Answer:
Explanation:
Given that,
Height of the bridge is 20m
Initial before he throws the rock
The height is hi = 20 m
Then, final height hitting the water
hf = 0 m
Initial speed the rock is throw
Vi = 15m/s
The final speed at which the rock hits the water
Vf = 24.8 m/s
Using conservation of energy given by the question hint
Ki + Ui = Kf + Uf
Where
Ki is initial kinetic energy
Ui is initial potential energy
Kf is final kinetic energy
Uf is final potential energy
Then,
Ki + Ui = Kf + Uf
Where
Ei = Ki + Ui
Where Ei is initial energy
Ei = ½mVi² + m•g•hi
Ei = ½m × 15² + m × 9.8 × 20
Ei = 112.5m + 196m
Ei = 308.5m J
Now,
Ef = Kf + Uf
Ef = ½mVf² + m•g•hf
Ef = ½m × 24.8² + m × 9.8 × 0
Ef = 307.52m + 0
Ef = 307.52m J
Since Ef ≈ Ei, then the rock thrown from the tip of a bridge is independent of the direction of throw
Answer:
The correct prediction will be:
The temperature of the surface of the ball bearing when rubbed over glass will be the least.
The incorrect prediction:
The tennis ball over the linoleum floor will have no friction, as the temperatures will not change.
Answer:
The net torque is 0.0372 N m.
Explanation:
A rotational body with constant angular acceleration satisfies the kinematic equation:
(1)
with ω the final angular velocity, ωo the initial angular velocity, α the constant angular acceleration and Δθ the angular displacement (the revolutions the sphere does). To find the angular acceleration we solve (1) for α:
Because the sphere stops the final angular velocity is zero, it's important all quantities in the SI so 2.40 rev/s = 15.1 rad/s and 18.2 rev = 114.3 rad, then:
The negative sign indicates the sphere is slowing down as we expected.
Now with the angular acceleration we can use Newton's second law:
(2)
with ∑τ the net torque and I the moment of inertia of the sphere, for a sphere that rotates about an axle through its center its moment of inertia is:
With M the mass of the sphere an R its radius, then:
Then (2) is:
Answer:
The most correct option is;
B. 10 km
Explanation:
Where:
y = Distance between the two headlights
d = Aperture of observers eye
λ = Wavelength of light
L = Distance between the observer and the headlight
Therefore, from the above solution, the distance between the observer and the headlights is 9386.066 km which is approximately 10 km.
Also we have
sinθ = y/L = 1.22 (λ/d)
sinθ = 1.22×10⁻⁴ rad
Answer:
The ball will have an upward velocity of 6 m/s at a height of 5.51 m.
Explanation:
Hi there!
The equations of height and velocity of the ball are the following:
y = y0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
y = height at time t.
y0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).
v = velocity of the ball at time t.
Placing the origin at the throwing point, y0 = 0.
Let´s use the equation of velocity to obtain the time at which the velocity is 12.0 m/s / 2 = 6.00 m/s.
v = v0 + g · t
6.00 m/s = 12.0 m/s -9.81 m/s² · t
(6.00 - 12.0)m/s / -9.81 m/s² = t
t = 0.612 s
Now, let´s calculate the height of the baseball at that time:
y = y0 + v0 · t + 1/2 · g · t² (y0 = 0)
y = 12.0 m/s · 0.612 s - 1/2 · 9.81 m/s² · (0.612 s)²
y = 5.51 m
The ball will have an upward velocity of 6 m/s at a height of 5.51 m.
Have a nice day!
