answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Rina8888 [55]
2 years ago
13

An air-track glider undergoes a perfectly inelastic collision with an identical glider that is initially at rest. what fraction

of the first glider's initial kinetic energy is transformed into thermal energy in this collision?

Physics
2 answers:
Mashutka [201]2 years ago
6 0

Half of the first glider's initial kinetic energy is transformed into thermal energy in this collision.

\texttt{ }

<h3>Further explanation</h3>

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

\large {\boxed {F = ma }

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

\texttt{ }

<u>Given:</u>

mass of first glider = m₁ = m

mass of second glider = m₂ = m

initial speed of first glider = u₁ = u

initial speed of second glider = u₂ = 0

final speed of both gliders = v₁ = v₂ = v <em>→ perfectly inelatic collision</em>

<u>Asked:</u>

change in kinetic energy = ΔEk = ?

<u>Solution:</u>

<em>Firstly , we will use </em><em>Conservation of Momentum Law </em><em>as follows:</em>

m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2

mu + m(0) = mv + mv

mu = 2mv

u = 2v

\boxed {v = \frac{1}{2}u}

\texttt{ }

<em>Next , we could calculate the change in kinetic energy of first glider:</em>

\Delta Ek : Ek_1 = ( Ek_1 - Ek ) : Ek_1

\Delta Ek : Ek_1 = ( \frac{1}{2}mu^2 - \frac{1}{2}(2mv^2)) : (\frac{1}{2}mu^2)

\Delta Ek : Ek_1 = ( mu^2 - 2mv^2 ) : (mu^2)

\Delta Ek : Ek_1 = ( mu^2 - 2m(\frac{1}{2}u)^2 ) : (mu^2)

\Delta Ek : Ek_1 = ( mu^2 - 2m(\frac{1}{4}u^2) ) : (mu^2)

\Delta Ek : Ek_1 = ( mu^2 - \frac{1}{2}mu^2 ) : (mu^2)

\Delta Ek : Ek_1 = ( \frac{1}{2}mu^2 ) : (mu^2)

\Delta Ek : Ek_1 = \frac{1}{2} : 1

\boxed {\Delta Ek = \frac{1}{2} Ek_1}

\texttt{ }

<h3>Conclusion:</h3>

Half of the first glider's initial kinetic energy is transformed into thermal energy in this collision.

\texttt{ }

<h3>Learn more</h3>
  • Impacts of Gravity : brainly.com/question/5330244
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
  • The Acceleration Due To Gravity : brainly.com/question/4189441
  • Newton's Law of Motion: brainly.com/question/10431582
  • Example of Newton's Law: brainly.com/question/498822

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

DiKsa [7]2 years ago
5 0
Refer to the diagram shown below.

The initial KE (kinetic energy) of the system is
KE₁ = (1/2)mu²

After an inelastic collision, the two masses stick together.
Conservation of momentum requires that
m*u = 2m*v
Therefore
v = u/2

The final KE is
KE₂ = (1/2)(2m)v²
       = m(u/2)²
       = (1/4)mu²
      = (1/2) KE₁

The loss in KE is
KE₁ - KE₂ = (1/2) KE₁.

Conservation of energy requires that the loss in KE be accounted for as thermal energy.

Answer:  1/2 

You might be interested in
If you were to triple the size of the Earth (R = 3R⊕) and double the mass of the Earth (M = 2M⊕), how much would it change the g
EastWind [94]

Answer:

Decreased by a factor of 4.5

Explanation:

"We have Newton formula for attraction force between 2 objects with mass and a distance between them:

F_G = G\frac{M_1M_2}{R^2}

where G =6.67408 × 10^{-11} m^3/kgs^2 is the gravitational constant on Earth. M_1, M_2 are the masses of the object and Earth itself. and R distance between, or the Earth radius.

So when R is tripled and mass is doubled, we have the following ratio of the new gravity over the old ones:

\frac{F_G}{f_g} = \frac{G\frac{M_1M_2}{R^2}}{G\frac{M_1m_2}{r^2}}

\frac{F_G}{f_g} = \frac{\frac{M_2}{R^2}}{\frac{m_2}{r^2}}

\frac{F_G}{f_g} = \frac{M_2}{R^2}\frac{r^2}{m_2}

\frac{F_G}{f_g} = \frac{M_2}{m_2}(\frac{r}{R})^2

Since M_2 = 2m_2 and r = R/3

\frac{F_G}{f_g} = \frac{2}{3^2} = 2/9 = 1/4.5

So gravity would have been decreased by a factor of 4.5  

8 0
2 years ago
A diffusion couple composed of two silver– gold alloys is formed; these alloys have compositions of 98 wt% Ag–2 wt% Au and 95 wt
jeyben [28]

Answer:

for this problem, 2.5 = (5+2/2)-(5-2/2)erf (50×10-6m/2Dt)

It now becomes necessary to compute the diffusion coefficient at 750°C (1023 K) given that D0= 8.5 ×10-5m2/s and Qd= 202,100 J/mol.

we have D= D0exp( -Qd/RT)

=(8.5×105m2/s)exp(-202,100/8.31×1023)

= 4.03 ×10-15m2/s

4 0
2 years ago
) A physics student wants to measure the stiffness of a spring (force required per cm stretched). He knows that according to Hoo
Maurinko [17]

Answer:

Explanation:

find the solution below

3 0
2 years ago
Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t
Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

M=-\frac{s'}{s}

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

M=-\frac{15 cm}{12 cm}=-1.25

D. Real and inverted

The magnification equation can be also rewritten as

M=\frac{y'}{y}

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

y'=My

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

M=\frac{5}{9}=-\frac{s'}{s}

which can be rewritten as

s'=-M s = -\frac{5}{9}s

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

s'=-Ms

and then we can substitute it into the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}

and we can solve for s:

\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0
2 years ago
A 125-g metal block at a temperature of 93.2 °C was immersed in 100. g of water at 18.3 °C. Given the specific heat of the metal
Nataly_w [17]

Answer:

34.17°C

Explanation:

Given:

mass of metal block = 125 g

initial temperature T_i = 93.2°C

We know

Q = m c \Delta T   ..................(1)

Q= Quantity of heat

m = mass of the substance

c = specific heat capacity

c = 4.19 for H₂O in J/g^{\circ}C

\Delta T = change in temperature

Now

The heat lost by metal = The heat gained by the metal

Heat lost by metal = 125\times 0.9\times (93.2-T_f)

Heat gained by the water = 100\times 4.184\times(T_f -18.3)

thus, we have

125\times 0.9\times (93.2-T_f) = 100\times 4.184\times(T_f -18.3)

10485-112.5T_f = 418.4T_f - 7656.72

⇒ T_f = 34.17^oC

Therefore, the final temperature will be = 34.17°C

6 0
2 years ago
Other questions:
  • According to the exercise principle of balance, a workout should __________.
    14·2 answers
  • Step 8: Observe How Changes in the Speed of the Bottle Affect Beanbag Height
    7·2 answers
  • A power washer is being used to clean the siding of a house. Water enters at 20 C, 1 atm, with a volumetric flow rate of 0.1 lit
    13·1 answer
  • A viscous liquid is sheared between two parallel disks of radius �, one of which rotates with angular speed Ω, while the other i
    14·1 answer
  • An 18-cm-long bicycle crank arm, with a pedal at one end, is attached to a 20-cm-diameter sprocket, the toothed disk around whic
    12·1 answer
  • A car enters a 300-m radius horizontal curve on a rainy day when the coefficient of static friction between its tires and the ro
    7·1 answer
  • What is the kinetic energy of a soccer ball which has a mass of 0.8 kg and is kicked at a velocity of 10 m/s
    8·1 answer
  • In this problem you are to consider an adiabaticexpansion of an ideal diatomic gas, which means that the gas expands with no add
    6·1 answer
  • When you drink liquid through a straw, is it more accurate to say the liquid is pushed up the straw rather than sucked up the st
    9·1 answer
  • In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 73.5 m/s. Th
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!