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3 years ago

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Two carts with masses 10.0kg and 2.5 kg move in opposite directions on a frictionless horizontal track with speeds of 6.0m/s and

3.0 m/a respectively. The carts stick together after colliding head on. Find the final speed.
Please help!!

1 answer:

Elis [28]3 years ago

8 0

Answer:

4.2

Explanation:

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Which of the following represent units of an electric field? select all that apply

zimovet [89]

Answer:

Volts/Meter

Newtons/Coulomb

Explanation:

Volts/ Meter and Newtons/Coulomb both are same and the units of Electric field intensity or electric field strength.

Electric field strength E is the force per unit charge. It is measured in Newton/Coulomb in SI unit. It is a vector quantity directed in the direction of force.

Mathematically,

Electric field strength = Force/Charge

E = F / q₀

= Newton / Coulomb = NC⁻¹ 1

We know that

Newton = Joule/meter 2

Also

Volt = Joule/Coulomb 3

So put 3 in 2 we get

Newton = (Volt Coulomb)/meter put in 1

E = (Volt Coulomb)/(meter Coulomb)

= Volt / meter

Hence

Newton / Coulomb = Volt / meter

8 0

3 years ago

Read 2 more answers

During the developing of a TLC plate, it is common to place a cover on the chromatography chamber and have a piece of moist filt

Artyom0805 [142]

Answer:

The moist filter paper is in charge of preventing evaporation and ensuring the proper saturation of the air of the chromatography chamber.

Explanation:

Thin-layer chromatography (TLC) is a chromatography technique used to separate non-volatile mixtures.

A strip of moist filter paper is put into the chromatography chamber so that its bottom touches the solvent and the paper lies on the chamber wall and reaches almost to the top of the container.

The container is closed and left for a few minutes to let the solvent vapors ascend the moist filter paper and saturate the air in the chamber.

The moist filter paper is in charge of preventing evaporation and ensuring the proper saturation of the air of the chamber.

6 0

3 years ago

The tension in cable da has a magnitude of tda=6.27 lb. find the cartesian components of tension tda, which is directed from d t

Oliga [24]

Complete Question

The Complete Question is attached below

We have that the Cartesian components of tension $T_{da}$ is

$T_{DA}=-4.433i-3.49j+2.735k$

From the Question we are told that

$M_{da}=6.27 lb\\\\w=9.50ft\\\\d=6.60ft\\\\h=4.50ft$

$\vec {DA}=-4.7i-3.7j+2.9k)ft$

$\vec {DB}=-1.9i-3.7j+1.9k)ft\\\\\vec {DC}=-1.9i+5.8j-1.6k)ft$

Generally the equation for $T_{DA}$ is mathematically given as

$T_{DA}=\phi_{DA}* M_{da}$

Where

$\phi_{DA}=\frac{-4.7i-3.7j+2.9k}{(-4.7)^2+(-3.7)^2+(2.9)^2}\\\\\phi_{DA}=\frac{-4.7i-3.7j+2.9k}{6.65}$

Therefore

$T_{DA}=\phi_{DA}* M_{da}$

$T_{DA}=\frac{-4.7i-3.7j+2.9k}{6.65}* 6.27$

$T_{DA}=-4.433i-3.49j+2.735k$

For more information on this visit

brainly.com/question/20746649?referrer=searchResults

5 0

3 years ago

Consider a cylindrical segment of a blood vessel 2.70 cm long and 3.10 mm in diameter. What additional outward force would such

Lyrx [107]

This question is incomplete, the complete question is;

- Calculate the difference in blood pressure between the feet and top of the head of a person who is 1.80m Tall

- Consider a cylindrical segment of a blood vessel 2.70 cm long and 3.10 mm in diameter. What additional outward force would such a vessel need to withstand in the person's feet compared to a similar vessel in her head

Answer:

- the difference in blood pressure is 18698.4 Pa

- the additional outward force F is 4.86 N

Explanation:

Given the data in the question;

we know that the expression for difference in blood pressure is;

ΔP = pgh

where p is density = 1060 kg/m³

g is acceleration due to gravity = 9.8 m/s²

and h is height = 1.80 m

now we substitute

ΔP = 1060 × 9.8 × 1.80

ΔP = 18698.4 Pa

therefore the difference in blood pressure is 18698.4 Pa

Also given that;

diameter of blood vessel d = 3.10 mm

radius r = 3.10 mm / 2 = 1.55 mm = 0.00155 m

length l = 2.70 cm = 0.027 m

Surface area of the cylindrical segment of a blood vessel is

A = 2πrl

we substitute

A = 2 × π × 0.00155 × 0.027

A = 2.6 × 10⁻⁴ m²

so

the required for will be;

F = PA

we substitute

F = 18698.4 Pa × 2.6 × 10⁻⁴ m²

F = 4.86 N

Therefore, the additional outward force F is 4.86 N

5 0

3 years ago

You push a desk with 245 N, but the desk doesn't move due to its friction with the ground. What is the magnitude of the friction

const2013 [10]

If the desk doesn't move, then it's not accelerating.

If it's not accelerating, then the net force on it is zero.

If the net force on it is zero, then any forces on it are balanced.

If there are only two forces on it and they're balanced, then they have equal strengths, and they point in opposite directions.

So the friction on the desk must be equal to your<em> 245N</em> .

7 0

3 years ago