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3 years ago

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A paper is placed in between two books can be quickly pulled out without movinf the books. whuch statement explains this phenome

non

2 answers:

garri49 [273]3 years ago

8 0

Answer:( D ) The books have a great deal of inertia and do not move easily.

Explanation:

Montano1993 [528]3 years ago

7 0

This is easily explained saying that the frictional force between the books and the paper isn't big enough to produce a displacement in the books. The displacement in the books doesn't happen because the frictional force between the books and the surface they are standing on is bigger than the paper's one.

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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

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<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Elasticity

8 0

3 years ago

Read 2 more answers

You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between

Murljashka [212]

Answer:

a) v = 0.0496 m / s , b) v = 0.0957 m / s

Explanation:

a) in this case we have an inelastic collision,

To solve these problems, the most important thing is to define the system formed by all the bodies, so that the forces during the crash have been internal and the amount of movement is conserved.

Therefore the system is made up of the ball and the player

initial instant. Before catching the ball

p₀ = m v₁

final moment. After you have grabbed the ball

$p_{f}$ = (m + M) v

the moment is preserved

po = p_{f}

m v₁ = (m + M) v

v = m / (m + M) v₁

let's calculate

v = 0.405 / (0.405 + 69) 8.50

v = 0.0496 m / s

in the same direction that the ball takes

b) In this case the ball bounces with a speed of V₂ = -7.80 m / s, the negative sign is because it is going in opposite directions

let's write the moment in two times

initial instant. Before the crash

p₀ = m v₁

try end. Right after the crash

p_{f} = m v₂ + M v

the moment is preserved

p₀ = p_{f}

m v₁ = m v₂ + M v

v = m (v₁ -v₂) / M

v = 0.405 /69 (8.50 - (7.80))

v = 0.0957 m / s

in the initial direction of the ball

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3 years ago

A 20×10⁹charge is moved between two points A andB that are 30mm apart and have an electric potential difference of 600v between

Alona [7]

Answer:

John Felix Anthony Cena is an American professional wrestler, actor, and television presenter. He is currently signed to WWE. Born and raised in West Newbury, Massachusetts, Cena moved to California in 1998 to pursue a career as a bodybuilder.

Explanation:

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3 years ago

Rama's weight is 4okg: She is carrying a load of 20kg up to a height of 20 meters.what work does she do?Also mention its type of

Maslowich

Answer:

rama is doing

Explanation:

work done=f×d×g

=60×20×9.8

=11760j

she is doing work against gravity

mark me

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3 years ago

An ideal Diesel cycle has a compression ratio of 18 and a cutoff ratio of 1.5. Determine the (1) maximum air temperature and (2)

weqwewe [10]

Answer:

(1) The maximum air temperature is 1383.002 K

(2) The rate of heat addition is 215.5 kW

Explanation:

T₁ = 17 + 273.15 = 290.15

$\frac{T_2}{T_1} =r_v^{k - 1} =18^{0.4} =3.17767$

T₂ = 290.15 × 3.17767 = 922.00139

$\frac{T_3}{T_2} =\frac{v_3}{v_2} = r_c = 1.5$

Therefore,

T₃ = T₂×1.5 = 922.00139 × 1.5 = 1383.002 K

The maximum air temperature = T₃ = 1383.002 K

(2)

$\frac{v_4}{v_3} =\frac{v_4}{v_2} \times \frac{v_2}{v_3} = \frac{v_1}{v_2} \times \frac{v_2}{v_3} = 18 \times \frac{1}{1.5} = 12$

$\frac{T_3}{T_4} =(\frac{v_4}{v_3} )^{k-1} = 12^{0.4} = 2.702$

Therefore;

$T_4 = \frac{1383.002}{2.702} =511.859 \ k$

$Q_1 = c_p(T_3-T_2)$

Q₁ = 1.005(1383.002 - 922.00139) = 463.306 kJ/jg

Heat rejected per kilogram is given by the following relation;

$c_v(T_4-T_1)$ = 0.718×(511.859 - 290.15) = 159.187 kJ/kg

The efficiency is given by the following relation;

$\eta = 1-\frac{\beta ^{k}-1}{\left (\beta -1 \right )r_{v}^{k-1}}$

Where:

β = Cut off ratio

Plugging in the values, we get;

$\eta = 1-\frac{1.5 ^{1.4}-1}{\left (1.5 -1 \right )18^{1.4-1}}= 0.5191$

Therefore;

$\eta = \frac{\sum Q}{Q_1}$

$\therefore 0.5191 = \frac{150}{Q_1}$

Heat supplied = $\frac{150}{0.5191} = 288.978 \ hp$

Therefore, heat supplied = 215491.064 W

Heat supplied ≈ 215.5 kW

The rate of heat addition = 215.5 kW.

7 0

4 years ago