Questions for 1.21 physics lab report

A 0.060 kg ball hits the ground with a speed of –32 m/s. The ball is in contact with the ground for 45 milliseconds and the grou

SCORPION-xisa [38]

Answer:

imma try and get it wrong

3 0

2 years ago

A mass m at the end of a spring oscillates with a frequency of 0.89 hz. when an additional 603 g mass is added to m, the frequen

MrMuchimi

The solution for this problem:

Given:

f1 = 0.89 Hz

f2 = 0.63 Hz

Δm = m2 - m1 = 0.603 kg

The frequency of mass-spring oscillation is:
f = (1/2π)√(k/m)
k = m(2πf)²

Then we know that k is constant for both trials, we have:
k = k

m1(2πf1)² = m2(2πf2)²

m1 = m2(f2/f1)²

m1 = (m1+Δm)(f2/f1)²

m1 = Δm/((f1/f2)²-1)

m 1 = 0.603/ (0.89/0.63)^2 – 1

= 0.609 kg or 0.61kg or 610 g

5 0

3 years ago

Organism would make the best index fossil. Answers a , b , c , d please help

aivan3 [116]

Answer: You didn't put what a, b, c, or d is so how is anyone supposed to know what the answer is?

5 0

3 years ago

Read 2 more answers

List an example of each of the four classes polymers that living things make and use

slamgirl [31]

The four classes of polymers are:

1. Nucleic acids. Examples are DNA and RNA

2. Protein. Examples are enzymes and hemoglobin

3. Carbohydrates. Examples as starch and glycogen

4. Lipids. Examples are triglycerides and phospholipids

The building blocks of nucleic acids are called bases and there are four types known as Guanine, Adenine, Thymine and Cytosine.

The building blocks of carbohydrates are glucose molecules.

The building blocks of protein are amino acids.

The building blocks of lipids are a combination of fatty acids and glycerol.

6 0

3 years ago

A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate

Olegator [25]

Answer:

See the answers below.

Explanation:

In order to solve this problem we must use the principle of energy conservation. Which tells us that the energy of a body will always be the same regardless of where it is located. For this case we have two points, point A and point B. Point A is located at the top at 120 [m] and point B is in the middle of the cliff at 60 [m].

$E_{A}=E_{B}$

The important thing about this problem is to identify the types of energy at each point. Let's take the reference level of potential energy at a height of zero meters. That is, at this point the potential energy is zero.

So at point A we have potential energy and since a velocity of 18 [m/s] is printed, we additionally have kinetic energy.

$E_{A}=E_{pot}+E_{kin}\\E_{A}=m*g*h+\frac{1}{2}*m*v^{2}$

At Point B the rock is still moving downward, therefore we have kinetic energy and since it is 60 [m] with respect to the reference level we have potential energy.

$E_{B}=m*g*h+\frac{1}{2}*m*v^{2}$

Therefore we will have the following equation:

$(6.5*9.81*120)+(0.5*6.5*18^{2} )=(6.5*9.81*60)+(0.5*6.5*v_{B}^{2} )\\3.25*v_{B}^{2} =4878.9\\v_{B}=\sqrt{1501.2}\\v_{B}=38.75[m/s]$

The kinetic energy can be easily calculated by means of the kinetic energy equation.

$KE_{B}=\frac{1}{2} *m*v_{B}^{2}\\KE_{B}=0.5*6.5*(38.75)^{2}\\KE_{B}=4878.9[J]$

In order to calculate the velocity at the bottom of the cliff where the reference level of potential energy (potential energy equal to zero) is located, we must pose the same equation, with the exception that at the new point there is only kinetic energy.

$E_{A}=E_{C}\\6.5*9.81*120+(0.5*9.81*18^{2} )=0.5*6.5*v_{C}^{2} \\v_{c}^{2} =\sqrt{2843.39}\\v_{c}=53.32[m/s]$

5 0

3 years ago