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loris [4]
3 years ago
5

Why is gas able to flow? a its particles have melted and can move around b its particles have high viscosity and can move around

c its particles have enough energy to move around d its particles have enough pressure to move around 3. which of the following is true about the viscosity of liquids? a liquids with high viscosity do not flow as well as liquids with low viscosity b liquids with high viscosity flow better than liquids with low viscosity c liquids with high viscosity do not resist flow any more than liquids with low viscosity d water is a good example of a liquid with high viscosity 4.in a(n_____, the particles form a regular, repeating pattern please help!?
Physics
1 answer:
ratelena [41]3 years ago
4 0
The answer to why is gas able to flow is answer c its particals have enough energy to move around.
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Which is a characteristic of centripetal acceleration?
djyliett [7]

Answer:

It is directed inward, toward the center of a circle.

5 0
3 years ago
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Three liquids are at temperatures of 6 ◦C, 23◦C, and 38◦C, respectively. Equal masses of the first two liquids are mixed, and th
kupik [55]

The equilibrium temperature is T13=3.12 ◦C

<u>Explanation:</u>

<u>Given </u>

The temperature of liquids: T1=6◦C, T2=23◦C, T3=38◦C

The temperature of 1+2 liquids mix: T12= 13◦C.

The temperature of 2+3 liquids mix: T23=26.8 ◦C.

The temperature of 1+3 liquids mix: T13= ??

<u>1.When the first two liquids are mixed:</u>

  • mC1(T1-T12)+mC2(T2-T12)=0
  • C1(6-13)=C2(23-13)=0
  • 7C1=10C2
  • C1=1.42C2

<u>2.When the second and third liquids are mixed</u><u>:</u>

  • mC2(T2-T23)+mC3(T3-T23)=0
  • C2(23-26.8)=C3(38-26.8)=0
  • 3.8C2=12.8C3
  • C2=3.36C3

<u>3.When the first and third liquids are mixed:</u>

  • mC1(T1-T13)+mC3(T3-T13)=0
  • C1(6-T13)+C3(38-T13)=0
  • C1=1.42C2  C2=3.36C3
  • C1=1.42C2(3.36C3)
  • C1=4.77C3
  • C1(6-T13)+C3(38-T13)=0
  • 4.77C3(6-T13)+C3(38-T13)=0
  • By solving the equation we get,
  • T13=3.12 ◦C
  • The equilibrium temperature is T13=3.12 ◦C

<u></u>

7 0
3 years ago
A 1 kg mass is attached to a spring with spring constant 7 Nt/m. What is the frequency of the simple harmonic motion? What is th
Scorpion4ik [409]

1. 0.42 Hz

The frequency of a simple harmonic motion for a spring is given by:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}

where

k = 7 N/m is the spring constant

m = 1 kg is the mass attached to the spring

Substituting these numbers into the formula, we find

f=\frac{1}{2\pi}\sqrt{\frac{7 N/m}{1 kg}}=0.42 Hz

2. 2.38 s

The period of the harmonic motion is equal to the reciprocal of the frequency:

T=\frac{1}{f}

where f = 0.42 Hz is the frequency. Substituting into the formula, we find

T=\frac{1}{0.42 Hz}=2.38 s

3. 0.4 m

The amplitude in a simple harmonic motion corresponds to the maximum displacement of the mass-spring system. In this case, the mass is initially displaced by 0.4 m: this means that during its oscillation later, the displacement cannot be larger than this value (otherwise energy conservation would be violated). Therefore, this represents the maximum displacement of the mass-spring system, so it corresponds to the amplitude.

4. 0.19 m

We can solve this part of the problem by using the law of conservation of energy. In fact:

- When the mass is released from equilibrium position, the compression/stretching of the spring is zero: x=0, so the elastic potential energy is zero, and all the mechanical energy of the system is just equal to the kinetic energy of the mass:

E=K=\frac{1}{2}mv^2

where m = 1 kg and v = 0.5 m/s is the initial velocity of the mass

- When the spring reaches the maximum compression/stretching (x=A=amplitude), the velocity of the system is zero, so the kinetic energy is zero, and all the mechanical energy is just elastic potential energy:

E=U=\frac{1}{2}kA^2

Since the total energy must be conserved, we have:

\frac{1}{2}mv^2 = \frac{1}{2}kA^2\\A=\sqrt{\frac{m}{k}}v=\sqrt{\frac{1 kg}{7 N/m}}(0.5 m/s)=0.19 m

5. Amplitude of the motion: 0.44 m

We can use again the law of conservation of energy.

- E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2 is the initial mechanical energy of the system, with x_0=0.4 m being the initial displacement of the mass and v_0=0.5 m/s being the initial velocity

- E_f = \frac{1}{2}kA^2 is the mechanical energy of the system when x=A (maximum displacement)

Equalizing the two expressions, we can solve to find A, the amplitude:

\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}kA^2\\A=\sqrt{x_0^2+\frac{m}{k}v_0^2}=\sqrt{(0.4 m)^2+\frac{1 kg}{7 N/m}(0.5 m/s)^2}=0.44 m

6. Maximum velocity: 1.17 m/s

We can use again the law of conservation of energy.

- E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2 is the initial mechanical energy of the system, with x_0=0.4 m being the initial displacement of the mass and v_0=0.5 m/s being the initial velocity

- E_f = \frac{1}{2}mv_{max}^2 is the mechanical energy of the system when x=0, which is when the system has maximum velocity, v_{max}

Equalizing the two expressions, we can solve to find v_{max}, the maximum velocity:

\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}mv_{max}^2\\v_{max}=\sqrt{\frac{k}{m}x_0^2+v_0^2}=\sqrt{\frac{7 N/m}{1 kg}(0.4 m)^2+(0.5 m/s)^2}=1.17 m/s m

4 0
3 years ago
Read 2 more answers
Please help! Will mark best answer!!
MA_775_DIABLO [31]
The answer is (A) hope it helps 
7 0
3 years ago
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3. What is the mass of a paratrooper who experiences an air resistance of 400 N and an acceleration of 4.5 m/s2
goblinko [34]

Answer:

88.89kg

Explanation:

The formula for mass is m=F/a. If we plug in the values, we get m=400N/4.5m/s^2. The mass is 88.89kg. We know that the unit is in kg because one newton (N) is 1kg*m/s^2. The m/s^2 is cancelled out by the acceleration, and we are left with kg.

4 0
4 years ago
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