Answer:
Sound travels through solids and liquids at the same speed
Explanation:
Because sound needs a dencer object to travel fast and since both liquid and solids are closer than gas sound travles faster in them.
Answer:
1.73 m/s²
3.0 cm
Explanation:
Draw a free body diagram of the yo-yo. There are two forces: weight force mg pulling down, and tension force T pulling up 10° from the vertical.
Sum of forces in the y direction:
∑F = ma
T cos 10° − mg = 0
T cos 10° = mg
T = mg / cos 10°
Sum of forces in the x direction:
∑F = ma
T sin 10° = ma
mg tan 10° = ma
g tan 10° = a
a = 1.73 m/s²
Draw a free body diagram of the sphere. There are two forces: weight force mg pulling down, and air resistance D pushing up. At terminal velocity, the acceleration is 0.
Sum of forces in the y direction:
∑F = ma
D − mg = 0
D = mg
½ ρₐ v² C A = ρᵢ V g
½ ρₐ v² C (πr²) = ρᵢ (4/3 πr³) g
3 ρₐ v² C = 8 ρᵢ r g
r = 3 ρₐ v² C / (8 ρᵢ g)
r = 3 (1.3 kg/m³) (100 m/s)² (0.47) / (8 (7874 kg/m³) (9.8 m/s²))
r = 0.030 m
r = 3.0 cm
Answer: Accoding to research "Triton is unique among all the large moons in the solar system because it orbits Neptune in a direction opposite to the planet's rotation (a "retrograde" orbit). It is unlikely to have formed in this configuration and was probably captured from elsewhere."
Explanation:
Answer:
(a) = -0.16%
(b) = smaller
Explanation:
given
power = 460 W
potential difference = 120 V
(a) what percentage will its heat output drop if the applied potential difference drops to 110 V ?
we know 
.....................(i)
we need to find change in power
..............(ii)
from equations we get
(b)
if we increase temperature resistance will increase and decrease with decrease in temperature and we know power is inversely proportional to resistance so if potential decrease and it would cause drop in power
and due to this increment of heating power resistance will decrease so actual drop in the power would be smaller
Answer: The height of the cloud = 394.55 m
Explanation:
The observer is 500m away from the spotlight.
Let x be the distance from the observer to the interception of the segment of the height, h with the floor. The equations are thus:
Tan 45° = h/x ... eq1
Tan 75° = h/(500- x ) ... eq2
From eq 1, Tan 45° = 1, therefore eq1 becomes:
h = x ... eq3
Put eq3 into eq2
Tan 75° = h/(500- h)
h = ( 500 - h ) Tan 75°
h = 500Tan 75° - hTan75°
h + h Tan 75° = 500 Tan 75°
h ( 1 + Tan 75° ) = 500 Tan75°
h = 500Tan75°/ (1 + Tan 75°)
h= 1866.02 / 4.73
h = 394.55m
