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3 years ago

8

Zeros that follow non-zero numbers and are also to the right of a decimal point are significant.

2 answers:

storchak [24]3 years ago

6 0

Yes that's correct.also zeros in between non-zero numbers are significant figures

Ivenika [448]3 years ago

5 0

Yes that's correct. Also zeros in between non-zero numbers are significant figures

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An airplane traveling at one third the speed of sound (i.e., 114 m/s) emits a sound of frequency 3.72 kHz. At what frequency doe

nlexa [21]

Answer:

f'=5.58kHz

Explanation:

This is an example of the Doppler effect, the formula is:

$f'=\frac{(v+v_o)}{(v+v_s)}f$

Where f is the actual frequency, $f'$ is the observed frequency, $v$ is the velocity of the sound waves, $v_o$ the velocity of the observer (which is negative if the observer is moving away from the source) and $v_s$ the velocity of the source (which is negative if is moving towards the observer). For this problem:

$f=3.72kHz\\v=342m/s\\v_o=0m/s\\v_s=-114m/s$

$f'=\frac{(342+0)}{(342-114)}3.72\times10^3\\f'=\frac{342}{228}3.72\times10^3\\f'=(1.5)3.72\times10^3\\f'=5580Hz=5.58kHz$

5 0

3 years ago

A girl standing on a bridge throws a stone vertically downward with an initial velocity of 15.0 m/s into the river below. If the

hodyreva [135]

Vi = 15 m/s
t = 2 s
a = 9.8 m/s^2
y = ?

The kinematic equation that has all of our variables is d = Vi*t + 0.5*a*t^2
y = 15*2 + 0.5*9.8*2^2 = 49.6 m

6 0

3 years ago

Based on what you know, please list at least 4 characteristics of weather. IDK IF RIGHT SUBJECT

Bas_tet [7]

Rain sleet hail and snow

5 0

3 years ago

A 5.00-kg object is attached to one end of a horizontal spring that has a negligible mass and a spring constant of 280 N/m. The

lina2011 [118]

Answer:

1) v = 0.45 m/s

2) v = 0.65 m/s

3) v = 0.75 m/s

Explanation:

1) We can find the speed of the object by conservation of energy:

$E_{i} = E_{f}$

$\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}$

Where:

k: is the spring constant = 280 N/m

v: is the speed of the object =?

m: is the mass of the object = 5.00 kg

x: is the displacement of the spring

$\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.08 m)^{2} + \frac{1}{2}5.00 kgv^{2}$

$v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.08 m)^{2}}{5.00 kg}} = 0.45 m/s$

2) When the object is 5.00 cm (0.050 m) from equilibrium, the speed of the object is:

$\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}$

$\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.05 m)^{2} + \frac{1}{2}5.00 kgv^{2}$

$v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.05 m)^{2}}{5.00 kg}} = 0.65 m/s$

3) When the object is at the equilibrium position, the speed of the object is:

$\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}$

$\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0 m)^{2} + \frac{1}{2}5.00 kgv^{2}$

$v = \sqrt{\frac{280N/m(0.10 m)^{2}}{5.00 kg}} = 0.75 m/s$

I hope it helps you!

8 0

3 years ago

NEED THIS NOW PLZ!!!

SVETLANKA909090 [29]

Answer: velocity

Explanation: Hope this helps :)

8 0

3 years ago