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3 years ago

7

A quantity of a gas is an absolute pressure of 400 K PA in the absolute temperature of 110° Kelvin when the temperature of the g

as is raised to 235° Kelvin what's the new pressure of the gas?

1 answer:

aliina [53]3 years ago

4 0

Given:

P1 = 400 kPa
T1 = 110 K

T2 = 235K

Required:

P2

Solution:

Apply Gay-Lussac’s law where P/T = constant

P1/T1 = P2/T2

P2 = T2P1/T1

P2 = (235K)(400kPa) / (110K)

P2 = 855 kPa

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An airplane accelerates at 16 m/s² from rest. If it lifts off from an aircraft carrier after 6.0 s, how far did it go down the r

Alona [7]

B is the answer
16 x 6.0 = 96
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1 year ago

Infer why the output force exerted by a rake must be less than input force?

adell [148]

<h3><u>Answer and Explanation</u>;</h3>

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2 years ago

An 80-kg hiker climbs to the top of a tall hill and builds up 470,000 J of gravitational potential energy. How high did the hike

nekit [7.7K]

Answer:

599 meters is the answer rounded to the nearest whole number and 599.489795918 meters is the complete answer

Explanation:

to find gravitational potential energy you multiply mass x acceleration due to gravity (always 9.8 on earth) x hight

since we know the gravitational potential energy and want to find out the hight, we take the gravitational potential energy (470,000) and divide it by the product of acceleration due to gravity x mass (9.8 x 80)

so how high the hiker climbed is equal to 470,000 divided by (9.8 x 80)

hight = 470,000 / (9.8 x 80)

hight = 470,000 / 784

hight = 599.489795918 meters

as for rounding, if the decimal is less than 5 you round "down" and keep the current whole number, if the decimal is 5 or greater you round "up" and add 1 to get your new number

5 0

2 years ago

Read 2 more answers

An object of mass m = 4.0 kg, starting from rest, slides down an inclined plane of length l = 3.0 m. The plane is inclined by an

kirill [66]

Answer:

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf= -20.4 J is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane:

Wg= 58.8 J is positive

Explanation:

Nomenclature

vf: final velocity

v₀ :initial velocity

a: acceleleration

d: distance

Ff: Friction force

W: weight

m:mass

g: acceleration due to gravity

Graphic attached

The attached graph describes the variables related to the kinetics of the object (forces and accelerations)

Calculation de of the components of W in the inclined plane

W=m*g

Wx₁ = m*g*sin30°

Wy₁= m*g*cos30°

Object kinematics on the inclined plane

vf₁²=v₀₁²+2*a₁*d₁

v₀₁=0

vf₁²=2*a₁*d₁

$v_{f1} = \sqrt{2*a_{1}*d_{1} }$ Equation (1)

Object kinetics on the inclined plane (μ= 0.2)

∑Fx₁=ma₁ :Newton's second law

-Ff₁+Wx₁ = ma₁ , Ff₁=μN₁

-μ₁N₁+Wx₁ = ma₁ Equation (2)

∑Fy₁=0 : Newton's first law

N₁-Wy₁= 0

N₁- m*g*cos30°=0

N₁ = m*g*cos30°

We replace N₁ = m*g*cos30 and Wx₁ = m*g*sin30° in the equation (2)

-μ₁m*g*cos30₁+m*g*sin30° = ma₁ : We divide by m

-μ₁*g*cos30°+g*sin30° = a₁

g*(-μ₁*cos30°+sin30°) = a₁

a₁ =9.8(-0.2*cos30°+sin30°)=3.2 m/s²

We replace a₁ =3.2 m/s² and d₁= 3m in the equation (1)

$v_{f1} = \sqrt{2*3.2*3} }$

$v_{f1} =\sqrt{2*3.2*3}$

$v_{f1} = 4.38 m/s$

Rough surface kinematics

vf₂²=v₀₂²+2*a₂*d₂ v₀₂=vf₁=4.38 m/s

0 =4.38²+2*a₂*d₂ Equation (3)

Rough surface kinetics (μ= 0.3)

∑Fx₂=ma₂ :Newton's second law

-Ff₂=ma₂

--μ₂*N₂ = ma₂ Equation (4)

∑Fy₂= 0 :Newton's first law

N₂-W=0

N₂=W=m*g

We replace N₂=m*g inthe equation (4)

--μ₂*m*g = ma₂ We divide by m

--μ₂*g = a₂

a₂ =-0.2*9.8= -1.96m/s²

We replace a₂ = -1.96m/s² in the equation (3)

0 =4.38²+2*-1.96*d₂

3.92*d₂ = 4.38²

d₂=4.38²/3.92

d₂=4.38²/3.92

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf = - Ff₁*d₁

Ff₁= μ₁N₁= μ₁*m*g*cos30°= -0.2*4*9.8*cos30° = 6,79 N

Wf= - 6.79*3 = 20.4 N*m

Wf= -20.4 J is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane

Wg=W₁x*d= m*g*sin30*3=4*9.8*0.5*3= 58.8 N*m

Wg= 58.8 J is positive

6 0

3 years ago

Help!!! I need it today <br> Thank you in advance

svlad2 [7]

Answer:

F = - k (x-xo) a graph of the weight or applied force against the elongation obtaining a line already proves Hooke's law.

Explanation:

The student wants to prove hooke's law which has the form

F = - k (x-xo)

To do this we hang the spring in a vertical position and mark the equilibrium position on a tape measure, to simplify the calculations we can make this point zero by placing our reference system in this position.

Now for a series of known masses let's get them one by one and measure the spring elongation, building a table of weight vs elongation,

we must be careful when hanging the weights so as not to create oscillations in the spring

we look for the mass of each weight

W = mg

m = W / g

and we write them in a new column, we make a graph of the weight or applied force against the elongation and it should give a straight line; the slope of this line is sought, which is the spring constant.

The fact of obtaining a line already proves Hooke's law.

5 0

2 years ago