What is the difference b/w cm,km,m,g.<br><br><br><br>

Find the object's speeds v1, v2, and v3 at times t1=2.0s, t2=4.0s, and t3=13s.

Burka [1]

Since this is a distance/time graph, the speed at any time is the slope
of the part of the graph that's directly over that time on the x-axis.

At time t1 = 2.0 s
That's in the middle of the first segment of the graph,
that extends from zero to 3 seconds.
Its slope is 7/3 . v1 = 7/3 m/s .

At time t2 = 4.0 s
That's in the middle of the horizontal part of the graph
that runs from 3 to 6 seconds.
Its slope is zero.
v2 = zero .

At time t3 = 13 s.
That's in the middle of the part of the graph that's sloping down,
between 11 and 16 seconds.
Its slope is -3/5 . v3 = -0.6 m/s .

7 0

3 years ago

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A child of mass m is standing at the edge of a carousel. Both the carousel and the child are initially stationary. The carousel

suter [353]

Answer:

the angular velocity of the carousel after the child has started running =

$\frac{2F}{mR} \delta t$

Explanation:

Given that

the mass of the child = m

The radius of the disc = R

moment of inertia I = $\frac{1}{2} mR^2$

change in time = $\delta \ t$

By using the torque around the inertia ; we have:

T = I×∝

where

R×F = I × ∝

R×F = $\frac{1}{2} mR^2$ ∝

F = $\frac{1}{2} mR$ ∝

∝ = $\frac{2F}{mR}$ ( expression for angular angular acceleration)

The first equation of motion of rotating wheel can be expressed as :

$\omega = \omega_0 + \alpha \delta t$

where ;

∝ = $\frac{2F}{mR}$

Then;

$\omega = 0+ \frac{2F}{mR} \delta t$

$\omega = \frac{2F}{mR} \delta t$

∴ the angular velocity of the carousel after the child has started running =

$\frac{2F}{mR} \delta t$

7 0

3 years ago

A mass weighting 48 lbs stretches a spring 6 inches. The mass is in a medium that exerts a viscous resistance of 27 lbs when the

Mademuasel [1]

Answer:

a)

$u(t)=0.499ft.e^{-\frac{144.76lb/s}{2(48lb)}t}cos(\omega t)\\\\u(t)=0.499ft.e^{-1.5t}cos(\omega t)$

b)

m = 48lb

c)

b = 144.76lb

Explanation:

The general equation of a damping oscillate motion is given by:

$u(t)=u_oe^{-\frac{b}{2m}t}cos(\omega t-\alpha)$ (1)

uo: initial position

m: mass of the block

b: damping coefficient

w: angular frequency

α: initial phase

a. With the information given in the statement you replace the values of the parameters in (1). But first, you calculate the constant b by using the information about the viscous resistance force:

$|F_{vis}|=bv\\\\b=\frac{|F_{vis}|}{v}\\\\|F_{vis}|=27lbs=27*32.17ft.lb/s^2=868.59ft.lb/s^2\\\\b=\frac{868.59}{6}lb/s=144.76lb/s$

Then, you obtain by replacing in (1):

6in = 0.499 ft

$u(t)=0.499ft.e^{-\frac{144.76lb/s}{2(48lb)}t}cos(\omega t)\\\\u(t)=0.499ft.e^{-1.5t}cos(\omega t)$

b.

mass, m = 48lb

c.

b = 144.76 lb/s

8 0

3 years ago

A group of Advanced Open Water Divers plans to make two dives. The first dive is on a reef in 90 feet of water for 20 minutes. T

Kaylis [27]

Answer:

The ending pressure is R.

Explanation:

This is Recreational scuba diving (R). According to standards, this diving has prescribed depth limit of 130 ft. This means that the limit must include a depth that is not greater than 130 ft while using only compressed air and not even requiring a decompression stop.

Now, from the question, we can see that that there was a prescribed limit of 60ft of water with a planned bottom time of 30 minutes.

Thus, the ending pressure is R.

5 0

3 years ago

A _________ is the time it takes for the Earth to rotate 360 degrees relative to the background stars.

Tom [10]

The answer is stellar day

8 0

2 years ago

Read 2 more answers

What is the difference b/w cm,km,m,g.​

What is the difference b/w cm,km,m,g.