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aleksandr82 [10.1K]

3 years ago

7

elastic wire extend by 1.ocm when a load on 20g range from It, what additional load will it be required Cause the futher extensi

on of 2.0cm

1 answer:

Citrus2011 [14]3 years ago

6 0

Answer:

40g

Explanation:

20g range > 1.0cm

Therefore,

40g range > 2.0cm

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Name the blood vessels that carry blood from the upper and lower parts of your body

aev [14]

Pulmonary Arteries: Blood vessels that carry deoxygenated blood from the heart to the lungs. Superior Vena Cava: A large vein that delivers deoxygenated blood from the upper body into the heart. Hope this helps

3 0

4 years ago

Read 2 more answers

As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a p

monitta

Answer:

force F = 1.66 × $10^{-13}$ N

Explanation:

given data

proton and an electron = 865 nm

solution

we get here force that is express as

force F = k q1 q2 ÷ r² ......................1

put here value and we get

force F = 9 × $10^{9}$ × $\frac{1.6\times (10^{-19})^{2}}{865 \times (10^{-9})^{2}}$

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4 0

3 years ago

A sample of chloroform is found to contain 12.0 g of carbon, 106.4 g of chlorine, and 1.01 g of hydrogen. If a second sample of

GREYUIT [131]

Given:

Sample 1:

Chloroform is $CHCl_{3}$

12 g Carbon

1.01 g Hydrogen

106.4 g Cl

Sample 2:

30.0 g of Carbon

Solution:

mass of chloroform from sample 1:

12 + 1.01 +106.4 =119.41 g

Now, for the total mass of chloroform in sample 2:

mass of chloroform $\times$ $\frac{given mass of carbon}{mass of carbon atom}$

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6 0

3 years ago

a 2100-kg car drives with a speed of 18 m/s onb a flat road around a curve that has a radius of curvature of 83m. The coefficien

Law Incorporation [45]

Answer:

<u>The magnitude of the friction force is 8197.60 N</u>

Explanation:

Using the definition of the centripetal force we have:

$\Sigma F=ma_{c}=m\frac{v^{2}}{R}$

Where:

m is the mass of the car
v is the speed
R is the radius of the curvature

Now, the force acting in the motion is just the friction force, so we have:

$F_{f}=m\frac{v^{2}}{R}$

$F_{f}=2100\frac{18^{2}}{83}$

$F_{f}=8197.60 \: N$

<u>Therefore the magnitude of the friction force is 8197.60 N</u>

I hope it helps you!

7 0

3 years ago

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andrey2020 [161]

Explanation:

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4 years ago