Answer: yo light bill finna be turnt off im just kidding 1`
Explanation:
Answer:
The materials with which the lab group are to use for the model includes;
i) A candle
ii) A match
iii) A strip of cloth
iv) Tongs
v) A metal can
The processes the lab group are to model are;
a) Conduction
b) Convection
c) Radiation
The procedure the lab group can use to do this is outlined as follows;
1) Tie the piece of cloth around the metal can with a strip of the cloth extending past the bottom of the can
2) Hold the metal can in with the aid of the tongs
3) Light the candle with the match
4) Place the metal can over over the burning candle so that it does not touch the flame
5) While holding the can with the tongs, ensure that the strip of cloth hanging by the side of the can does not come in contact with the flame
Conduction
Conduction heat transfer is observed by the rising temperature of the tongs that is in the contact with the can
Convection
Convection heat transfer is observed by the rising temperature of the can that is placed in the path of the rising convection current from the candle wax
Radiation
Radiation heat transfer is observed by the shrinking of the piece of cloth placed beside the candle flame
Explanation:
Answer:
a=12 m/s²
Explanation:
Newton's second law of motion states that the acceleration of a body is directly proportional to the force applied and takes place in the direction of force.
This can be summarized as: F=ma, where m is the mass of the object on which force F acts. a is the acceleration due to the force applied.
12N= 1kg×a
a=12N/1kg
a=12m/s²
Explanation:
Elongation of the wire is:
ΔL = F L₀ / (E A)
where F is the force,
L₀ is the initial length,
E is Young's modulus,
and A is the cross sectional area.
ΔL = T (0.5 m) / ((2.0×10¹¹ Pa) (0.02 cm²) (1 m / 100 cm)²)
ΔL = T (1.25×10⁻⁶ m/N)
T = (80,000 N/m) ΔL
Draw a free body diagram of the mass at the bottom of the circle. There are two forces: tension force T pulling up and weight force mg pulling down.
Sum of forces in the centripetal direction:
∑F = ma
T − mg = mv²/r
T − mg = mω²r
T − (15 kg) (9.8 m/s²) = (15 kg) (2 rev/s × 2π rad/rev)² (0.5 m + ΔL)
T − 147 N = (2368.7 N/m) (0.5 m + ΔL)
Substitute:
(80,000 N/m) ΔL − 147 N = (2368.7 N/m) (0.5 m + ΔL)
(80,000 N/m) ΔL − 147 N = 1184.35 N + (2368.7 N/m) ΔL
(797631.3 N/m) ΔL = 1331.35 N
ΔL = 0.00167 m
ΔL = 1.67 mm
The answer to the question you have asked is 1.8 miles east.
