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3 years ago

Which is one way scientists indicate how precise and accurate there experimental measurements are

1 answer:

8 0

They do the method 3 times to be sure. Because if you do it once, that could mean anything. If you do it twice, it may or may not have the same result. If you do it 3 times and it matches one of the previous answers, then it's likely that it's correct.

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A mass is placed at the end of a spring. It has starting velocity of V & allowed to oscillate freely. If the mass has a star

LiRa [457]

Answer:

Equation for SHM can be written

V = w A cos w t where w is the angular frequency and the velocity is a maximum at t = 0

V1 = w1 A cos w1 t

V2 = w2 A cos w2 t

V2 / V1 = w2 / w1 since cos X t = 1 if t = zero

V2 / V1 = 2 pi f2 / (2 pi f1) = f2 / f1 = T1 / T2

If the velocity is twice as large the period will be 1/2 long

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2 years ago

Why do sound waves move faster through the ground than through the air? a. Particles of matter are packed more loosely in the gr

lbvjy [14]

B. is it i just got done this in class like two weeks ago hope it helps

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3 years ago

Read 2 more answers

How much force is required to accelerate a 22Kg mass at 6 m/s?

GuDViN [60]

Answer:

F = 132N

Explanation:

7 0

3 years ago

Read 2 more answers

A 54.0 kg ice skater is gliding along the ice, heading due north at 4.10 m/s . The ice has a small coefficient of static frictio

lesya [120]

Answer:

a. 2.668 m/s

b. 0.00494

Explanation:

The computation is shown below:

a. As we know that

$W = F\times d$

$KE = 0.5\times m\times v^2$

As the wind does not move the skater to the east little work is performed in this direction. All the work goes in the direction of the N-S. And located in that direction the component of the Force.

F = 3.70 cos 45 = 2.62 N

$W = F \times d = 2.62 N \times 100 m$

$W = 261.6 N\times m$

We know that

KE1 = Initial kinetic energy

KE2 = kinetic energy following 100 m

The energy following 100 meters equivalent to the initial kinetic energy less the energy lost to the work performed by the wind on the skater.

So, the equation is

KE2 = KE1 - W

$0.5 m\times v2^2 = 0.5 m\ v1^2 - W$

Now solve for v2

$v2 = \sqrt{v1^2 - {\frac{2W}{M}}}$

$= \sqrt{4.1 m/s)^2 - \frac{2 \times 261.6 N\times m}{54.0 kg}}$

= 2.668 m/s

b. Now the minimum value of Ug is

As we know that

Ff = force of friction

Us = coefficient of static friction

N = Normal force = weight of skater

So,

$Ff = Us\times N$

Now solve for Us

$= \frac{Ff}{N}$

$= \frac{3.70 N \times cos 45 }{54.0 kg \times 9.81 m/s^2}$

= 0.00494

4 0

3 years ago

A 56 kg sprinter, starting from rest, runs 49 m in 7.0 s at constant acceleration.what is the sprinter's power output at 2.0 s,

alexgriva [62]

The sprinter is in uniform accelerated motion, and its initial velocity is zero, so the relationship betwen space (S) and time (t) is
$S= \frac{1}{2} a t^2$
where a is the acceleration. Using the data of the problem, we can find a:
$a= \frac{2S}{t^2} = \frac{2 \cdot 49 m}{(7.0 s)^2} =2.0 m/s^2$
So now we can solve the 3 parts of the problem.

a) power output at t=2.0 s
The velocity at t=2.0 s is
$v(t)=at=(2.0 m/s^2)(2.0 s)=4.0 m/s$

the kinetic energy of the sprinter is
$K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(4.0 m/s)^2=448 J$

and so the power output is
$P= \frac{E}{t} = \frac{448 J}{2.0 s} =224 W$

b) power output at t=4.0s
The velocity at t=4.0 s is
$v(t)=at=(2.0 m/s^2)(4.0 s)=8.0 m/s$

the kinetic energy of the sprinter is
$K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(8.0 m/s)^2=1792 J$

and so the power output is
$P= \frac{E}{t} = \frac{1792 J}{4.0 s} =448 W$

c) Power output at t=6.0 s
The velocity at t=2.0 s is
$v(t)=at=(2.0 m/s^2)(6.0 s)=12.0 m/s$

the kinetic energy of the sprinter is
$K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(6.0 m/s)^2=4032 J$

and so the power output is
$P= \frac{E}{t} = \frac{4032 J}{6.0 s} =672 W$

8 0

3 years ago