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alexandr402 [8]
3 years ago
6

When at rest, a ruler has a length of 12.0 inches. How fast (in m/s) must the ruler fly past you in order for it to shorten to 6

.00 inches in length? Hint: to use scientific notation within Canvas, use e+ instead of 10^ (ex. 3.00×108 = 3.00e+8).
Physics
1 answer:
PIT_PIT [208]3 years ago
3 0

Answer:

2.598×10^8 m/sec

Explanation:

The length of the contraction is given by l=l_0\sqrt{1-\frac{v^2}{c^2}} here l_0 is length of the ruler at rest which is given as 12 inches

l is the length of the ruler in moving condition which is given as 6 inches c is the speed of the light and v is the velocity of the ruler.

l=l_0\sqrt{1-\frac{v^2}{c^2}}

\frac{l}{l_0}=\sqrt{1-\frac{v^2}{c^2}}

1-(\frac{l}{l_0})^2=\frac{v^2}{c^2}

1-(\frac{6}{12})^2=\frac{v^2}{c^2}

0.866=\frac{v}{c}

v=0.866\times 3\times 10^8=2.598\times 10^8 m/sec

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matrenka [14]

Answer:

The answer to your question is

Explanation:

Data

mass = 0.5kg

T1 = 35

T2 = ?

Q = - 6.3 x 10⁴ J  = - 63000 J

Cp = 4184 J / kg°C

Formula

                        Q = mCp(T2 - T1)

                         T2 = T1 + Q/mCp    

Substitution

                       T2 = 35 - 63000/(0.5 x 4184)

                        T2 = 35 - 63000/2092

                        T2 = 35 - 30.1

                         T2 = 4.9 °C

6 0
2 years ago
What is the magnitude of your displacement when you follow directions that tell you to walk 225 m in one direction, make a 90° t
Gekata [30.6K]

Start by facing East. Your first displacement is the vector

<em>d</em>₁ = (225 m) <em>i</em>

Turning 90º to the left makes you face North, and walking 350 m in this direction gives the second displacement,

<em>d</em>₂ = (350 m) <em>j</em>

Turning 30º to the right would have you making an angle of 60º North of East, so that walking 125 m gives the third displacement,

<em>d</em>₃ = (125 m) (cos(60º) <em>i</em> + sin(60º) <em>j</em> )

<em>d</em>₃ ≈ (62.5 m) <em>i</em> + (108.25 m) <em>j</em>

The net displacement is

<em>d</em> = <em>d</em>₁ + <em>d</em>₂ + <em>d</em>₃

<em>d</em> ≈ (287.5 m) <em>i</em> + (458.25 m) <em>j</em>

and its magnitude is

|| <em>d</em> || = √[ (287.5 m)² + (458.25 m)² ] ≈ 540.973 m ≈ 541 m

7 0
3 years ago
[TRUE/FALSE] If an object is moving at constant velocity it must be in equilibrium.
DENIUS [597]

Answer:

yes

Explanation:

objects with constant velocity also have zero net external force. this means the forces on the object are balanced. this mean they are in equilibrium

4 0
3 years ago
Help me find the acceleration
ANEK [815]

a = 3.09 m/s²

<h3>Explanation</h3>

This question doesn't tell anything about how long it took for the car to go through 105 meters. As a result, the <em>timeless </em>suvat equation is likely what you need for this question.

In the <em>timeless</em> suvat equation,

a = \dfrac{v^2 - u^2}{2\; x}

where

  • a is the acceleration of the car;
  • v is the <em>final</em> velocity of the car;
  • u is the <em>initial</em> velocity of the car; and
  • x is the displacement of the car.

Note that <em>v</em> and <em>u</em> are velocities. Make sure that you include their signs in the calculation.

In this question,

  • a is the unknown;
  • v = -10.9 \; \text{m} \cdot \text{s}^{-2};
  • u = -27.7 \; \text{m} \cdot \text{s}^{-2}; and
  • x = - 105 \; \text{m}.

Apply the <em>timeless</em> suvat equation:

a = \dfrac{v^{2} - u^{2}}{2\; x}\\\phantom{a} = \dfrac{(-10.9)^{2} - (-27.7)^{2}}{2 \times (-105)}\\\phantom{a} = 3.09 \; \text{m} \cdot \text{s}^{-2}.

The value of a is greater than zero, which is reasonable. Velocity of the car is negative, meaning that the car is moving backward. The car now moves to the back at a slower speed. Effectively it accelerates to the front. Its acceleration shall thus be positive.

7 0
3 years ago
A block is attached to a horizontal spring. On top of this block rests another block. The two-block system slides back and forth
Minchanka [31]

Answer:

Angular frequency will increase

No change in the amplitude

Explanation:

At extreme end of the SHM the energy of the SHM is given by

E = \frac{1}{2} (m_1 + m_2)\omega^2 A^2

here we know that

\omega^2 = \frac{k}{m_1 + m_2}

now at the extreme end when one of the mass is removed from it

then in that case the angular frequency will change

\omega'^2 = \frac{k}{m_1}

So angular frequency will increase

but the position of extreme end will not change as it is given here that the top block is removed without disturbing the lower block

so here no change in the amplitude

6 0
2 years ago
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