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1 year ago

Multiply the number 4.48E-8 by 5.2E-4 using Google. What is the correct answer in scientific notation?

Physics

1 answer:

Oksi-84 [34.3K]1 year ago

8 0

Answer:

$2.32\times 10^{-11}$

Explanation:

First number is $4.48\times 10^{-8}$

Second number is $5.2\times 10^{-4}$

We need to multiply the two numbers.

$4.48\times 10^{-8}\times 5.2\times 10^{-4}=(4.48\times 5.2)\times 10^{(-8-4)}\\\\=23.296\times 10^{-12}$

In scientific notation : $2.32\times 10^{-11}$

Hence, this is the required solution.

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A 1500 kg car traveling at 20 m/s suddenly runs out of gas while approaching the valley shown in the figure. The alert driver im

geniusboy [140]

Answer:

v_f = 17.4 m / s

Explanation:

For this exercise we can use conservation of energy

starting point. On the hill when running out of gas

Em₀ = K + U = ½ m v₀² + m g y₁

final point. Arriving at the gas station

Em_f = K + U = ½ m v_f ² + m g y₂

energy is conserved

Em₀ = Em_f

½ m v₀ ² + m g y₁ = ½ m v_f ² + m g y₂

v_f ² = v₀² + 2g (y₁ -y₂)

we calculate

v_f ² = 20² + 2 9.8 (10 -15)

v_f = √302

v_f = 17.4 m / s

8 0

2 years ago

A variable-length air column is placed just below a vibrating wire that is fixed at both ends. The length of the air column, ope

pogonyaev

Answer:

The speed of transverse waves in the wire is 234.26 m/s.

Explanation:

Given that,

Length of wire = 129 cm

Speed of sound in air = 340 m/s

First position of resonance = 31.2 cm

We need to calculate the wavelength

For pipe open at one end and closed at other, there is node at closed end and an anti node at open end for 1st resonance.

At 1st resonance,

$L=\dfrac{\lambda}{4}$

$\lambda=4\times L$

Put the value into the formula

$\lambda=4\times31.2\times10^{-2}$

$\lambda=1.248\ m$

We need to calculate the frequency of sound in pipe

Using formula of frequency

$f=\dfrac{v}{\lambda}$

Put the value into the formula

$f=\dfrac{340}{1.248}$

$f=272.4\ Hz$

We need to calculate the node distance

For the wire, there are 3 segments, so 4 nodes

Node-node distance ,

$L=\dfrac{l}{3}$

$L = \dfrac{129}{3}$

$L=43\ cm$

We need to calculate the wavelength of the wire

Using formula of length

$L=\dfrac{\lambda}{2}$

$\lambda=L\times 2$

Put the value into the formula

$\lambda=43\times2$

$\lambda=86\ cm$

We need to calculate the speed of the wave

Using formula of frequency

$f=\dfrac{v}{\lambda}$

$v=f\times\lambda$

Put the value into the formula

$v=272.4\times86\times10^{-2}$

$v=234.26\ m/s$

Hence, The speed of transverse waves in the wire is 234.26 m/s.

4 0

2 years ago

In which situation is the object experiencing unbalanced forces? A) A box resting on a horizontal floor. B) A car slowing as it

evablogger [386]

"Unbalanced forces" show themselves as a change in the speed
or direction of an object's motion.

The only choice where the speed or direction of motion is changing
is the car that's slowing down for the light.

3 0

1 year ago

Read 2 more answers

A capacitor, C1, consists of two parallel circular plates with radius R and separation of d. A second capacitor, C2, consists of

qaws [65]

Answer:

$\frac{E_1}{E_2}= 4$

Explanation:

Capacitance C is given by

$C= \frac{\epsilon_0A}{d}$

A= area of capacitor cross section

d= distance

therefore,

$C_1= \frac{\epsilon_0A_1}{d_1}$

A_1= πR^2

d_1= d

$C_2= \frac{\epsilon_0A_2}{d_2}$

A_= π(2R)^2

d_2 = 2d

$q= \frac{\epsilon_0A_1}{d_1}V_1$

threfore

$V_1= \frac{qd_1}{\epsilon_0A_1}$

and

$V_2= \frac{qd_2}{\epsilon_0A_2}$

also we know that E= V/d

⇒ $\frac{E_1}{E_2}= \frac{V_1}{V_2}\times\frac{d_2}{d_1}$

⇒ $\frac{E_1}{E_2}= \frac{qd_1}{\epsilon_0A_1}\times\frac{\epsilon_0A_2}{qd_2}\times\frac{d_2}{d_1}$

= A_1/A_2= $\frac{4R^2}{R^2}$ =4

therefore,

$\frac{E_1}{E_2}= 4$

5 0

1 year ago

Consider a wave along the length of a stretched slinky toy, where the distance between coils increases and decreases. What type

GrogVix [38]

Answer:

"Longitudinal wave" is the appropriate answer.

Explanation:

Generating waves whenever the form of communication being displaced in a similar direction as well as in the reverse way of the wave's designated points, could be determined as Longitudinal waves.
A wave running the length of something like a Slinky stuffed animal, which expands as well as reduces the spacing across spindles, produces a fine image or graphic.

3 0

1 year ago

Read 2 more answers