You run 100 meters in 15 seconds. What is your speed in m/s? and show your work

what is the force constant, in newtons per meter, needed to produce a period of 0.45 s for a 0.011-kg mass on the spring?

topjm [15]

The force constant is 2.145 N/m.

<h3>What is spring constant?</h3>

The spring constant is the force required to stretch or compress a spring divided by the distance traveled by the spring. It is used to determine whether a spring is stable or unstable.

K is the proportionality constant, also known as the 'spring constant.' Hooke's law (F = -kx) specifies stiffness and strength via the k variable. The greater the value of k, the greater the force required to stretch an object to a given length.

Using the relation;

T = 2π√m/k

T = time period = 0.45 s

m = mass of object in kilograms = 0.011kg

k = spring constant

To find k based on the formula,

k = 4 × (3.142)^2 × 0.011 / (0.45 )^2

k = 2.145 N/m

Therefore the force constant is 2.145 N/m.

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8 0

1 year ago

3.

Flura [38]

Answer:

about 4 km

Explanation:

15 minutes is a quarter of an hour, so you divide 16km by 4 to get your answer

8 0

2 years ago

A mortar is like a small cannon that launches shells at steep angles. A mortar crew is positioned near the top of a steep hill.

Elena-2011 [213]

1) Distance down the hill: 1752 ft (534 m)

2) Time of flight of the shell: 12.9 s

3) Final speed: 326.8 ft/s (99.6 m/s)

Explanation:

1)

The motion of the shell is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion of the shell is a uniformly accelerated motion, so the vertical position is given by the following equation:

$y=(u sin \theta)t-\frac{1}{2}gt^2$ (1)

where:

$u sin \theta$ is the initial vertical velocity of the shell, with $u=156 ft/s$ and $\theta=49.0^{\circ}$

$g=32 ft/s^2$ is the acceleration of gravity

At the same time, the horizontal motion of the shell is a uniform motion, so the horizontal position of the shell at time t is given by the equation

$x=(ucos \theta)t$

where $u cos \theta$ is the initial horizontal velocity of the shell.

We can re-write this last equation as

$t=\frac{x}{u cos \theta}$ (1b)

And substituting into (1),

$y=xtan\theta -\frac{1}{2}gt^2$ (2)

where we have choosen the top of the hill (starting position of the shell) as origin (0,0).

We also know that the hill goes down with a slope of $\alpha=-41.0^{\circ}$ from the horizontal, so we can write the position (x,y) of the hill as

$y=x tan \alpha$ (3)

Therefore, the shell hits the slope of the hill when they have same x and y coordinates, so when (2)=(3):

$xtan\alpha = xtan \theta - \frac{1}{2}gt^2$

Substituting (1b) into this equation,

$xtan \alpha = x tan \theta - \frac{1}{2}g(\frac{x}{ucos \theta})^2\\x (tan \theta - tan \alpha)-\frac{g}{2u^2 cos^2 \theta} x^2=0\\x(tan \theta - tan \alpha-\frac{gx}{2u^2 cos^2 \theta})=0$

Which has 2 solutions:

x = 0 (origin)

and

$tan \theta - tan \alpha=\frac{gx}{2u^2 cos^2 \theta}=0\\x=(tan \theta - tan \alpha) \frac{2u^2 cos^2\theta}{g}=1322 ft$

So, the distance d down the hill at which the shell strikes the hill is

$d=\frac{x}{cos \alpha}=\frac{1322}{cos(-41.0^{\circ})}=1752 ft=534 m$

2)

In order to find how long the mortar shell remain in the air, we can use the equation:

$t=\frac{x}{u cos \theta}$

where:

x = 1322 ft is the final position of the shell when it strikes the hill

$u=156 ft/s$ is the initial velocity of the shell

$\theta=49.0^{\circ}$ is the angle of projection of the shell

Substituting these values into the equation, we find the time of flight of the shell:

$t=\frac{1322}{(156)(cos 49^{\circ})}=12.9 s$

3)

In order to find the final speed of the shell, we have to compute its horizontal and vertical velocity first.

The horizontal component of the velocity is constant and it is

$v_x = u cos \theta =(156)(cos 49^{\circ})=102.3 ft/s$

Instead, the vertical component of the velocity is given by

$v_y=usin \theta -gt$

And substituting at t = 12.9 s (time at which the shell strikes the hill),

$v_y=(156)(cos 49^{\circ})-(32)(12.9)=-310.4ft/s$

Therefore, the final speed of the shell is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(102.3)^2+(-310.4)^2}=326.8 ft/s=99.6 m/s$

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5 0

3 years ago

In a weekly project update meeting, Liza asks the following questions of one of her employees: "Why were you late meeting your l

Aleonysh [2.5K]

Answer: This type of questions are called probing questions. The correct option is D.

Explanation:

Probing questions are the type of questions that are asked to investigate an ongoing event. It helps the investigator to know more about what is happening and how to obtain conclusive decisions through the personal opinions of the respondent . For example from the question, Liza wanted to know more about the project updates which was held in the weekly meetings. She asked her employees questions like:

- Why were you late meeting your last deadline?

-Were there external factors that delayed your work?

-Did other coworkers get their part of the assignment to you on time?

- Do you need more help from me?".

7 0

3 years ago

An object experiences an acceleration of -6.8 m/s2. As a result, it accelerates from 54 m/s to a complete stop. How much dista

inessss [21]

Answer:

The distance traveled during its acceleration, d = 214.38 m

Explanation:

Given,

The object's acceleration, a = -6.8 m/s²

The initial speed of the object, u = 54 m/s

The final speed of the object, v = 0

The acceleration of the object is given by the formula,

a = (v - u) / t m/s²

∴ t = (v - u) / a

= (0 - 54) / (-6.8)

= 7.94 s

The average velocity of the object,

V = (54 + 0)/2

= 27 m/s

The displacement of the object,

d = V x t meter

= 27 x 7.94

= 214.38 m

Hence, the distance the object traveled during that acceleration is, a = 214.38 m

3 0

3 years ago