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ladessa [460]
4 years ago
11

10. How do you use a triple-beam balance? Fill in the blanks.

Physics
1 answer:
wolverine [178]4 years ago
5 0

Answer:

1st – Place the film canister on the <u>scale</u>.

2nd – Slide the large <u>weight </u>to the right until the arm drops below the line and then move it back one notch.

3rd – Repeat this process with the <u>top</u> weight. When the arm moves below the line, back it up one groove.

4th – Slide the <u>small </u>weight on the front beam until the <u>lines</u> match up.

5th – Add the amounts on each beam to find the total <u>mass </u>to the nearest tenth of a gram.

Explanation:

The triple beam balance is an instrument that is used in measuring the mass of substances to a very high degree of precision. The reading error is given by ±0.05 grams. The triple beam balance as the name implies has three beams that measure substances of different mass levels.

The beams are categorized as small, medium, and large. There is a balance on which the substance to be weighed is placed directly upon. To use this measuring device, the procedures mentioned above are followed.

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a sealed bottle of water was taken from a refrigerator and left in a warm kitchen. explain in the terms of its molecules what ha
DiKsa [7]

Answer:

Explanation:

The sealed bottle serves as an open system for the water molecules, thus heat energy can be exchanged with its surroundings. When the bottle is placed in a warm environment, there would be a gradual increase in the thermal energy of the molecules of water due to a change in the temperature of its surroundings.

Thus, the molecules of water closer to the boundary of the bottle absorbs heat, becomes lighter and changes position. This gradual process continues until all molecules attains an average temperature which equals that of the surroundings of the bottle.

Therefore, both the thermal energy and temperature of the water increases.

4 0
3 years ago
The inductor in the RLC tuning circuit of an AM radio has a value of 250 mHmH . You may want to review (Pages 857 - 860) . Part
ANEK [815]

Answer:

The value of variable capacitor is 1.89 \times 10^{-13} F

Explanation:

Given :

Inductance L = 250 \times 10^{-3} H

Frequency f = 731 \times 10^{3} Hz

According to the cutoff frequency,

   f = \frac{1}{2\pi \sqrt{LC} }

Now we find the value of capacitance,

  C = \frac{1}{4\pi ^{2} f^{2}  L }

  C = \frac{1}{4\times 9.85 \times (731 \times 10^{3} )^{2} \times 250 \times 10^{-3}  }

  C = 1.89 \times 10^{-13} F

Therefore, the value of variable capacitor is 1.89 \times 10^{-13} F

6 0
3 years ago
A mass of 15 kg of air in a piston-cylinder device is heated from 25 o C to 77 o C by passing current through a resistance heate
Sophie [7]

Answer:

The electrical energy supplied is 0.233 kWh

Explanation:

Given :

Mass of air m = 15 kg

Initial temperature T _{1} = 298 K

Final temperature T_{2} = 350 K

Heat loss Q _{out} = 60 kJ

From the first law of thermodynamics,

     \Delta U = W_{in} - W_{out} - Q_{out}

We know that internal energy is proportional to difference of temperature,

    W_{in} = m (T_{2} - T_{1}  ) + 60

    W_{in} = 15 \times  (350-298) + 60 }

Now we need answer into kWh so ( 1 kWh = 3600 kJ ).

So we need to multiply ( \frac{1}{3600} )

     W_{in} = [15 \times  (350-298) + 60 ] \times \frac{1}{3600} kWh

     W_{in} = 0.233 kWh

Therefore, the electrical energy supplied 0.233 kWh

3 0
3 years ago
n order better to map the surface features of the Moon, a 361 kg361 kg imaging satellite is put into circular orbit around the M
ollegr [7]

Answer:

Explanation:

Mass of satellite

M_s = 361 kg

Distance of satellite from moon

h = 147 km = 147,000m

Radius of the moon is

R_m = 1740 km = 1740,000m

Mass of the moon is

M_m = 7.36 × 10²² kg.

The kinetic energy is equal to the potential energy of the body to the surface of the moon from the conservation of energy.

K.E = P.E = mgh

Gravity on moon is g = 1.62 m/s²

K.E = 361 × 1.62 × 147,000

K.E = 8.597 × 10^7 J.

B. The gravitational potential energy can be calculated using

U = G•M_s × M_m (1/R_s - 1 / R)

R is the total distance from the centre of the moon to the satellite

R = h + R_m = 147 + 1740 = 1887km

R = 1,887,000 m

U = 6.67 × 10^-11 × 361 × 7.36 × 10²² (1/1,740,000 - 1/1,887,000)

U = 6.67 × 10^-11 × 361 × 7.36 × 10²² × 4.48 × 10^-8

U = 7.93 × 10^7 J

Then,

The total energy becomes

E = K.E + U

E= 8.597 × 10^7 + 7.93 × 10^7 J

E = 1.653 × 10^8 J

3 0
4 years ago
How does the magnitude of the horizontal velocity change as the human cannonball passes through points a
Ainat [17]

Answer:

The question has the following answer options:

A. It decreases.

B. It increases.

C. It decreases then increases.

D. It does not change.

The correct answer is: D. It does not change.

Explanation:

The trajectory of a projectile is the path that a body follows when it has been thrown. It can be depicted in a coordinate system. Typically, the horizontal x axis stands for the distance that the object travels (in direction x) and the vertical and axis stands for the height (in direction y) of the throw. This very common form of movement is surprisingly simple to analyze if the following two assumptions are made:

1. The acceleration of free fall, g, is constant throughout the range of motion and is directed downward.

2. The effect of air resistance can be ignored.

With these assumptions, it is found that the curve that describes a projectile, and that it is known as its trajectory, is always a parable.

The horizontal velocity remains constant because in that direction the acceleration is zero.

7 0
4 years ago
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