What is the function of a landfill

This question is incomplete, the complete question is;

Two blocks, M1M1 and M2,M2, are connected by a massless string that passes over a massless pulley as shown in the figure. M2,M2, which has a mass of 25.0 kg,25.0 kg, rests on a long ramp of angle =25.0∘.θ=25.0∘. Friction can be ignored in this problem.

Determine the mass 1m1 of block M1M1 for which the two blocks are in equilibrium (no acceleration).

Answer:

the mass m1m1 of block M1M1 for which the two blocks are in equilibrium (no acceleration) is 10.57 kg

Explanation:

Given that;

m₂ = 25 kg

θ = 25°

Now at equilibrium, T = m₁g ------------------lets say equ 1

and also T = m₂gsinθ

therefore

m₁g = m₂gsinθ

m₁ = m₂sinθ

so we substitute

m₁ = 25 × sin(25)

m₁ = 25 × 0.4226

m₁ = 10.565 ≈ 10.57 kg

therefore the mass m1m1 of block M1M1 for which the two blocks are in equilibrium (no acceleration) is 10.57 kg

Answer:

I don't do physics but I think the answer would be -3674

Explanation:

Newton's 2 law of motion

Answer: 10.58 C has flowed during the lightning bolt

Explanation:

Given that;

Time of flow t = 1.2 × 10⁻³

perpendicular distance r = 21 m

Magnetic field B = 8.4 x 10⁻⁵ T

Now lets consider the expression for magnetic field;

B = u₀I / 2πr

the current flow is;

I = ( B × 2πr ) / u₀

so we substitute

I = ( (8.4 x 10⁻⁵) × 2 × 3.14 × 21 ) / 4π ×10⁻⁷

=  0.01107792 / 0.000001256

= 8820 A

Hence the charge flows during lightning bolt  will be;

q = It

so we substitute

q = 8820 × 1.2 × 10⁻³

q = 10.58 C

therefore 10.58 C has flowed during the lightning bolt

Answer:

Yes that is almost always the problem!

Explanation:

Answer:

the answer is Acid reflux hope this helped!

Answer:

The time is 1.4 sec

The initial velocity is 10.7 m/s.

Explanation:

Given that,

Height = 10 m

Distance = 15 m

We need to calculate the time

Using equation of motion

[tex]s=ut-\dfrac{1}{2}gt^2[/tex]

Put the value into the formula

[tex]10=0+\dfrac{1}{2}\times9.8\times t^2[/tex]

[tex]t^2=\dfrac{2\times10}{9.8}[/tex]

[tex]t=\sqrt{\dfrac{2\times10}{9.8}}[/tex]

[tex]t=1.4\ sec[/tex]

We need to calculate the initial velocity

Using formula of velocity

[tex]v=\dfrac{d}{t}[/tex]

Put the value into the formula

[tex]v=\dfrac{15}{1.4}[/tex]

[tex]v=10.7\ m/s[/tex]

Hence, The time is 1.4 sec

The initial velocity is 10.7 m/s.

Answer:

Option C.

Explanation:

To know which velocity-time graph best describes the motion of the ball, let us calculate the velocity of the ball and the time taken for the ball to get the ground. This can be obtained as follow:

1. Determination of the velocity.

Initial velocity (u) = 0 m/s

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

Height (h) = 100 m

Final velocity (v) =.?

v² = u² + 2gh

v² = 0² + (2 × 9.8 × 100)

v² = 0 + 1960

v² = 1960

Take the square root of both side.

v = √(1960)

v = 44.27 m/s

2. Determination of the time taken.

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

Height (h) = 100 m

Time (t) =.?

h = ½gt²

100 = ½ × 9.8 × t²

100 = 4.9 × t²

Divide both side by 4.9

t² = 100 / 4.9

Take the square root of both side

t = √(100 / 4.9)

t = 4.52 s

From the above illustration,

Initial time (t1) = 0 s

Final time (t2) = 4.52 s

Initial velocity (u) = 0 m/s

Final velocity (v) = 44.27 m/s

Thus, we can see that as the time increase, the velocity also increase. Therefore, option C gives the correct answer to the question.

Answer:

A. decantation and screening

Explanation:

Decantation is the one of the process of separating the mixture. In this process the precipitated liquid is separated from the solid. According to the given instruction for the recipe, the fat which is in liquid state is separated from meat. In the process of screening, more liquid is separated by placing the mixture on the screen. Here, the gravity plays an important role for the process of separation.  

Answer:

a

Explanation:

Answer:

B. 12

Explanation:

4 x 3 = 12

Answer:

The Speed of the vehicles is 9.34m/s

Explanation:

For an elastic collision the two bodies move with similar velocities after collision

Given

M1=1330kg

V1=15m/s

M2=805kg

V2=0(the car is parked on neutral)

The formula is

M1V1+M2V2=(M1+M2)V

1330*15+805*0=(1330+805)V

19950+0=2135V

2135V=19950

divide both sides by 2135

V=19950/2135

V=9.34m/s

v² - u² = 2 a ∆x

where u = initial velocity (27 m/s), v = final velocity (0), a = acceleration (-8 m/s², taken to be negative because we take direction of movement to be positive), and ∆x = stopping distance.

So

0² - (27 m/s)² = 2 (-8 m/s²) ∆x

∆x = (27 m/s)² / (16 m/s²)

∆x ≈ 45.6 m

The stopping distance of car achieved during the braking is of 45.56 m.

Given data:

The initial speed of car is, u = 27 m/s.

The final speed of car is, v = 0 m/s. (Because car comes to stop finally)

The magnitude of deacceleration is, [tex]a = 8\;\rm m/s^{2}[/tex].

In order to find the stopping distance of the car, we need to use the third kinematic equation of motion. Third kinematic equation of motion is the relation between the initial speed, final speed, acceleration and distance covered.

Therefore,

[tex]v^{2}=u^{2}+2(-a)s[/tex]

Here, s is the stopping distance.

Solving as,

[tex]0^{2}=27^{2}+2(-8)s\\\\s = 45.56 \;\rm m[/tex]

Thus, we can conclude that the stopping distance of car achieved during the braking is of 45.56 m.

Learn more about the kinematic equation of motion here:

https://brainly.com/question/11298125

Answer:

The speed will be decrease

Answer:

Explanation:

Find the complete question attached

Using the principle of moment

Clockwise moment = Anticlockwise moment

AntiClockwise moment = M × 2.0

ACW moment = 2M

Clockwise moment = 40×4

Clockwise moment = 160kgcm

Equate both expression and calculate M

2M = 160

M = 160/2

M = 80kg

Hence the mass of his friend is 80kg

Answer:

The appropriate response is "Optical printer ".

Explanation:

Answer:

9 ft*lb

Explanation:

super simple but you just have to understand that the integral is going with the curve

work = integral a to b of f(x)dx = integral 0 to 9 of 10/(1+x)^2dx = 9ft*lb

Explanation:

the link enjoy

Answer:

Correlational research can be used to see if two variables are related and to make predictions based on this relationship.

Answer:

Covalent Bond

Explanation:

i took the test , mark me brainliest.

Answer: Electrical

Explanation: Atoms are tied together by electrical bonding forces.

Answer:

The linear velocity is represented by the following expression: [tex]v = \frac{s}{t}[/tex]

Explanation:

From Rotation Physics we know that linear velocity of a point moving with uniform circular motion is:

[tex]v = r\cdot \omega[/tex] (Eq. 1)

Where:

[tex]r[/tex] - Radius of rotation of the particle, measured in meters.

[tex]\omega[/tex] - Angular velocity, measured in radians per second.

[tex]v[/tex] - Linear velocity of the point, measured in meters per second.

But we know that angular velocity is also equal to:

[tex]\omega = \frac{\theta}{t}[/tex] (Eq. 2)

Where:

[tex]\theta[/tex] - Angular displacement, measured in radians.

[tex]t[/tex] - Time, measured in seconds.

By applying (Eq. 2) in (Eq. 1) we get that:

[tex]v = \frac{r\cdot \theta}{t}[/tex] (Eq. 3)

From Geometry we must remember that circular arc ([tex]s[/tex]), measured in meters, is represented by:

[tex]s = r\cdot \theta[/tex]

[tex]v = \frac{s}{t}[/tex]

The linear velocity is represented by the following expression: [tex]v = \frac{s}{t}[/tex]

Answer:

Hello your question is incomplete below is the complete question

Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000

answer : V = 1.624* 10^-5 m/s

Explanation:

First we have to calculate the value of a

a = 93 * 10^6 mile/m  * 1609.344 m

  = 149.668 * 10^8 m

next we will express the distance between the earth and the sun

[tex]r = \frac{a(1-E^2)}{1+Ecos\beta }[/tex]   --------- (1)

a = 149.668 * 10^8

E (eccentricity ) = ( 1/60 )^2

[tex]\beta[/tex] = 90°

input the given values into equation 1 above

r = 149.626 * 10^9 m

next calculate the Earths velocity of approach towards the sun using this equation

[tex]v^2 = \frac{4\pi^2 }{r_{c} }[/tex]   ------ (2)

Note :

Rc = 149.626 * 10^9 m

equation 2 becomes

([tex]V^2 = (\frac{4\pi^{2} }{149.626*10^9})[/tex]

therefore : V = 1.624* 10^-5 m/s

Answer:

[tex]E=7.28\times 10^6\ J[/tex]

Explanation:

Given that,

Mass of a person, m = 84 kg

The person is standing at a top of Mt. Everest at an altitude of 8848 m

We need to find the gravitational potential energy of the person. We know that the gravitational potential energy is possessed due to the position of an object. It is given by :

E = mgh, g is the acceleration due to gravity

[tex]E=84\ kg\times 9.8\ m/s^2\times 8848\ m\\\\E=7283673.6\ J\\\\E=7.28\times 10^6\ J[/tex]

So, the gravitational potential energy of the person is [tex]7.28\times 10^6\ J[/tex]

Answer:

303 Ω

Explanation:

Given

Represent the resistors with R1, R2 and RT

R1 = 633

RT = 205

Required

Determine R2

Since it's a parallel connection, it can be solved using.

1/Rt = 1/R1 + 1/R2

Substitute values for R1 and RT

1/205 = 1/633 + 1/R2

Collect Like Terms

1/R2 = 1/205 - 1/633

Take LCM

1/R2 = (633 - 205)/(205 * 633)

1/R2 = 428/129765

Take reciprocal of both sides

R2 = 129765/428

R2 = 303 --- approximated

Answer:

300000 Joules

Explanation:

Recall that the unit Watts is Joules per second, derived from the quotient of energy per unite of time. Therefore, to calculate the energy used on the given time, we need to multiply the power used (1000 W) times the time used. Then we need to express the 5 minutes in seconds to get our answer in Joules:

5 minutes = 5 * 60 seconds = 300 seconds

Final energy calculation gives:

E = 1000 W * 300 sec= 300000 Joules

Complete Question

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

Answer:

The expression is  [tex]W_c =  P_o V_o ln (R_v)[/tex]  

Explanation:

Generally smallest workdone done by  a gas is mathematically represented as

          [tex]dW  =  PdV[/tex]

Generally for an isothermal process

    [tex]PV  =  nRT = constant [/tex]

=>   [tex]P = \frac{nRT}{V}[/tex]

Generally the total workdone is mathematically represented as

   [tex]W_c =  \int\limits^{v_f}_{V_o} {\frac{nRT}{V} } \, dV[/tex]

=> [tex]W_c = nRT  \int\limits^{V_f}_{V_o} {\frac{1}{V} } \, dV[/tex]

=>  [tex]nRT [lnV]   | \left \ {V_f}} \atop {V_o}} \right.[/tex]

=>  [tex]W_c = nRT [ln(V_f) - ln(V_o)][/tex]

=>  [tex]W_c = nRT ln \frac{V_f}{V_o}[/tex]

From the question [tex]\frac{V_f}{V_o }  =  R_v[/tex]

=> [tex]W_c =  P Vln (R_v)[/tex]

at initial  state

[tex]W_c =  P_o V_o ln (R_v)[/tex]  

Answer: 59.8 or 299/5​- 55 x 4.8=264

Explanation:

Or if you wish to multiply it then the answer is above with the addition version.

Hope this helped :)

Answer:

I=10.0A

Explanation:

Answer:

the acceleration of gravity.

Answer:

g stand for the acceleration of gravity .

Explanation:

Answer:

(a) 3.43 m/s

(b) 3.43 m/s

(c) 95.8 kPa

Explanation:

Use Bernoulli equation:

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

where P is pressure (either absolute or gauge), ρ is density, v is velocity, g is acceleration due to gravity, and h is elevation.

(a) Let's choose point 1 at the surface of the fluid in the container, and point 2 at point Z at the exit of the tube.  I'll say 0 elevation is at point Z, and I'll use gauge pressure.

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

0 Pa + ½ ρ (0 m/s)² + ρ (9.8 m/s²) (0.60 m) = 0 Pa + ½ ρ v² + 0

ρ (9.8 m/s²) (0.60 m) = ½ ρ v²

5.88 m²/s² = ½ v²

v = 3.43 m/s

(b) The tube's cross section is constant, so the fluid's speed is the same at all points in the tube. v = 3.43 m/s.

(c) Use Bernoulli equation again, choosing point 2 to be at Y.  I'll say 0 elevation is at the surface of the fluid, and again use gauge pressure.

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

0 + 0 + 0 = P + ½ (700 kg/m³) (3.43 m/s)² + (700 kg/m³) (9.8 m/s²) (0.20 m)

0 = P + 4116 Pa + 1372 Pa

P = -5488 Pa

The gauge pressure is -5488 Pa, so the absolute pressure is 101,300 Pa + -5488 Pa = 95812 Pa, or approximately 95.8 kPa.