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Sveta_85 [38]
3 years ago
11

Which would hold more water- a teaspoon (tsp) or a milliliter?

Physics
2 answers:
ki77a [65]3 years ago
4 0
A tsp

1 milliliter equals 0.202. US teaspoons. boi did that help
salantis [7]3 years ago
3 0

Answer:

A teaspoon

Explanation:

As we know that the capacity of hold something by a standard teaspoon is related with milliliters as,

1 teaspoon is equivalent to 4.92892 milliliters

From this it can be see that the 1 teaspoon is equivalent to approximately 5 milliliters.

Therefore, the water hold by 1 teaspoon is more than the water hold by a milliliter.

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Two objects, m1 = 0.6 kg and m2 = 4.4 kg undergo a one-dimensional head-on collision
Sidana [21]

a) The velocity after the collision.is 11.456 m/s.

b) The kinetic energy lost due to the collision is 44.564 J.

<h3>What is conservation of momentum principle?</h3>

When two bodies of different masses move together each other and have head on collision, they travel to same or different direction after collision.

The external force is not acting here, so the initial momentum is equal to the final momentum. For inelastic collision, final velocity is the common velocity for both the bodies.

m₁u₁ +m₂u₂ =(m₁ +m₂) v

Given are the two objects, m1 = 0.6 kg and m2 = 4.4 kg undergo a one-dimensional head-on collision. Their initial velocities along the one-dimension path are vi1 = 32.4 m/s [right] and vi2 = 8.6 m/s [left].

(a) Substitute the values, then the final velocity will be

0.6 x32.4 +4.4 x 8.6 = (0.6+4.4)v

v = 11.456 m/s

Thus, the velocity after collision is 11.456 m/s.

(b) Kinetic energy lost due to collision will be the difference between the kinetic energy before and after collision.

= [1/2m₁u₁² +1/2m₂u₂² ] - [1/2(m₁ +m₂) v²]

Substitute the value, we have

= [1/2 x 0.6 x32.4² + 1/2 x4.4 x 8.6²] - [1/2 x(0.6+4.4)11.456²]

= 44.564 J

Thus, the kinetic energy lost due to the collision is 44.564 J.

Learn more about conservation of momentum principle

brainly.com/question/14033058

#SPJ2

4 0
1 year ago
Read 2 more answers
An iron wire has a cross-sectional area equal to 5.00×10⁻⁶ m² . Carry out the following steps to determine the drift speed of th
Doss [256]
  1. In mass, there are 55.85 × 10⁻³ kg/mol in in 1 mole of iron.
  2. The molar density of iron is equal to 1.41 × 10⁵ mol/m³.
  3. The density of iron atoms is equal to 8.49 × 10²⁸ atoms/m³.
  4. The number density of conduction electrons is equal to 1.70 × 10²⁹ conduction electrons/m³.
  5. The drift speed of conduction electrons is equal to 2.21 × 10⁻⁴ m/s.

<h3>How to calculate the drift speed of the conduction electrons?</h3>

Mathematically, the drift speed of the conduction electrons can be calculated by using this formula:

V = (m × σ × V)/ρ × e × f × l)

V = I/(n × A × Q)

Where:

  • U represents the drift speed of the conduction electrons, in m/s.
  • m represents the molecular mass of the metal, in kg.
  • e represents the elementary charge, in C.
  • f represents the number of free electrons per atom.
  • σ represents the electric conductivity of the medium at a particular temperature in S/m.
  • ρ represents the density of the conductor, in kg/m³.
  • ℓ represents the length of the conductor, in m.
  • ΔV represents the voltage applied or potential difference across the conductor in V.

<h3>How many kilograms are there in 1 mole of iron? </h3>

Molar mass of iron = 55.85 g/mol.

In Kilograms, we have:

Mass = 55.85 × 1/1000

Mass = 55.85 × 10⁻³ kg/mol.

For the molar density of iron, we have:

Molar density = density/molar mass

Molar density = 7874/0.056

Molar density = 1.41 × 10⁵ mol/m³.

For the density of iron atoms, we have:

Density of iron atoms = Avogadro's constant × molar density

Density of iron atoms = 6.023 × 10²³ × 1.406 × 10⁵

Density of iron atoms = 8.49 × 10²⁸ atoms/m³.

For the number density of conduction electrons, we have:

Fe ---> Fe²⁺ + 2e⁻

Number density of conduction electrons = 2 conduction electrons/1 atom of iron

Number density of conduction electrons = 2 × 8.49 × 10²⁸

Number density of conduction electrons = 1.70 × 10²⁹ conduction electrons/m³.

For the drift speed of conduction electrons, we have:

V = I/(n × A × Q)

V = 30/(1.70 × 10²⁹ × 1.602 × 10⁻¹⁹ × 5 × 10⁻⁶)

Drift speed, V = 2.21 × 10⁻⁴ m/s.

Read more on drift speed here: brainly.com/question/15219891

#SPJ4

Complete Question:

An iron wire has a cross-sectional area of 5.00 x 10-6 m2. Carry out steps (a) through (e) to compute the drift speed of the conduction electrons in the wire.

(a) How many kilograms are there in 1 mole of iron?

(b) Starting with the density of iron and the result of part (a), compute the molar density of iron (the number of moles of iron per cubic meter).

(c) Calculate the number density of iron atoms using Avogadro’s number.

(d) Obtain the number density of conduction electrons given that there are two conduction electrons per iron atom.

(e) If the wire carries a current of 30.0 A, calculate the drift speed of conduction electrons.

4 0
1 year ago
Suppose that a teacher driving a 1972 LeMans zooms out of a darkened tunnel at 34.5 m/s. He is momentarily blinded by the sunshi
valkas [14]

Answer:

489.19m

Explanation:

To find the minimum distance you first calculate the time in which the teacher stops:

v=v_o-at\\\\t=\frac{v_o-v}{a}=\frac{34.5m/s-0m/s}{2.5m/s^2}=13.8s

however, the reaction of the teacher is 0.31s later, then you use

t=13.8-0.31s=13.49s

during this time the camper has traveled a distance of:

x=vt=(15.1m/s)(13.49s)=203.69m   (1)

Next you calculate the distance that teacher has traveled for 13.6s:

x=x_o+v_ot+\frac{1}{2}at^2\\\\x=0m+34.5m/s(13.49s)+\frac{1}{2}(2.5m/s^2)(13.49s)^2=692.88m  (2)

The minimum distance between the driver and the camper will be the difference between (2) and (1):

x_{min}=692.88m-203.69m=489.19m

5 0
3 years ago
I drop a meter stick , and catch it as it falls. If it fell exactly 22 cm, how much did it take me to catch the meter stick (wha
statuscvo [17]
D=Vot+1/2at^2

In this case, there is no initial y velocity so the term Vot=0 so d=1/2at^2

acceleration=acceleration due to gravity=-9.8m/s^2

It falls - 22cm or -0.22m

We have - 0.22=1/2(-9.8)t^2

t^2=(-0.44)/(-9.8)

t=sqrt[0.44/9.8]
3 0
2 years ago
____ are formed where bumps from two surfaces come into contact.
NemiM [27]

the answer is Friction.

hope this heslp

6 0
3 years ago
Read 2 more answers
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