Elements that typically give up electrons

Derive the dimension of coefficient of linear expansivity

Reil [10]

Answer:

The SI unit of coefficient of linear expansion can be expressed as °C-1 or °K-1. ... The dimension of coefficient of linear expansion will be M0L0T0K−1.

3 0

2 years ago

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A 920-kg sports car collides into the rear end of a 2300-kg SUV stopped at a red light. The bumpers lock, the brakes are locked,

GrogVix [38]

Answer:21.45 m/s

Explanation:

Given

Mass of sport car=920 kg

Mass of SUV=2300 kg

distance to which both car skid is 2.4 m

coefficient of friction ( $\mu$ )=0.8

Let u be the initial velocity of both car at the starting of skidding

and they finally come to zero velocity

$v^2-u^2=2as$

$acceleration=\mu g=0.8\times 9.8=7.84 m/s^2$

s=2.4 m

$0-(u)^2=2\times (-7.84)\times 2.4$

u=6.13 m/s

so before colliding sport car must be travelling at a speed of

$920\times v=(920+2300)\times 6.13$ (conserving momentum)

v=21.45 m/s

7 0

2 years ago

How can a small human retina detect objects larger than itself?

aliya0001 [1]

Exactly the same way that you can photograph a mountain or a skyscraper
with the itty bitty camera in your smartphone.

Lenses are used to form a tiny image of a gigantic object.

7 0

2 years ago

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A CD spins at a constant angular velocity of 5.0 revolutions per second clockwise.

Lera25 [3.4K]

The true statement about the CD is:

<h3><em>b. No net torque acts on it at all.</em></h3>

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<h3>Further explanation</h3>

Centripetal Acceleration can be formulated as follows:

$\large {\boxed {a = \frac{ v^2 } { R } }$

<em>a = Centripetal Acceleration ( m/s² )</em>

<em>v = Tangential Speed of Particle ( m/s )</em>

<em>R = Radius of Circular Motion ( m )</em>

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Centripetal Force can be formulated as follows:

$\large {\boxed {F = m \frac{ v^2 } { R } }$

<em>F = Centripetal Force ( m/s² )</em>

<em>m = mass of Particle ( kg )</em>

<em>v = Tangential Speed of Particle ( m/s )</em>

<em>R = Radius of Circular Motion ( m )</em>

Let us now tackle the problem !

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<em>Complete Question:</em>

<em>A CD spins at a constant angular velocity of 5.0 revolutions per second clockwise. Which of the following statements about the CD is true?</em>

<em>a. A net torque acts on it clockwise to keep it moving</em>

<em>b. No net torque acts on it at all.</em>

<em>c. A net torque acts on it counterclockwise to keep it moving</em>

<u>Given:</u>

angular velocity = ω = 5.0 revolutions per second

<u>Asked:</u>

net torque = Στ = ?

<u>Solution:</u>

Constant angular velocity → angular acceleration = α = 0 rad/s²

$\Sigma \tau = I \alpha$

$\Sigma \tau = I (0)$

$\Sigma \tau = 0 \texttt{ Nm}$

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<h3>Conclusion:</h3>

The true statement about the CD is:

<em>b. No net torque acts on it at all.</em>

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<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

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<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Circular Motion

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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

#LearnWithBrainly

8 0

3 years ago

A test rocket is launched vertically from ground level (y=0), at time t=0.0s. The rocket engine provides constant upward acceler

Luden [163]

From the definition of average velocity,

$\bar v=\dfrac{\Delta y}{\Delta t}=\dfrac{49\,\mathrm m}t$ ,

and the fact that constant acceleration means

$\bar v=\dfrac{v_f+v_0}2=\dfrac{30\,\frac{\mathrm m}{\mathrm s}}2=15\,\dfrac{\mathrm m}{\mathrm s}$

we can solve for the time $t$ :

$15\,\dfrac{\mathrm m}{\mathrm s}=\dfrac{49\,\mathrm m}t\implies t=3.3\,\mathrm s$

7 0

2 years ago