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stira [4]
3 years ago
12

A bicyclist passing through a city accelerates after he passes the sign post marking the city limits at x=0. His acceleration is

constant at 5.0 m/s^2 east. At time t1= 0.0s, he is located at x1 = 6.0m and has a velocity v1=4.0 m/s east.
A) Find his position and velocity at t2= 2.0s.

B) How far is he from the sign post when his velocity is 6.0 m/s.
Physics
1 answer:
Anni [7]3 years ago
8 0

Answer:

A. 24 m, 14 m/s

B. 8.0 m

Explanation:

Given:

x₀ = 6.0 m

v₀ = 4.0 m/s

a = 5.0 m/s²

t = 2.0 s

A. Find: x and v

x = x₀ + v₀ t + ½ at²

x = (6.0 m) + (4.0 m/s) (2.0 s) + ½ (5.0 m/s²) (2.0 m/s)²

x = 24 m

v = at + v₀

v = (5.0 m/s²) (2.0 s) + (4.0 m/s)

v = 14 m/s

B. Find x when v = 6.0 m/s.

v² = v₀² + 2a (x − x₀)

(6.0 m/s)² = (4.0 m/s)² + 2 (5.0 m/s²) (x − 6.0 m)

x = 8.0 m

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Explanation:

7 0
3 years ago
An ideal spring hangs from the ceiling. A 1.25-kg mass is hung from the spring. After all vibrations have died away, the spring
ch4aika [34]

The kinetic energy of the mass at the instant it passes back through its equilibrium position is about 1.20 J

\texttt{ }

<h3>Further explanation</h3>

Let's recall Elastic Potential Energy formula as follows:

\boxed{E_p = \frac{1}{2}k x^2}

where:

<em>Ep = elastic potential energy ( J )</em>

<em>k = spring constant ( N/m )</em>

<em>x = spring extension ( compression ) ( m )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

mass of object = m = 1.25 kg

initial extension = x = 0.0275 m

final extension = x' = 0.0735 - 0.0275 = 0.0460 m

<u>Asked:</u>

kinetic energy = Ek = ?

<u>Solution:</u>

<em>Firstly , we will calculate the spring constant by using </em><em>Hooke's Law</em><em> as follows:</em>

F = k x

mg = k x

k = mg \div x

k = 1.25(9.8) \div 0.0275

k = 445 \frac{5}{11} \texttt{ N/m}

\texttt{ }

<em>Next , we will use </em><em>Conservation of Energy</em><em> formula to solve this problem:</em>

Ep_1 + Ek_1 = Ep_2 + Ek_2

\frac{1}{2}k (x')^2 + mgh + 0 = \frac{1}{2}k x^2 + Ek

Ek = \frac{1}{2}k (x')^2 + mgh - \frac{1}{2}k x^2

Ek = \frac{1}{2}k ( (x')^2 - x^2 ) + mgh

Ek = \frac{1}{2}(445 \frac{5}{11}) ( 0.0460^2 - 0.0275^2 ) + 1.25(9.8)(0.0735)

\boxed {Ek \approx 1.20 \texttt{ J}}

\texttt{ }

<h3>Learn more</h3>
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302
  • Young Modulus : brainly.com/question/9202964
  • Simple Harmonic Motion : brainly.com/question/12069840

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Elasticity

8 0
3 years ago
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Why do cells have receptors? Receptors make mRNA. Receptors let the cell know when to let things in and out of the cell. Recepto
Pavlova-9 [17]

Answer: Cells have receptors because Receptors let the cell know when to let things in and out of the cell.

Explanation:

Cell receptors also called transmembrane receptors are proteins located on the surface of a cell (extracellularly) or inside the cell which receive signals that alters the functions of the cell. The functions of the cells which can be altered includes the alteration in gene transcription and the cell morphology.

Cell receptors are generally categorizes into the following groups:

--> Internal receptors

--> cell surface receptors

--> ion channel receptors

--> G protein coupled receptors

--> enzyme linked receptors

Interaction of cell membrane receptors with specific ligands that bonds to the receptors causes conformational changes in the receptor protein. This in turn, enzymatically activates the intracellular part of the protein or induces interactions between the receptor and the proteins in the cytoplasm that act as second messengers, thereby relaying the signal from the extracellular part of the receptor to the interior of the cell. This enables the cell to know when to let things in or out of it through the information conveyed.

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3 years ago
An 800 kg fishing boat going south at 12 m/s runs into a stopped pontoon boat (1400 kg). The boats stick together and move south
kykrilka [37]

Answer:

the velocity of the boats after the collision is 4.36 m/s.

Explanation:

Given;

mass of fish, m₁ = 800 kg

mass of boat, m₂ = 1400 kg

initial velocity of the fish, u₁ = 12 m/s

initial velocity of the boat, u₂ = 0

let the final velocity of the fish-boat after collision = v

Apply the principle of conservation of linear momentum for inelastic collision;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

800 x 12    +   1400 x 0 = v(800 + 1400)

9600 = 2200v

v = 9600/2200

v = 4.36 m/s

Therefore, the velocity of the boats after the collision is 4.36 m/s.

7 0
3 years ago
If the coefficient of static friction between your coffee cup and the horizontal dashboard of your car is µs = 0.800, how fast c
Cloud [144]

Answer:

Before start of slide velocity will be 14.81 m/sec

Explanation:

We have given coefficient of static friction \mu =0.8

Angle of inclination is equal to \Theta =tan^{-1}\mu

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tan{38.65^{\circ}}=0.8

Radius is given r = 28 m

Acceleration due to gravity g=9.8m/sec^2

We know that tan\Theta =\frac{v^2}{rg}

0.8=\frac{v^2}{28\times 9.8}

v^2=219.52

v=14.816m/sec

So before start of slide velocity will be 14.81 m/sec

3 0
3 years ago
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