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3 years ago

select the correct scientific notation form of this numeral 541 5 * 10^2 5.4 * 10^2 5.4 * 10^2 5.0 * 10^2

1 answer:

7 0

To find the scientific notation, you need to divide at the decimal by the power of 10. So since there are 2 powers of 10, what you want to do is move the decimal 2 places to the left which will give you: .054

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A sound is recorded at 19 decibels. What is the intensity of the sound?

sp2606 [1]

$1 \times 10^{-10.1} \mathrm{Wm}^{-2}$ is the intensity of the sound.

Answer: Option B

<u>Explanation:</u>

The range of sound intensity that people can recognize is so large (including 13 magnitude levels). The intensity of the weakest audible noise is called the hearing threshold. (intensity about $1 \times 10^{-12} \mathrm{Wm}^{-2}$ ). Because it is difficult to imagine numbers in such a large range, it is advisable to use a scale from 0 to 100.

This is the goal of the decibel scale (dB). Because logarithm has the property of recording a large number and returning a small number, the dB scale is based on a logarithmic scale. The scale is defined so that the hearing threshold has intensity level of sound as 0.

$\text { Intensity }(d B)=(10 d B) \times \log _{10}\left(\frac{I}{I_{0}}\right)$

Where,

I = Intensity of the sound produced

$I_{0}$ = Standard Intensity of sound of 60 decibels = $1 \times 10^{-12} \mathrm{Wm}^{-2}$

So for 19 decibels, determine I as follows,

$19 d B=(10 d B) \times \log _{10}\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)$

$\log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=\frac{19}{10}$

$\log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=1.9$

When log goes to other side, express in 10 to the power of that side value,

$\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)=10^{1.9}$

$I=1 \times 10^{-12} \mathrm{Wm}^{-2} \times 10^{1.9}=1 \times 10^{-12-1.9}=1 \times 10^{-10.1} \mathrm{Wm}^{-2}$

5 0

3 years ago

What is a newton equal to in terms of units of mass and acceleration

artcher [175]

1 newton is the force needed to accelerate 1 kilogram of mass
at the rate of 1 meter per second² .

1 N = 1 kg-m/s² .

It's a force equal to roughly 3.6 ounces.

7 0

3 years ago

Read 2 more answers

The engine of a 1520-kg automobile has a power rating of 75 kW. Determine the time required to accelerate this car from rest to

LUCKY_DIMON [66]

Answer:

t=15.68 s

Explanation:

Given that

m = 1520 kg

P =75 KW

We know that

Power ,P = F .v

F=force

v=velocity

v= 100 km/h

$v=\dfrac{1000}{3600}\times 100\ m/s$

v=27.77 m/s

75 x 1000 = F x 27.77

$F= \dfrac{75000}{27.77}\ N$

F= 2700.75 N

F= m a

m=mass

a=acceleration

2700.75 = 1520 x a

a=1.77 m/s²

time t given as

v= u + a t

27.77 = 0 + 1.77 x t

t=15.68 s

3 0

4 years ago

Which question can be used to identify a physical property of an element?

BabaBlast [244]

Answer:C

Explanation:

Do bubbles form when the element is mixed with an acid

3 0

3 years ago

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You have a two-wheel trailer that you pull behind your ATV. Two children with a combined mass of 76.2 kg hop on board for a ride

marin [14]

a) The spring constant is 12,103 N/m

b) The mass of the trailer 2,678 kg

c) The frequency of oscillation is 0.478 Hz

d) The time taken for 10 oscillations is 20.9 s

Explanation:

When the two children jumps on board of the trailer, the two springs compresses by a certain amount

$\Delta x = 6.17 cm = 0.0617 m$

Since the system is then in equilibrium, the restoring force of the two-spring system must be equal to the weight of the children, so we can write:

$2mg = k'\Delta x$ (1)

where

m = 76.2 kg is the mass of each children

$g=9.8 m/s^2$ is the acceleration of gravity

$k'$ is the equivalent spring constant of the 2-spring system

For two springs in parallel each with constant k,

$k'=k+k=2k$

Substituting into (1) and solving for k, we find:

$2mg=2k\Delta x\\k=\frac{mg}{\Delta x}=\frac{(76.2)(9.8)}{0.0617}=12,103 N/m$

The period of the oscillating system is given by

$T=2\pi \sqrt{\frac{m}{k'}}$

where

And for the system in the problem, we know that

T = 2.09 s is the period of oscillation

m is the mass of the trailer

$k'=2k=2(12,103)=24,206 N/m$ is the equivalent spring constant of the system

Solving the equation for m, we find the mass of the trailer:

$m=(\frac{T}{2\pi})^2 k'=(\frac{2.09}{2\pi})^2 (24,206)=2,678 kg$

The frequency of oscillation of a spring-mass system is equal to the reciprocal of the period, therefore:

$f=\frac{1}{T}$

where

f is the frequency

T is the period

In this problem, we have

T = 2.09 s is the period

Therefore, the frequency of oscillation is

$f=\frac{1}{2.09}=0.478 Hz$

The period of the system is

T = 2.09 s

And this time is the time it takes for the trailer to complete one oscillation.

In this case, we want to find the time it takes for the trailer to complete 10 oscillations (bouncing up and down 10 times). Therefore, the time taken will be the period of oscillation multiplied by 10.

Therefore, the time needed for 10 oscillations is:

$t=10T=10(2.09)=20.9 s$

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7 0

3 years ago