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lana [24]
1 year ago
7

Complete the following hypothesis the pressure lncrease as

Physics
1 answer:
Nitella [24]1 year ago
8 0

Temperature is increased the pressure also increases due to fast movement of molecules because of more energy.

<h3>How temperature is related to pressure?</h3>

The temperature of the gas is directly proportional to the pressure. Faster moving particles will collide with the walls of the container that put force on the wall which leads to increase in pressure.

So we can conclude that temperature is increased the pressure also increases due to fast movement of molecules because of more energy.

Learn more about pressure here: brainly.com/question/25736513

#SPJ1

You might be interested in
Air expands adiabatically in a piston–cylinder assembly from an initial state where p1 = 100 lbf/in.2, v1 = 3.704 ft3/lb, and T1
lutik1710 [3]

Answer:

Final temperature is equal to 1291.63°R  

Explanation:

given,

p₁ = 100 lb f/in²,               v₁ = 3.704 ft³/lb,           and T₁ = 1000 °R

p₂ = 30 lb f/in²                 n = 1.4

Δ u = 0.171(T₂ - T₁)

we know for poly tropic process

p vⁿ = constant

p₁ v₁ⁿ = p₂ v₂ⁿ

100 × 3.704¹°⁴ = 30 × v₂¹°⁴

v₂ = 8.753 ft³/lb

work done for poly tropic process

W = \dfrac{p_1v_1-p_2v_2}{n-1}

    = \dfrac{100\times 3.704-30\times 8.753}{1.4-1}

    = 269.525 lbf/in².ft³

W = \dfrac{269.525}{5.40395} Btu/lb

   = 49.87 Btu/lb

in the piston cylinder arrangement air is expanding acrobatically

Δ q = Δu + w

Δ u = - w

0.171(T₂ - T₁) = -49.87

0.171(T₁ - T₂) = -49.87

0.171 T₂ = 0.171 × 1000 + 49.87

T₂ = 1291.63 °R

Final temperature is equal to 1291.63°R  

5 0
2 years ago
A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A cons
Virty [35]

Complete Question

A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.What will be the equilibrium height of the mass?

Answer:

H_m=1.65m

H_E=1.16307m

Explanation:

From the question we are told that

Mass of ball M=2kg

Length of string L= 2m

Wind force F=13.2N

Generally the equation for \angle \theta is mathematically given as

tan\theta=\frac{F}{mg}

\theta=tan^-^1\frac{F}{mg}

\theta=tan^-^1\frac{13.2}{2*2}

\theta=73.14\textdegree

Max angle =2*\theta= 2*73.14=>146.28\textdegree

Generally the equation for max Height H_m is mathematically given as

H_m=L(1-cos146.28)

H_m=0.9(1+0.8318)

H_m=1.65m

Generally the equation for Equilibrium Height H_E is mathematically given as

H_E=L(1-cos73.14)

H_E=0.9(1+0.2923)

H_E=1.16307m

8 0
2 years ago
A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)
Phantasy [73]

Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is U  =  \int\limits^T_0 {P(t)} \, dt

Compute U if the bulb remains on for 5h

Answer:

The value is  U  =  7.563 *10^{5} \  J

Explanation:

From the question we are told that

   The power rating of the bulb is P  =  100 \  W

   The resistance is   R =  143 \ \Omega

   The  voltage is  V  =  V_o  sin [2 \pi ft]

   The  energy expanded is U  =  \int\limits^T_0 {P(t)} \, dt

   The  voltage  V_o  =  110 \  V

   The frequency is  f =  60 \  Hz

    The  time considered is  t =  5 \  h  =  18000 \  s

Generally power is mathematically represented as

             P =  \frac{V^2}{ R}

=>          P =  \frac{( 110  sin [2 \pi * 60t])^2}{ 144}

=>           P =  \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}

So  

     U  =  \int\limits^T_0 { \frac{ 110^2*  [sin [120 \pi t])^2}{ 144}} \, dt

=>  U  =  \frac{110^2}{144} \int\limits^T_0 { (   sin^2 [120 \pi t]} \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | T} \atop {0}} \right.

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | 18000} \atop {0}} \right.

U =  \frac{110^2}{144} [\frac{18000}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi (18000))}{240 \pi} ] ]

=>   U  =  7.563 *10^{5} \  J

7 0
3 years ago
A net force of 125 N accelerates a 25.0 kg mass. What is the resulting acceleration?
Neko [114]

Answer: a=5 m/s^2

Explanation:

The acceleration of an object can be calculated by using Newton's second law:

F=ma

where

F is the net force applied on the object

m is the mass of the object

a is its acceleration

In this problem, we have F=125 N and m=25.0 kg, so we can rearrange the equation to calculate the acceleration:

a=\frac{F}{m}=\frac{125 N}{25.0 kg}=5 m/s^2

5 0
3 years ago
Read 2 more answers
A single point on a distance time graph tells the
Sonbull [250]

Answer: Instantaneous speed.

Explanation:

4 0
3 years ago
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