1. Which of the following provides a correct answer for a problem that can be solved using the kinematic equations?

so I know you can solve this either by using Vox or Voy. I'm getting 3.08s when using Vox and 3.14s for Voy way. For Voy I'm usi

Tatiana [17]

The person's horizontal position is given by

$x=v_0\cos40^\circ t$

and the time it takes for him to travel 56.6 m is

$56.6\,\mathrm m=\left(24.0\,\dfrac{\mathrm m}{\mathrm s}\right)\cos40^\circ t\implies t=3.08\,\mathrm s$

so your first computed time is the correct one.

The question requires a bit of careful reading, and I think there may be a mistake in the problem. The person's vertical velocity $v_y$ at time $t$ is

$v_y=v_{0y}-gt$

which tells us that he would reach the ground at about $t=3.15\,\mathrm s$ . In this time, he would have traveled

$x=v_{0x}(3.15\,\mathrm s)=57.9\,\mathrm m$

But we're told that he is caught by a net at 56.6 m, which would mean that the net cannot have been placed at the same height from which he was launched. However, it's possible that the moment at which he was launched doesn't refer to the moment the cannon went off, but rather the moment at which the person left the muzzle of the cannon a fraction of a second after the cannon was set off. After this time, the person's initial vertical velocity $v_{0y}$ would have been a bit smaller than $\left(24.0\,\frac{\mathrm m}{\mathrm s}\right)\sin40^\circ$ .

7 0

3 years ago

Suppose you are standing at the exact center of a park surrounded by a circular road. An ambulance drives completely around this

vladimir2022 [97]

Answer: The pitch of the sound does not change.

Explanation: The pitch of sound wave is dependent on the frequency of the sound wave. The frequency of sound wave when their is a relative motion between an observer and a source is given by Doppler effect.

Doppler effect is a mathematical equation that gives the relationship between the observed frequency by an observer from any sound source and the relative motion between the observer and the source.

Mathematically,

f'= (v+v') /(v-vs) * f

Where f' = observed frequency of sound wave

v= speed of sound in air

v'= velocity of observer relative to sound source

vs= velocity of sound source relating to observer

f= frequency of sound produced by source

The observer is at the center, thus the distance between the observer and the source is constant (according to mensuration, the radius is constant for any given circle and since the car is moving along a circular path and the observer is at the center, thus the distance between them is constant), thus making the relative velocity between the observer and the source constant (vs=constant).

Also the frequency of sound wave produce by the source is a constant (f=constant)

The speed of sound in air is also a constant (v=336m/s)

The observer is standing at the center thus he is not moving, hence the relative motion between observer and source is also constant (v'=constant)

Since all parameters are constant, then the observed frequency will be constant too.

4 0

3 years ago

One of the most important properties of materials in many applications is strength. Two of the qualitative measures of the stren

katrin2010 [14]

To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.

In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.

By definition we know that the tensile strength is defined as

$\sigma = \frac{F}{A}$

Where,

$\sigma =$ Tensile strength

F = Tensile Force

A = Cross-sectional Area

In the other hand we have that the shear strength is defined as

$\sigma_y = \frac{F_y}{A}$

where,

$\sigma_y =$ Shear strength

$F_y =$ Shear Force

$A_0 =$ Parallel Area

PART A) Replacing with our values in the equation of tensile strenght, then

$311*10^6 = \frac{F}{(15*10^{-6})(30*10^{-2})}$

Resolving for F,

$F= 1399.5N$

PART B) We need here to apply the shear strength equation, then

$\sigma_y = \frac{F_y}{A}$

$210*10^6 = \frac{F_y}{15*10^{-6}30*10^{-2}}$

$F_y = 945N$

In such a way that the material is more resistant to tensile strength than shear force.

6 0

3 years ago

In the eastern US, hurricanesusually form over the warm waters of the Caribbean Sea and travel north. When they are over the col

garik1379 [7]

Answer:

Less powerful

Explanation:

Hurricanes rely on warm water. It sucks heat energy from the water to use for fuel. Warmer water means more moisture, which also mean a bigger and/or stronger hurricane. The North Atlantic is definitely much colder than the Caribbean so the hurricane will not have much fuel.

Have a great day!

8 0

3 years ago

A railroad car of mass 2,000 kg traveling at a velocity v = 10 m/s is stopped at the end of the tracks by a spring-damper system

Murrr4er [49]

Stopped at the end of the tracks by a spg-damper system, as shown in fig. 1

6 0

3 years ago