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weeeeeb [17]
3 years ago
15

A velocity vs. time graph for an object is shown.

Physics
1 answer:
Afina-wow [57]3 years ago
7 0

Answer:

W  & X

Explanation:

gradient of a velocity - time graph indicates the acceleration.

straight lines ,

gradient = \frac{difference...in..Y..direction}{difference..in..X..direction}

  • positive gradient ----> acceleration
  • negative gradient-----> deceleration
  • gradient = 0 ------> constant velocity

curved lines indicate changing acceleration/deceleration

If we consider the gradient values here ,

W ----- 15/1        = 15

X-------(20-15)/1 = 5

Y ------(0/2)       = 0

Z-------(0-20)/1 = (-20)

looking at these we can say

Y motion has no acceleration : moves in a constant velocity

Z motion has a negative acceleration (deceleration)

W & X have accelerating motion.

W_(acceleration)> X_(acceleration)

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Two car horns are sounded creating two sound waves with frequencies that differ by a factor of three. How does the speed of the
Andrews [41]

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.

7 0
3 years ago
A 1100 kg car rounds a curve of radius 68 m banked at an angle of 16 degrees. If the car is traveling at 95 km/h, will a frictio
Mariulka [41]

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 F = \frac{mv^2}{R} = \frac{1100\times (26.3)^2}{68} = 1.1\times 10^4~N

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.

F = mg\sin(16) = 1100(9.8)\sin(16) = 2.97\times10^3

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 8.21\times 10^3~N

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.

5 0
4 years ago
Describe the agronomic significance of the upper and lower plastic limits, and of the plasticity index?​
GarryVolchara [31]

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.

4 0
3 years ago
What is true weightlessness ?​
alukav5142 [94]

Answer:

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3 0
3 years ago
The 2.50 kg cube in the figure has edge lengths d = 6.50 cm and is mounted on an axle
kozerog [31]

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.

7 0
3 years ago
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