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arlik [135]
3 years ago
15

sugar solution does not change the colour of green universal indicator what does this tell you about sugar solution

Physics
1 answer:
Hunter-Best [27]3 years ago
6 0

Answer:

Sugar solution is a neutral solution.

Explanation:

-The universal indicator is used to tests the acidity or alkalinity of a solution.

-It turns  red when added to a strong acid solution.

--It turns purple when added to a strong base.

-It remains green when added to a neutral solution.

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A weight of 30.0 N is suspended from a spring that has a force constant of 220 N/m. The system is undamped and is subjected to a
Nimfa-mama [501]

Answer:

F_0 = 393 N

Explanation:

As we know that amplitude of forced oscillation is given as

A = \frac{F_0}{ m(\omega^2 - \omega_0^2)}

here we know that natural frequency of the oscillation is given as

\omega_0 = \sqrt{\frac{k}{m}}

here mass of the object is given as

m = \frac{W}{g}

\omega_0 = \sqrt{\frac{220}{\frac{30}{9.81}}}

\omega_0 = 8.48 rad/s

angular frequency of applied force is given as

\omega = 2\pi f

\omega = 2\pi(10.5) = 65.97 rad/s

now we have

0.03 = \frac{F_0}{3.06(65.97^2 - 8.48^2)}

F_0 = 393 N

6 0
3 years ago
The weights in atwoods machine, starting at rest, attain a velocity of 2ft/sec in one sec. Find the ratio of the masses
Orlov [11]
Refer to the figure shown below.
Let m₁ and m₂ e the two masses.
Let a = the acceleration.
Let T =  tension over the frictionless pulley.

Write the equations of motion.
m₂g - T = m₂a            (1)
T - m₁g = m₁a            (2)

Add equations (1) and (2).
m₂g - T + T - m₁g = (m₁ + m₂)a
(m₂ - m₁)g = (m₁ + m₂)a

Divide through by m₁.
(m₂/m₁ - 1)g = (1 + m₂/m₁)a

Define r = m₂/m₁ as the ratio of the two masses. Then
(r - 1)g = (1 +r)a
r(g-a) = a + g
r = (g - a)/(g + a)

With  = 2 ft/s from rest, the acceleration is
a = 2/32.2 = 0.062 ft/s²
Therefore
r = (32.2 - 0.062)/(32.2 + 0.062) = 0.9962

Answer:
The ratio of masses is 0.9962 (heavier mass divided by the lighter mass).

8 0
3 years ago
The spring of a toy car is wound by pushing the car
Zepler [3.9K]

The elastic potential energy stored in the car's spring during the process is 3.75 J

<h3>Determination of the spring constant</h3>

From the question given above, the following data were obtained:

  • Force (F) = 15 N
  • Extention (e) = 0.5 m
  • Spring constant (K) =?

K = F/e

K = 15 / 0.5

K = 30 N/m

<h3>Determination of the potential energy</h3>
  • Spring constant (K) = 30 N/m
  • Extention (e) = 0.5 m
  • Potential energy (PE) =?

PE = ½Ke²

PE = ½ × 30 × 0.5²

PE = 15 × 0.25

PE = 3.75 J

Therefore, the elastic potential energy stored in the car's spring during the process is 3.75 J

Learn more about energy stored in spring:

brainly.com/question/4280346

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2 years ago
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