Answer:
1. A non example of a constant in physics is a value that changes. For example, the temperature of Hong Kong, let's say, is not a constant, since it changes and does not stay at the same value. It gets higher in the summer and lower in the winter.
2. An example of a constant in physics would be the gravitational acceleration of the Earth. Since the Earth's mass' changes is negligible when compared with the overall mass of the Earth (unless something big happens, of course, like a meteor or asteroid), the gravitational acceleration on Earth is largely unchanged, even constant from our viewpoint.
Explanation:
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Answer:
C
Explanation:
Spoon decrease, water increase
By using the coefficient of linear expansion, the increase in the length of the metal plate is by 0.015m and in the area is by 0.3074.
The rate of change in length of a metal per degree change in temperature is known as the coefficient of linear expansion.
Given:
Coefficient of linear expansion, α = 29 x /k
Length, L1 = 10m
T1 = 25℃
T2 = 78℃
ΔT = 78 – 25 = 53℃
To find:
Change in length (ΔL) and area (ΔA) of metal plate = ?
Formula:
ΔL = α L1 ΔT
ΔA = A1 2 α ΔT
Calculations:
ΔL = 29 x x 10 x 53
ΔL = 0.01537m
ΔA = 100 x 2 x 29 x x 53
ΔA = 0.3074
A2 = 100.3074
Result:
The increase in the length and area is by 0.015m and 0.3074 respectively.
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The magnitude of the acceleration is 0.1038 m/s²
Explanation:
Net force is given by the square root of the sum of squares of the forces in y and x coordinates
Absolute Net Force will be;
However, Fnet can be calculated as the product of force and acceleration
Fnet=m*a
Fnet= 80.45 N
m= 775 kg
a=?
80.45=775*a
80.45/775= a
0.1038 m/s²=a
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Keywords : mass, force, magnitude, acceleration
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Answer:
a) 269.23 N
b) Fr = 70.77 N
c) Work output = 5250 J
d) Work Input = 6630 J
e) Mechanical Advantage
- ideal mechanical advantage = 3.9
- Actual Mechanical Advantage = 3.09
Explanation:
a) In an ideal machine, all the work input is converted to work output.
Work output = F₀ × d₀ = 1050 × 5 = 5250 J
Work input = Fᵢ × dᵢ = 19.5 Fᵢ
19.5 Fᵢ = 5250
Fᵢ = 269.23 N
b) Real effort = 340 N
Since 269.23 N is the force required to lift the 1050 N load in this pulley system, the rest of the 340 N force services frictional forces.
Frictional force = 340 - 269.23 = 70.77 N
c) Output work = Work done by the load = F₀ × d₀ = 1050 × 5 = 5250 J
d) Input Work = Work done by the Effort = Fᵢ × dᵢ = 340 × 19.5 = 6630 J
e) Ideal Mechanical advantage = distance travelled by effort/distance travelled by load
ideal mechanical advantage = 19.5/5 = 3.9
Actual Mechanical Advantage = Load/Effort = 1050/340 = 3.09
Efficiency = Work output/Work input = 5250/6630 = 0.792