Answer:
Remains the same
Explanation:
The speed of waves of higher and lower frequency both will be same.
the speed of sound in a medium is constant and independent of it's frequency. Moreover, when the frequency changes wavelength changes accordingly, such that their product remains constant.
we know that
υ×λ = constant = velocity
υ= frequency
λ= wavelength.
Answer:
Yes. Towards the center. 8210 N.
Explanation:
Let's first investigate the free-body diagram of the car. The weight of the car has two components: x-direction: towards the center of the curve and y-direction: towards the ground. Note that the ground is not perpendicular to the surface of the Earth is inclined 16 degrees.
In order to find whether the car slides off the road, we should use Newton's Second Law in the direction of x: F = ma.
The net force is equal to 
Note that 95 km/h is equal to 26.3 m/s.
This is the centripetal force and equal to the x-component of the applied force.

As can be seen from above, the two forces are not equal to each other. This means that a friction force is needed towards the center of the curve.
The amount of the friction force should be 
Qualitatively, on a banked curve, a car is thrown off the road if it is moving fast. However, if the road has enough friction, then the car stays on the road and move safely. Since the car intends to slide off the road, then the static friction between the tires and the road must be towards the center in order to keep the car in the road.
Answer:It shows the size of the range of the moisture contents at which the soil remains plastic. In general, the plasticity index depends only on the amount of clay present. It indicates the fineness of the soil and its capacity to change shape without altering its volume.
Answer:
weightlessness, condition experienced while in free-fall, in which the effect of gravity is canceled by the inertial (e.g., centrifugal) force resulting from orbital flight. ... Excluding spaceflight, true weightlessness can be experienced only briefly, as in an airplane following a ballistic (i.e., parabolic) path.
Answer:
0.191 s
Explanation:
The distance from the center of the cube to the upper corner is r = d/√2.
When the cube is rotated an angle θ, the spring is stretched a distance of r sin θ. The new vertical distance from the center to the corner is r cos θ.
Sum of the torques:
∑τ = Iα
Fr cos θ = Iα
(k r sin θ) r cos θ = Iα
kr² sin θ cos θ = Iα
k (d²/2) sin θ cos θ = Iα
For a cube rotating about its center, I = ⅙ md².
k (d²/2) sin θ cos θ = ⅙ md² α
3k sin θ cos θ = mα
3/2 k sin(2θ) = mα
For small values of θ, sin θ ≈ θ.
3/2 k (2θ) = mα
α = (3k/m) θ
d²θ/dt² = (3k/m) θ
For this differential equation, the coefficient is the square of the angular frequency, ω².
ω² = 3k/m
ω = √(3k/m)
The period is:
T = 2π / ω
T = 2π √(m/(3k))
Given m = 2.50 kg and k = 900 N/m:
T = 2π √(2.50 kg / (3 × 900 N/m))
T = 0.191 s
The period is 0.191 seconds.