Temperature is a measure of the average _________________ of each particle within an object.

A) kinetic energy
B) potential energy
C) heat

Answer:

The speed and direction of the tangled players is 2.89 m/s towards right.

Explanation:

Given;

mass of the player heading right, m₁ = 40 kg

initial velocity of the player heading right, u₁ = 9.0 m/s

mass of the player heading left, m₂ = 50 kg

initial velocity of the player heading left, u₂ = -2.0 m/s

let the speed of the tangled players = v

Apply the law of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

(40 x 9) + (50 x - 2) = v(40 + 50)

360 - 100 = v(90)

260 = v(90)

v = 260 / 90

v = 2.89 m/s (towards right, since it is in position direction)

Therefore, the speed and direction of the tangled players is 2.89 m/s towards right.

Answer:

375 J

Explanation:

Work done is the product of force and distance moved by the object.

W=F*d

Given that ;

Force= F= 250 N

Distance= d =1.5 m

W= 250 * 1.5 =375 J

Answer: [tex]T=26.36^{\circ}C[/tex]

Explanation:

Given

mass of the first object [tex]m_1=5\ kg[/tex]

mass of the second object [tex]m_2=0.5\ kg[/tex]

specific heat of the two is [tex]c_1=c_2=4200\ J/kg/^{\circ}C[/tex]

The temperature of the first [tex]T_1=20^{\circ}[/tex]

The temperature of the second [tex]T_2=90^{\circ}[/tex]

Heat flows from high temperature to low temperature

Suppose T is the common temperature

[tex]m_1c_1(T-T_1)=m_2c_2(T_2-T)\\5\times 4200\times (T-20)=0.5\times 4200\times (90-T)\\5T-100=45-0.5T\\5.5T=145\\T=26.36^{\circ}C[/tex]

Answer:

BMI is a simple indicator of weight for height and can't differentiate between muscle mass and fat mass. So BMI tends to overestimate the health risk for adults with a high muscle mass, such as some athletes, and underestimate the risk for adults with a low muscle mass, as can occur with sedentary lifestyles.

Explanation:

Hope it is helpful...

Answer:

1.5 m/s2

Explanation:

Gizmo Explanation: The acceleration of an object in uniform circular motion is inversely proportional to the radius of the motion. This means that, if the radius is multiplied by a number, the acceleration is divided by that same number. In this case, the radius of the path doubles, from 5 m to 10 m. From this you can conclude that the acceleration must be half of the original value, or 1.5 m/s2.

Answer:

Explanation:

The gravitational potential energy of a body can be found by using the formula

GPE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 9.8 m/s²

From the question we have

GPE = 200 × 9.8 × 50

We have the final answer as

Hope this helps you

Answer:

The force acting on the shaft is 1324.75 N

Explanation:

Given that,

Cross sectional area of shaft [tex]A_{sh}=0.8\ cm^{2}[/tex]

Gas pressure [tex]P_{gas}=3\ bar=3\times10^5\ Pa[/tex]

Total mass M=24.5+0.5=25 kg

Diameter of piston, d=10 cm

We need to calculate the cross section area of piston

Using formula of area

[tex]A_{p}=\pi\times\dfrac{d^2}{4}[/tex]

Put the value into the formula

[tex]A_{p}=\pi\times\dfrac{0.1^2}{4}[/tex]

[tex]A_{p}=0.00785\ m^{2}[/tex]

We need to calculate the weight of the piston and shaft

Using formula of weight

[tex]W=mg[/tex]

Put the value into the formula

[tex]W=25\times9.81[/tex]

[tex]W=245.25\ N[/tex]

We need to calculate the force due to gas pressure

Using formula of force

[tex]F_{gas}=P_{gas}\times A_{p}[/tex]

Put the value into the formula

[tex]F_{gas}=3\times10^{5}\times0.00785[/tex]

[tex]F_{gas}=2355\ N[/tex]

We need to calculate the force due to atmospheric pressure,

Using formula of force

[tex]F_{atm}=P_{atm}\times A_{p}[/tex]

Put the value into the formula

[tex]F_{atm}=1\times10^5\times0.00785[/tex]

[tex]F_{atm}=785\ N[/tex]

We need to calculate the force acting on the shaft

from free body diagram

[tex]F_{gas}=F+F_{atm}+W[/tex]

Put the value into the formula

[tex]2355=F+785+245.25[/tex]

[tex]F=2355-785-245.25[/tex]

[tex]F=1324.75\ N[/tex]

Hence, The force acting on the shaft is 1324.75 N

The complete question is :

The figure (attached) shows a gas contained in a vertical piston-cylinder assembly. A vertical shaft whose cross-sectional area is 0.8 cm2 is attached to the top of the piston. Determine the magnitude, F, of the force acting on the shaft, in N, required if the gas pressure is 3 bar. The masses of the piston and attached shaft are 24.5kg and 0.5kg respectively. The piston diameter is 10cm. The local atmospheric pressure is 1 bar. The piston moves smoothly in the cylinder and g=9.81 m/s2.

Solution :

Given

The cross sectional area of the shaft,  [tex]$A_{s} = 0.8 \ cm^2$[/tex]

Gas pressure, [tex]$P_{gas} = 3 \ bar = 3 \times 10^5 \ Pa$[/tex]

The total mass, m = 24.5 + 0.5

                             = 25 kg

Diameter of the piston, d = 10 cm

The cross sectional area of the piston, [tex]$A_{p} = \frac{\pi}{4} \times (0.1)^2$[/tex]

Weight of the piston and shaft, W = mg

                                                        = 25 x 9.81

                                                       = 245.25 N

Force due to the gas pressure,

[tex]$F_{gas} = P_{gas} \times A_{p}$[/tex]

[tex]$F_{gas} = 3 \times 10^5 \times 0.00785 $[/tex]

       = 2355 N

Force due to atmospheric pressure,

[tex]$F_{atm} = P_{atm} \times A_{p}$[/tex]

        [tex]$= 1 \times 10^5 \times 0.00785$[/tex]

        = 785 N

Now [tex]$F_{gas} = F + F_{atm} + W$[/tex]

     ⇒ 2355 = F + 785 + 245.25

     ⇒ F = 1324.75 N

Answer:

1. 57.99V

2. 37.5Hz

3. 0.7072A

4. 82 ohms

5. 5.18x10^-5F

Explanation:

In answer to this question, we have the Standard equation of AC emf to be

V = V0 x sin ωt

We have

V0 = 82V,

ω = 75π

1.

RMS Voltage =

V0/√2 = 82 /√2

= 82/1.414

= 57.99V

2.

ω = 2π* f

75π = 2πf

Frequency,f = 75π/2π

= 235.5/6.28

= 37.5 Hz

3.

RMS current

= Imax/√2

= 1.00/1.414

= 0.7072A

4.

Reactance

= Vrms/ Irms

= 57.99/0.7072

= 81.999

= 82.0 Ω

5.

Reactance = 1/ ω x C

Reactance = 82

ω = 75π

We put these values into the equation above and make c the subject of the formula

C = 1/82.0 x 75π

C = 1/ 82.0 x 75 x 3.14

C = 1/19311

Capacitance = 5.18x10^-5F

Answer:

We know that the gravitational acceleration in the surface of the Earth can be written as:

g = G*M/r^2

Where:

M = mass of the Earth

r = radius of the Earth.

G = gravitational constant.

The weight of an object of mass m, is written as:

W = m*g = m*(G*M/r^2)

Now, if we move our object to a place that has a mass equal to 1/6 times the mass of the Earth, and 1/3 the radius of the earth.

The gravitational acceleration on this planet is written as:

g' = G*(M/6)/(r/3)^2 = (1/6)*(G*M)/(r^2/9) = (9/6)*(G*M/r^2) = (3/2)*g

then the weight on this planet is:

W' = m*g' = m*(3/2)*g = (3/2)*(m*g)

and m*g was the weight on Earth, then:

W' = (3/2)*(m*g) = (3/2)*W

The new weight is 3/2 times the weight on Earth.

Answer:

a) [tex]a=33.73mm/s^{2}[/tex]

b) mg>N

c) [tex]\%_{change}=0.343\%[/tex]

d) [tex]a=24.07mm/s^{2}[/tex]

Explanation:

In order to solve part a) of the problem, we can start by drawing a free body diagram of the presented situation. (see attached picture).

In this case, we know the centripetal acceleration is given by the following formula:

[tex]a_{c}=\omega ^{2}r[/tex]

where:

[tex]\omega=\frac{2\pi}{T}[/tex]

we know the period of rotation of the earth is about 24 hours, so:

[tex]T=24hr*\frac{3600s}{1hr}=86400s[/tex]

so we can now find the angular speed:

[tex]\omega=\frac{2\pi}{86400s}[/tex]

[tex]\omega=72.72x10^{-6} rad/s^{2}[/tex]

So the centripetal acceleration will be:

[tex]a_{c} =(72.72x10^{-6} rad/s^{2})^{2}(6478x10^{3}m)[/tex]

which yields:

[tex]a_{c}=33.73mm/s^{2}[/tex]

b)

In order to answer part b, we must draw a free body diagram of us sitting on a chair. (See attached picture.)

So we can do a sum of forces in equilibrium:

[tex]\sum F=0[/tex]

so we get that:

[tex]N-mg+ma_{c} = 0[/tex]

and solve for the normal force:

[tex]N=mg-ma_{c}[/tex]

In this case, we can clearly see that:

[tex]mg>mg-ma_{c}[/tex]

therefore mg>N

This is because the centripetal acceleration is pulling us upwards, that will make the magnitude of the normal force smaller than the product of the mass times the acceleration of gravity.

c)

So let's calculate our weight and normal force:

Let's say we weight a total of 60kg, so:

[tex]mg=(60kg)(9.81m/s^{2})=588.6N[/tex]

and let's calculate the normal force:

[tex]N=m(g-a_{c})[/tex]

[tex]N=(60kg)(9.81m/s^{2}-33.73x10^{-3}m/s^{2})[/tex]

N=586.58N

so now we can calculate the percentage change:

[tex]\%_{change} = \frac{mg-N}{mg}x100\%[/tex]

so we get:

[tex]\%_{change} = \frac{588.6N-586.58N}{588.6N} x 100\%[/tex]

[tex]\%_{change}=0.343\%[/tex]

which is a really small change.

d) In order to find this acceleration, we need to start by calculating the radius of rotation at that point of earth. (See attached picture).

There, we can see that the radius can be found by using the cos function:

[tex]cos \theta = \frac{AS}{h}[/tex]

In this case:

[tex]cos \theta = \frac{r}{R_{E}}[/tex]

so we can solve for r, so we get:

[tex]r= R_{E}cos \theta[/tex]

in this case we'll use the average radius of earch which is 6,371 km, so we get:

[tex]r = (6371x10^{3}m)cos (44.4^{o})[/tex]

which yields:

r=4,551.91 km

and now we can calculate the acceleration at that point:

[tex]a=\omega ^{2}r[/tex]

[tex]a=(72.72x10^{-6} rad/s)^{2}(4,551.91x10^{3}m[/tex]

[tex]a=24.07 mm/s^{2}[/tex]

In this exercise we have to use the knowledge of mechanics to solve the magnitude of the acceleration and the acceleration of gravity, in this way we find that:

a)[tex]a=33.73 mm/s^2[/tex]

b)[tex]mg>N[/tex]

c) [tex]Change=0.343%[/tex]

d) [tex]a=24.07 mm/s^2[/tex]

Then calculating from the information given in the text;

a)In this case, we know the centripetal acceleration is given by the following formula:

[tex]a_c=w^2r\\w=\frac{2\pi}{T} \\T=24*\frac{3600}{60} = 86400 s[/tex]

so we can now find the angular speed:

[tex]w=\frac{2\pi}{86400} \\w=72.72*10^{-6} rad/s^2[/tex]

So the centripetal acceleration will be:

[tex]a_c=(72.72*10^{-6}rad/s^2)^2(6478*10^3m)\\a_c=33.73mm/s^2[/tex]

b)So we can do a sum of forces in equilibrium:

[tex]\sum F=o\\N-mg+ma_c=0\\N-mg+ma_c=0\\N=mg-ma_c\\mg>mg-ma_c\\mg>N[/tex]

This exist cause the centripetal increasing speed happen pulling united states of america upwards, that will create the size of the normal force tinier than the result or goods created of the bulk times the increasing speed of importance.

c) So let's calculate our weight and normal force:

[tex]mg=(60)(9.81)=588.6N\\N=m(g-a_c)\\N=(60)(9.81-33.73*10^{-3})\\N=586.58 N[/tex]

So now we can calculate the percentage change:

[tex]\%change= \frac{mg-N}{mg}*100\% \\=\frac{588.6-586.58}{588.6}*100\% \\=0.343\%[/tex]

d) There, we can see that the radius can be found by using the cos function:

[tex]cos\theta=\Delta S/h\\cos\theta=r/R_E\\r=R_E cos\theta\\r=(6371*10^3)cos(44.4)\\r=4,551.91 km[/tex]

And now we can calculate the acceleration at that point:

[tex]a=w^2r\\a=(72.72*10^{-6})^2(4,551.91*10^3)\\a=24.07 mm/s^2[/tex]

See more about acceleration at brainly.com/question/2437624

Answer:

a. 2936.2 N b. 773.8 N

Explanation:

Let m = mass of car = 1020 kg and m' = mass of trailer and a = acceleration = 2.12 m/s².

a. Determine the net force on the car. N forward

Since the car pulls both itself and the trailer, the combined mass is m + m' and the net force F on the car is F = (m + m')a

= (1020 kg + 365 kg)2.12 m/s²

= 1385 kg × 2.12 m/s²

= 2936.2 N

b. Determine the net force on the trailer. N forward

The net force F' on the trailer is F = m'a

= 365 kg × 2.12 m/s²

= 773.8 N

Answer:

vf = v₁/3 + 2v₂/3

Explanation:

Using the law of conservation of linear momentum,

momentum before impact = momentum after impact

So, Mv₁ + 2Mv₂ = 3Mv (since the railroad cars combine) where v₁ = initial velocity of first railroad car, v₂ = initial velocity of the other two coupled railroad cars, and vf = final velocity of the three railroad cars after impact.

Mv₁ + 2Mv₂ = 3Mvf

dividing through by 3M, we have

v₁/3 + 2v₂/3 = vf

vf = v₁/3 + 2v₂/3

Answer:

v = 37.8 m/s

Explanation:

       [tex]F_{cent} = F_{g} - F_{n} (1)[/tex]

       [tex]v_{max} = \sqrt{g*r} =\sqrt{9.8 m/s2*146 m} = 37.8 m/s (5)[/tex]

Answer:

I believe the answer is sea floor spreading

Answer:

The answer is 5.0

Explanation:

Hope that helped :D

Answer:

9.81 m/s²

Explanation:

It is given that the upward speed of the object is 8.9 m/s.

An object is released.

Now, we know that if the object is released then it will move under the control of gravity which means that the initial velocity of the rocket will not affect the acceleration of the object.

Hence, the downward acceleration of the object will be equal to g i.e. 9.81 m/s².

Answer:

B or C

Explanation:

i don't if I'm right or not

Answer:

Explanation:

The first case relates to open end pipe with fundamental frequency . Let the length of flute be l . The length vibrating column  is l .

λ = v / n where n is frequency , v is velocity of light and λ is wavelength of sound produced .

λ = 343 / 261.6

= 1.31 m .

For fundamental frequency

λ / 2 = l

l = λ / 2

= 1.31 / 2

= .655 m

= 65.5 cm .

b )

Nearby colder room will create lower frequency because velocity of sound will be smaller .

frequency of note produced = 261.6 - 3 = 258.6 Hz .

velocity of sound v = n λ

= 258.6 x  1.31 m

= 338.76 m /s

decrease of velocity = 343 - 338.76 = 4.24 m /s

1 degree change in temperature produces a change of .61 m/s change in velocity .

decrease in temperature = 4.24 / .61

= 7⁰C

Temperature of colder room = 20 - 7 = 13⁰C .

Explanation:

Helium is the type of atom, with 2 protons in the nucleus.

He-3 (which is the proper form), is an isotope of helium that has an atomic number of 3 (so 2 protons and 1 electron).

The most abundant helium isotope is He-4, just an extra fact.

Answer:

The answer is below

Explanation:

The equation for a linear line graph is given by:

y = mx + b, where y and x are variables, m is the slope of the graph and b is the y intercept (that is value of y when x is zero).

The slope (m) of a line passing through two points [tex](x_1,y_1)\ and\ (x_2,y_2)[/tex] is given by:

[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]

A) The first line passes through the point (0, 0) and (10, 60). It is represented as (time, velocity). Hence the slope is:

[tex]m=\frac{60 -0}{10-0}=6\ m/min^2[/tex]

B) The second line passes through the point (10, 60) and (15, 60). Hence the slope is:

[tex]m=\frac{60 -60}{15-10}=0\ m/min^2[/tex]

C) The third line passes through the point (15, 60) and (40, -40). Hence the slope is:

[tex]m=\frac{-40 -60}{40-15}=-4\ m/min^2[/tex]

D) The third line passes through the point (40, -40) and (55, 0). Hence the slope is:

[tex]m=\frac{0 -(-40)}{55-40}=2.67\ m/min^2[/tex]

Answer:

adbce i think

Explanation:

Answer:

Friction between two objects causes a transfer of electrons from one object to the other.

Explanation:

Answer:

[tex]P=18,933.3Pa=18.9kPa[/tex]

Explanation:

Hello!

In this case, since we can compute the volume of the brick as shown below:

[tex]V=0.1m*0.1m*0.2m=0.002m^3[/tex]

Next, we can compute the mass of the brick given its density:

[tex]\rho =m/V\\\\m=V*\rho\\\\m=0.002m^3*19,300kg/m^3\\\\m=38.6kg[/tex]

Now, since the force exerted on the table corresponds to the weight of the brick, we use the gravity to obtain:

[tex]W=38.6kg*9.81m/s^2=378.7N[/tex]

Finally, since the surface of the brick in contact with the table corresponds to the 0.1x0.2 area (length and width), the area on which the weight force is exerted is:

[tex]A=0.1m*0.2m=0.02m^2[/tex]

Therefore, the pressure is:

[tex]P=\frac{F}{A}=\frac{W}{A}=\frac{378.7N}{0.02m^2}\\\\P=18,933.3Pa=18.9kPa[/tex]

Best regards!