec A}{\mid \vec A\mid}[/tex]

[tex]\displaystyle \vec{U_A}=\frac{12\hat i-4\hat j+6\hat k}{7}[/tex]

[tex]\displaystyle \vec{U_A}=12/7\hat i-4/7\hat j+6/7\hat k[/tex]

c) A vector opposite to A with magnitude 4 m. We assume the original vector is also expressed in m.

The opposite vector to A is obtained simply by multiplying the unit vector by -1. To make its magnitude equal to 4, also multiply by 4. In all, we multiply the unit vector by -4:

[tex]-4\vec{U_A}=-4(12/7\hat i-4/7\hat j+6/7\hat k)[/tex]

[tex]-4\vec{U_A}=-48/7\hat i+16/7\hat j-24/7\hat k[/tex]

Answer:

h = 2 m

Explanation:

Mechanical advantage of a ramp is given by :

MA = length of incline/height of incline

Length of the ramp, l = 12 m

MA = 6

We need to find the height of the ramp.

So,

height of ramp = length of incline/MA

h = 12/6

h = 2 m

So, the height of the ramp is 2 m.

Answer:

v = 1.74 m/s

Explanation:

Given that,

Diameter of a circular track, d = 100 m

Distance covered for the 4 laps,

[tex]D=4\pi d\\\\D=4\pi \times 100\\\\D=1256.63\ m[/tex]

Time, t = 12 minutes = 720 s

We need to find the velocity of the peter. It can be calculated as follows :

[tex]v=\dfrac{D}{t}\\\\v=\dfrac{1256.63\ m}{720\ s}\\\\v=1.74\ m/s[/tex]

So, the speed is running with a velocity of 1.74 m/s.

Peter is running at 1.7453 m/sec.

Given to us,

Diameter of the circular track, D = 100 meters,

Number of laps Peter run, L = 4 laps,

Time taken by Peter, t = 12 minutes,

1 lap = circumference of the circle,

4 laps = 4 x circumference of the circle,

As we know, the circumference of a circle is given by πD.

So, 4 laps = 4 x circumference of the circle,

[tex]\begin{aligned}4 laps &= 4\times \pi \times D\\&= 4 \times \pi \times 100\\& = 1,256.6370\ meters\\\end{aligned}[/tex]

Also, we know that 1 minute has 60 sec.

so, 4 minutes = (4 x 60) seconds

Further, speed is given [tex]\bold{(\dfrac{Distance}{Time} )}[/tex]

Thus,

[tex]\begin{aligned}speed &= \dfrac{Distance\ coverd\ by\ Peter}{Time\ taken\ by\ Peter}\\&=\dfrac{1,256.6370}{12\times 60}\\&=1.7453\ m/sec \end{aligned}[/tex]

Hence, Peter is running at 1.7453 m/sec.

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Answer: The options are not given.

Here are the options.

a) There is an additional force lifting up on you.

(b) At the top you continue going straight and the seat moves out from under you.

(c) You press on the seat less than when the coaster is at rest.Thus the seat presses less on you. (

d) Both b and c are correct.

(e) a, b, and c are correct.

The correct option Is D.

B.At the top you continue going straight and the seat moves out from under you. C.At the same time, you press on the seat less than when the coaster is at rest because the normal force expirienced will be less.

Explanation:

At the top you continue going straight and the seat moves out from under you.At the same time, you press on the seat less than when the coaster is at rest because the normal force expirienced will be less because it is as a result of a phenomenon called Weightlessness. This occur when there is no force or little force is acting on your body. At the top you continue going straight and the seat moves out from under you because there is no force acting on your body and when the body is in free fall i.e acceleration due to gravity , the person is not supported by any thing at.

That is the scenarion that occur...

Answer:

The one with the faster velocity is the one with a velocity of -10m/s

Answer:

His weight would be zero on the scale i.e he is weightless at that instance.

Explanation:

weight = mg

where m is the mass of the object, and g is the acceleration of gravity.

⇒ 810 = mg

During free fall, the weight of an object can be determined by:

W = mg - ma (provided that acceleration of gravity is greater than acceleration of the object)

where a is the acceleration of the object.

But since James fall at the acceleration of gravity, then:

g = a

mg = ma = 810 N

So that;

W = 810 - 810

    = 0 N

Therefore though the weight of James is 810 N, but the scale reads 0 N. this condition is referred to as weightlessness.

Answer:

I think it's = 12 seconds

Explanation:

the formula for speed is:

speed=[tex]\frac{distance}{time}[/tex] SO, time is equal to:

time=[tex]\frac{distance}{speed}[/tex]

(sub the numbers in the formula)

distance=1200m, speed=100m/s

time=[tex]\frac{1200}{100}[/tex]

=12 seconds

Answer:

6.7 seconds

Explanation:

d=(1/2)at^2

equation

1000=(1/2)45t^2.

substitute

2000=45t^2.

multiply by 2 for both sides

44.44=t^2.

divide both sides by 45

6.7=t

take the square root of both sides

We are given:

Mass of object (m) = 12 kg

acceleration (a) = 6 m/s²

Solving for the Force:

From newton's second law of motion:

F = ma

replacing the variables

F = 12*6

F = 72N

Answer:

Explanation:

mass of water coming out per second = A x v where A is area of cross section of the nozzle and v is velocity of water

A = 3.14 x .005²

= 785 x 10⁻⁷ m²

mass of water coming out per second = 785 x 10⁻⁷ x 30 = 23.55 x 10⁻⁴ kg

momentum of this mass = 23.55 x 10⁻⁴ x 30 = 706.5 x 10⁻⁴ kg m /s .

Rate of change of momentum  = 706.5 x 10⁻⁴

Let force be F

F - mg = 706.5 x 10⁻⁴

F = mg + 706.5 x 10⁻⁴

F =  23.55 x 10⁻⁴  x 9.8 + 706.5 x 10⁻⁴

= 937.3 x 10⁻⁴ N .

Answer:

x = 73.71 [m]

Explanation:

In order to solve this problem we must use two formulas of kinematics. It is important to make it clear that these formulate are for uniformly accelerated motion, i.e. with constant acceleration.

[tex]v_{f }= v_{i}-(a*t)[/tex]

where:

Vf = fnal velocity = 0

Vi = initial velocity = 32.4 [m/s]

t = time = 4,55 [s]

a = acceleration or desacceleration [m/s^2]

0 = 32.4 - (a*4.55)

a = 7.12 [m/s^2]

Note: it is important to clarify that the negative sign in the above equation is because the car stops and decreases its speed to zero, thus its final velocity is equal to zero.

Now using the following equation:

[tex]x = x_{o} + (v_{i}*t)-(\frac{1}{2} )*a*t^{2}[/tex]

where:

xo = initial distance = 0

x = final distance [m]

Therefore we have:

x = 0 + (32.4*4.55) - (0.5*7.12*4.55^2)

x = 73.71 [m]

Answer:

Explanation:

The frequency of wave is directly proportional to velocity

f = kV

k = f/V

f1/V1 = f2/V2

Given

f1 = 450Hz

V1 = 343m/s

f2 = ?

V2 = 20m/s

Substitute into the formula

450/343 = f2/20

Cross multiply

343f2 = 450×20

343f2 = 9000

f2 = 9000/343

f2 = 26.24Hz

Answer:

Mg (atomic number 12)

K (atomic number 19)

Explanation:

The size of an atom is estimated in terms of its atomic radius.

The atomic radius is taken as half of the inter-nuclear distance between two covalently bonded atoms of non-metallic elements or half of the distance (d) between two nuclei in the solid - state of metals.

Cl is further right of Mg in the third period

K is below Na in the first group

Answer:

it obeys the law of vector addition

Explanation:

In the analytical method, we use the Pythagorean theorem to find the magnitude of the resultant because it obeys the law of vector addition.

The resultant of two vectors A and B is given by;

R' = A' +  B'

[tex]R = \sqrt{A^2 + B^2}[/tex]

Therefore, Pythagorean theorem is used if the two quantities are vectors and vector quantities obey the law of vector addition.

The cyclist accelerates at 1.2 m / s² after 2 seconds he reaches a speed of 2.4 m / s, then the momentum at this point would be 211.2 kg-m/s, therefore the correct answer is option D.

It can be defined as the product of the mass and the speed of the particle, it represents the combined effect of mass and the speed of any particle, and the momentum of any particle is expressed in Kg m/s unit.

As given in the problem a cyclist and his bicycle has a combined mass of 88 kg and a combined weight of 862.4 N. The cyclist accelerates at 1.2 m/s2. After 2 seconds he reaches a speed of 2.4 m/s.

The momentum of the cyclist = 88 × 2.4

                                                 = 211.0 kgm/s

Thus, the momentum of the cyclist would be  211.0 kgm/s.

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Complete Question

The complete question is shown on the first and second uploaded image

Answer:

There is a change in the electric field by this factor [tex]\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]}[/tex]

Explanation:

From the question we are told that

  The electric field is  [tex]E(r)_1 =  [\frac{V_o}{ln(b) -ln(a)} ] * \frac{1}{r}[/tex]

    Now when the outer diameter is doubled, the radius(b) is also doubled

So

    [tex]E(r)_2 =  [\frac{V_o}{ln(2b) -ln(a)} ] * \frac{1}{r}[/tex]

Now

      [tex]\frac{E(r)_2}{E(r)_1}  =  \frac{\frac{V_o}{ln(b) -ln(a)} ] * \frac{1}{r}}{\frac{V_o}{ln(2b) -ln(a)} ] * \frac{1}{r}}[/tex]

=>   [tex]\frac{E(r)_2}{E(r)_1}  =  \frac{V_o}{ln(b) -ln(a)} ] * \frac{1}{r} * \frac{ ln(2b) -ln(a)}{V_o} ] * \frac{r}{1}[/tex]

=>  [tex]\frac{E(r)_2}{E(r)_1}  =\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]}[/tex]

[tex]=> E(r)_2 =\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]} }{E(r)_1}[/tex]

Here we see that the electric field changes by the factor [tex]\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]}[/tex]

Answer:

35.35 m

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 20 m/s

Angle of projection (θ) = 30°

Acceleration due to gravity (g) = 9.8 m/s²

Range (R) =.?

The range (i.e how far away) of the ball can be obtained as follow:

R = u² Sine 2θ /g

R = 20² Sine (2×30) / 9.8

R = 400 Sine 60 / 9.8

R = (400 × 0866) / 9.8

R = 346.4 / 9.8

R = 35.35 m

Therefore, the range (i.e how far away) of the ball is 35.35 m