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3 years ago

Does cobalt gains or loses electrons in bonding

Physics

2 answers:

11111nata11111 [884]3 years ago

7 0

It could gain or lose 4 too reach nobility <span />

ki77a [65]3 years ago

6 0

To what i have learned cobalt gains in bonding

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Suppose a distant world with surface gravity of 5.20 m/s2 has an atmospheric pressure of 7.16 ✕ 104 Pa at the surface. (a) What

Nesterboy [21]

Answer:

899752.13598 N

Explanation:

p = Pressure on the plate = $7.16\times 10^4\ Pa$

A = Area = $\pi r^2=\pi 2^2=\pi 4$

F = Force on the region

When force is divided by area we get pressure

$p=\frac{F}{A}\\\Rightarrow F=p\times A\\\Rightarrow F=7.16\times 10^4\times \pi 4\\\Rightarrow F=899752.13598\ N$

Force is exerted by the atmosphere on a disk-shaped region is 899752.13598 N

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3 years ago

A student conduct an experiment to determine habitation of salt to water affect the density of the water the student feels three

Nikolay [14]

Any like answer choices would be helpful '-'

8 0

3 years ago

A car is traveling at 120 km/h (75 mph). When applied the braking system can stop the car with a deceleration rate of 9.0 m/s2.

Bumek [7]

Answer:

the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

Explanation:

Given that;

speed of car V = 120 km/h = 33.3333 m/s

Reaction time of an alert driver = 0.8 sec

Reaction time of an alert driver = 3 sec

extra time taken by sleepy driver over an alert driver = 3 - 0.8 = 2.2 sec

now, extra distance that car will travel in case of sleepy driver will be'

S_d = V × 2.2 sec

S_d = 33.3333 m/s × 2.2 sec

S_d = 73.3333 m

hence, number of car of additional car length n will be;

n = S_n / car length

n = 73.3333 m / 5m

n = 14.666 ≈ 15

Therefore, the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

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3 years ago

A brass alloy is known to have a yield strength of 240 MPa (35,000 psi), a tensile strength of 310 MPa (45,000 psi), and an elas

Karo-lina-s [1.5K]

Answer:

Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.

Explanation:

Given that

Yield strength ,Sy= 240 MPa

Tensile strength = 310 MPa

Elastic modulus ,E= 110 GPa

L=380 mm

ΔL = 1.9 mm

Lets find strain:

Case 1 :

Strain due to elongation (testing)

ε = ΔL/L

ε = 1.9/380

ε = 0.005

Case 2 :

Strain due to yielding

$\varepsilon' =\dfrac{S_y}{E}$

$\varepsilon' =\dfrac{240}{110\times 1000}$

ε '=0.0021

Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.

For computation of load strain due to testing should be less than the strain due to yielding.

4 0

3 years ago

Do geochemists need to have a knowledge of physical science? Explain your<br> answer. please help:))

saul85 [17]

No they don’t lol I hate school

6 0

4 years ago