Answer:
5.52 m
Explanation:
Using the equation;
Vy = Vo sin θ
Where; Vo = 2.25 m/s and θ = 35°
We get;
= 2.25 sin 35°
= 1.29 m/s
Thus; Vy = 1.29 m/s
But a = -9.81 m/s (against gravity)
t = 1.20 s and
y = Vyt + 1/2 at²
where y is the height above the water
= 1.29 × 1.2 + (1/2 × -9.81 × 1.2²)
= 1.548 + (-7.0632)
= - 5.5152 m
Thus; the height of the girl above the water is 5.52 m
Answer:
Say you are holding a thread to the end of which is tied a stone. Now when you start whirling it around you will notice that two forces have to be applied simultaneously. One which pulls the thread inwards and the other which throws it sideways or tangentially.
Both these forces will generate their respective accelerations.
The one pointed inwards will generate centripetal or radial acceleration.
The one pointing sideways will generate tangential acceleratio
Explanation:
A major difference between tangential acceleration and centripetal acceleration is their direction
Centripetal means “center seeking”. Centripetal acceleration is always directed inward.
Tangential acceleration is always directed tangent to the circle.
Tangential acceleration results from the change in magnitude of the tangential velocity of an
object. An object can move in a circle and not have any tangential acceleration. No tangential
acceleration simply means the angular acceleration of the object is zero and the object is moving
with a constant angular velocity
<em>Answer:</em>
<em>Melting</em>
<em>Explanation:</em>
<em>I took the quiz and got it right so I am 100 percent sure the answer is, A.melting :)</em>
Answer:
C
Explanation:
To solve this question, we will need to develop an expression that relates the diameter 'd', at temperature T equals the original diameter d₀ (at 0 degrees) plus the change in diameter from the temperature increase ( ΔT = T):
d = d₀ + d₀αT
for the sphere, we were given
D₀ = 4.000 cm
α = 1.1 x 10⁻⁵/degrees celsius
we have D = 4 + (4x(1.1 x 10⁻⁵)T = 4 + (4.4x10⁻⁵)T EQN 1
Similarly for the Aluminium ring we have
we were given
d₀ = 3.994 cm
α = 2.4 x 10⁻⁵/degrees celsius
we have d = 3.994 + (3.994x(2.4 x 10⁻⁵)T = 3.994 + (9.58x10⁻⁵)T EQN 2
Since @ the temperature T at which the sphere fall through the ring, d=D
Eqn 1 = Eqn 2
4 + (4.4x10⁻⁵)T =3.994 + (9.58x10⁻⁵)T, collect like terms
0.006=5.18x10⁻⁵T
T=115.7K
