Answer:
Height covered after crossing window top = 0.75 m
Explanation:
In the question,
The height of the window pane = 1.8 m
Time for which the flowerpot is in view = 0.66 s
So,
The time for which it was in view while going up is = 0.33 s
Time for which it was in view while going down = 0.33 s
So,
Let us say,
The initial velocity of the flowerpot = u m/s
So,
Using the equation of the motion,
So,
Velocity at the top of the window pane is given by,
Now,
Let us say the height to which the flowerpot goes after crossing the window pane is = h
So,
Using the equation of motion,
Therefore, the height covered by the flowerpot after window is = 0.75 m
I think that your answer should be D) the state should make use of satellites and radars because they detect when a storm is coming
Answer:
The car must travel 51.34 m for its speed to reach 2.7 m/s
Explanation:
Mass of car = 2.9 x 10³ kg
Forward force = 1148 N
Resistive force = 941 N
Total force = 1148 - 941 = 207 N
We know
Force = Mass x Acceleration
207 = 2.9 x 10³ x Acceleration
Acceleration = 0.071 m/s²
Now we have equation of motion, v² = u² + 2as
Initial velocity, u = 0 m/s
Final velocity, v = 2.7 m/s
Acceleration, a = 0.071 m/s²
Substituting
v² = u² + 2as
2.7² = 0² + 2 x 0.071 x s
s = 51.34 m
The car must travel 51.34 m for its speed to reach 2.7 m/s
a). We could read off the period if we had the graph.
The frequency is 1/period.
b). The velocity is the first derivative of the displacement,
so it should lag the displacement by 90° .
The acceleration is the derivative of the velocity,
so that puts it 180° behind the displacement.
c). If we could see the graph, then, knowing the pendulum's
length and the period of its swing, we could calculate the
acceleration of gravity on the planet at which he is at.
The number of 'G-forces' is
9.8 m/s² / (acceleration of gravity where he is at) .
The number is a ratio ... without units.
Answer:
x = 41.2 m
Explanation:
The electric force is a vector magnitude, so it must be added as vectors, remember that the force for charges of the same sign is repulsive and for charges of different sign it is negative.
In this case the fixed charges (q₁ and q₂) are positive and separated by a distance (d = 100m), the charge (q₃ = -1.0 10⁻³ C)) is negative so the forces are attractive, such as loads q₃ must be placed between the other two forces subtract
F = F₁₃ - F₂₃
let's write the expression for each force, let's set a reference frame on the charge q1
F₁₃ =
F₂₃ =
they ask us that the net force be zero
F = 0
0 = F₁₃ - F₂₃
F₁₃ = F₂₃
k \frac{q_1 q_3}{x^2} =k \frac{q_2 q_3}{(d-x)^2}
q1 / x2 = q2 / (d-x) 2
(d-x)² = x²
we substitute
(100 - x)² = 2/1 x²
100- x = √2 x
100 = 2.41 x
x = 41.2 m