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daser333 [38]
2 years ago
11

1. Write down the readings on the side of Figs. 6 (a), (b) and (c) respectively. What is the least count of instrument scale for

each of the three measurement tools? Fig. 6 (a) Meterstick (cm)
Fig. 6 (b) Vernier Caliper (cm)
Fig. 6 (c) Micrometer (mm)

2. What is the difference between the measured values 1.05 m and 1.050 m? What factor of a
measurement tool determines the significant figures of a measured value?

Physics
1 answer:
alexandr402 [8]2 years ago
3 0

Explanation:

This should be right. if any doubt post a comment.

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What is the magnitude of the force that is exerted on a 20 kg mass to give it an acceleration of 10.0
ANEK [815]

Answer:Mass of the body = 20 kg.

Final Velocity = 5.8 m/s.

Initial velocity = 0

Time = 3 seconds.  

Using the Formula,  

Acceleration = (v - u)/ t

= (5.8 - 0)/ 3

= 1.6 m /s².

Now, Using the Formula,  

Force = mass × acceleration

= 20 × 1.6

=

Explanation: I REALLY  HOPE THIS HELPS I'M KINDA NEW AT THIS :] :]

8 0
2 years ago
A sealed container holding 0.0255 L of an ideal gas at 0.981 atm and 65 ∘ C is placed into a refrigerator and cooled to 41 ∘ C w
user100 [1]

Answer:

0.911 atm

Explanation:

In this problem, there is no change in volume of the gas, since the container is sealed.

Therefore, we can apply Gay-Lussac's law, which states that:

"For a fixed mass of an ideal gas kept at constant volume, the pressure of the gas is proportional to its absolute temperature"

Mathematically:

p\propto T

where

p is the gas pressure

T is the absolute temperature

For a gas undergoing a transformation, the law can be rewritten as:

\frac{p_1}{T_1}=\frac{p_2}{T_2}

where in this problem:

p_1=0.981 atm is the initial pressure of the gas

T_1=65^{\circ}+273=338 K is the initial absolute temperature of the gas

T_2=41^{\circ}+273=314 K is the final temperature of the gas

Solving for p2, we find the final pressure of the gas:

p_2=\frac{p_1 T_2}{T_1}=\frac{(0.981)(314)}{338}=0.911 atm

3 0
3 years ago
Does coefficient of thermal expansion vary with temperature?
N76 [4]

Answer:

yes

Explanation:

8 0
3 years ago
The force between two charges, q, and 92, is F. If the distance between the
Sholpan [36]

Answer:

4F

Explanation:

F = kQ₁Q₂/d²

F' = k2Q₁2Q₂/d²

F' = 4(kQ₁Q₂/d²)

F' = 4F

4 0
2 years ago
A 9.0-kg bowling ball on a horizontal, frictionless surface experiences a net force of 6.0 n. what will be its acceleration?
Vladimir [108]

This question involves the concepts of Newton's Second Law of Motion.

The acceleration of the bowling ball will be "0.67 m/s²".

<h3>Newton's Second Law of Motion</h3>

According to Newton's Second Law of Motion, when an unbalanced force is applied on an object, it produces an acceleration in it, in the direction of the applied force. This acceleration is directly proportional to the force applied and inversely proportional to the mass of the object. Mathematically,

F=ma\\\\a=\frac{F}{m}

where,

  • a = acceleration = ?
  • F = Magnitude of the applied force = 6 N
  • m = Mass of the ball = 9 kg

Therefore,

a=\frac{6\ N}{9\ kg}

a = 0.67 m/s²

Learn more about Newton's Second Law of Motion here:

brainly.com/question/13447525

#SPJ1

7 0
2 years ago
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