What is the scientific notation of 65000?

Choose all the answers that apply.

Elena-2011 [213]

Answer:

a.is the same as science

c. influences science

e. helps scientists observe fast phenomena

Explanation:

3 0

3 years ago

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What would be the density if a nugget of gold’s mass of 965g and a volume of 50cm

olchik [2.2K]

p = 19.3 g/cm3

V√3=3.6840315cm :))

3 0

4 years ago

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What type of reaction occurs between Mg and HCl and explain why?

lyudmila [28]

Bonding is ionic for sure
Mg has lost 2e,s to become Mg2+
and there are 2 Cl-
the strong attraction between the Mg2+ and 2Cl- form the ionic bonding

6 0

3 years ago

When a metal was exposed to light at a frequency of 4.07× 1015 s–1, electrons were emitted with a kinetic energy of 3.30× 10–19

Licemer1 [7]

Answer : The maximum number of electrons released = $1.432\times 10^{12}electrons$

Explanation : Given,

Frequency = $4.07\times 10^{15}s^{-1}$

Kinetic energy = $3.30\times 10^{-19}J$

Total energy = $3.39\times 10^{-7}J$

First we have to calculate the work function of the metal.

Formula used :

$K.E=h\nu -w$

where,

K.E = kinetic energy

h = Planck's constant = $6.626\times 10^{-34}J/s$

$\nu$ = frequency

w = work function

Now put all the given values in this formula, we get the work function of the metal.

$3.30\times 10^{-19}J=(6.626\times 10^{-34}J/s\times 4.07\times 10^{15}s^{-1})-w$

By rearranging the terms, we get

$w=2.367\times 10^{-18}J$

Therefore, the works function of the metal is, $2.367\times 10^{-18}J$

Now we have to calculate the maximum number of electrons released.

The maximum number of electrons released = $\frac{\text{ Total energy}}{\text{ work function}}$

The maximum number of electrons released = $\frac{3.39\times 10^{-7}J}{2.367\times 10^{-19}J}=1.432\times 10^{12}electrons$

Therefore, the maximum number of electrons released is $1.432\times 10^{12}electrons$

8 0

3 years ago

Practice Problem: True Stress and Strain A cylindrical specimen of a metal alloy 49.7 mm long and 9.72 mm in diameter is stresse

amm1812

Answer:

The true stress required = 379 MPa

Explanation:

True Stress is the ratio of the internal resistive force to the instantaneous cross-sectional area of the specimen. True Strain is the natural log to the extended length after which load applied to the original length. The cold working stress – strain curve relation is as follows,

σ(t) = K (ε(t))ⁿ, σ(t) is the true stress, ε(t) is the true strain, K is the strength coefficient and n is the strain hardening exponent

True strain is given by

Epsilon t =㏑ (l/l₀)

Substitute㏑(l/l₀) for ε(t)

σ(t) = K(㏑(l/l₀))ⁿ

Given values l₀ = 49.7mm, l =51.7mm , n =0.2 , σ(t) =379Mpa

379 x 10⁶ = K (㏑(51.7/49.7))^0.2

K = 379 x 10⁶/(㏑(51.7/49.7))^0.2

K = 723.48 MPa

Knowing the constant value would be same as the same material is being used in the second test, we can find out the true stress using the above formula replacing the value of the constant.

σ(t) = K(㏑(l/l₀))ⁿ

l₀ = 49.7mm, l = 51.7mm, n = 0.2, K = 723.48Mpa

σ(t) = 723.48 x 106 x (㏑(51.7/49.7))^0.2

σ(t) = 379 MPa

The true stress necessary to plastically elongate the specimen is 379 MPa.

6 0

3 years ago

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