Answer:
(a) 2 (b) 4 (c) 4
Explanation:
Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.
Rules for significant figures:
- Digits from 1 to 9 are always significant and have infinite number of significant figures.
- All non-zero numbers are always significant. For example: 654, 6.54 and 65.4 all have three significant figures.
- All zero’s between integers are always significant. For example: 5005, 5.005 and 50.05 all have four significant figures.
- All zero’s preceding the first integers are never significant. For example: 0.0078 has two significant figures.
- All zero’s after the decimal point are always significant. For example: 4.500, 45.00 and 450.0 all have four significant figures.
- All zeroes used solely for spacing the decimal point are not significant. For example : 8000 has one significant figure.
As per question,
0.000054 has 2 significant figures.
3.001 x 10⁵ has 4 significant figures.
5.600 has 4 significant figures.
Answer:
point of support on which a lever rotates.
Explanation:
The fulcrum is the point of support on which a lever rotates. Fulcrum is a pivotal part of simple machines.
The fulcrum provides the platform for a lever to torque.
- The force that opposes motion by the applied force is termed the frictional force.
- Friction is a force that opposes motion.
- The stored energy of an object is its potential energy.
- The potential energy is the energy due to the position of a body.
- The distance an object moves when doing work is termed its displacement.
Answer:
the two vehicles will be moving at a speed of 6.16 m/s
Explanation:
This is a case of completely inelastic collision, therefore, the conservation of momentum can be written as:
which given the information provided results into:
R=U/I so
U=RxI
U= 10 x 42
U= 420 volts
Answer:
Explanation:
Time taken to accelerate to 28 m /s
= 28 / 2 = 14 s
a ) Total length of time in motion
= 14 + 41 + 5
= 60 s .
b )
Distance covered while accelerating
s = ut + 1/2 at²
= 0 + .5 x 2 x 14²
= 196 m .
Distance covered while moving in uniform motion
= 28 x 41
= 1148 m
distance covered while decelerating
v = u - at
0 = 28 - a x 5
a = 5.6 m / s²
v² = u² - 2 a s
0 = 28² - 2 x 5.6 x s
s = 28² / 2 x 5.6
= 70 m .
Total distance covered
= 196 + 1148 + 70
= 1414 m
total time taken = 60 s
average velocity
= 1414 / 60
= 23.56 m /s .
