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NARA [144]
2 years ago
9

A 25 kg box on a horizontal frictionless surface is moving to the right at a speed of 5.0 m s. The box hits and remains attached

to one end of a spring of negligible mass whose other end is attached to a wall. As a result, the spring compresses a maximum distance of 0.60 m. (a) i. The spring does work on the box from the moment the box first hits the spring to the moment the spring first reaches its maximum compression. Indicate whether the work done by the spring is positive, negative, or zero.
Physics
1 answer:
seropon [69]2 years ago
6 0

Answer:

Explanation:

Work done by the spring is negative .

Work done by force F creating displacement  d is given by the following expression .

Work = F x d

Both force and displacement are vector quantity .

When direction of force and direction of displacement is same , work is positive . When direction of force and direction of displacement is opposite , work is negative .

When spring is compressed , it exerts a restoring or opposing force in a direction opposite to the direction of displacement of box . Hence here force is opposite to displacement . Restoring force acts opposite to displacement . Hence work done by spring on box is negative .

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. Which of the following is NOT a property of water? A. Ability to dissolve many substances B. Constant volume upon freezing C.
ss7ja [257]

Answer:

a

Explanation:

water is only able to desolve two substances , salt , sugar,

7 0
3 years ago
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A very narrow beam of white light is incident at 40.80° onto the top surface of a rectangular block of flint glass 11.6 cm thick
DerKrebs [107]
Dispersion angle = 0.3875 degrees. 
Width at bottom of block = 0.09297 cm 
Thickness of rainbow = 0.07038 cm 
 Snell's law provides the formula that describes the refraction of light. It is:
 n1*sin(θ1) = n2*sin(θ2)
 where
 n1, n2 = indexes of refraction for the different mediums
 Î¸1, θ2 = angle of incident rays as measured from the normal to the surface. 
 Solving for θ2, we get
 n1*sin(θ1) = n2*sin(θ2)
 n1*sin(θ1)/n2 = sin(θ2)
 asin(n1*sin(θ1)/n2) = θ2 
 The index of refraction for air is 1.00029, So let's first calculate the angles of the red and violet rays.
 Red:
 asin(n1*sin(θ1)/n2) = θ2
 asin(1.00029*sin(40.80)/1.641) = θ2
 asin(1.00029*0.653420604/1.641) = θ2
 asin(0.398299876) = θ2
 23.47193844 = θ2 
 Violet:
 asin(n1*sin(θ1)/n2) = θ2
 asin(1.00029*sin(40.80)/1.667) = θ2
 asin(1.00029*0.653420604/1.667) = θ2
 asin(0.39208764) = θ2
 23.08446098 = θ2 
 So the dispersion angle is:
 23.47193844 - 23.08446098 = 0.38747746 degrees. 
 Now to determine the width of the beam at the bottom of the glass block, we need to calculate the difference in the length of the opposite side of two right triangles. Both triangles will have a height of 11.6 cm and one of them will have an angle of 23.47193844 degrees, while the other will have an angle of 23.08446098 degrees. The idea trig function to use will be tangent, where
 tan(θ) = X/11.6
 11.6*tan(θ) = X
 So for Red:
 11.6*tan(θ) = X
 11.6*tan(23.47193844) = X
 11.6*0.434230136 = X
 5.037069579 = X 
 And violet:
 11.6*tan(θ) = X
 11.6*tan(23.08446098) = X
 11.6*0.426215635 = X
 4.944101361 = X 
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 The actual width of the beam after it exits the flint glass block will be thinner. The beam will exit at an angle of 40.80 degrees and we need to calculate the length of the sides of a 40.80/49.20/90 right triangle. If you draw the beams, you'll realize that:
 cos(θ) = X/0.092968218
 0.092968218*cos(θ) = X 
 0.092968218*cos(40.80) = X
 0.092968218*0.756995056 = X
 0.070376481 = X 
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7 0
3 years ago
The ancient Greek recommendation of rest in the treatment of abnormal behavior soon gave rise to ______. A. Trephining B. The as
Lilit [14]
C. Is the correct answer
7 0
3 years ago
You drop a rock down a well that is 5.4 m deep. How long does it take the rock to hit the bottom of the well?
Natali5045456 [20]
Equation of motion:

y_{f}=y_{o}+v_{o}t+ \frac{1}{2} at^{2}

Since initial velocity is zero, the second term goes away:

y_{f}=y_{o}+0+ \frac{1}{2} at^{2}

y_{f}=y_{o}+\frac{1}{2} at^{2}

y_{f}-y_{o}= \frac{1}{2} at^{2}

y_{f}-y_{o}=5.4m

5.4m= \frac{1}{2} at^{2}

\frac{2(5.4m) }{a} = t^{2}

a = g = 9.81  \frac{m}{ s^{2}}

\frac{2(5.4m) }{9.81 \frac{m}{ s^{2} } } = t^{2}

1.1 s^{2} = t^{2}\sqrt{1.1 s^{2}} =  \sqrt{t^{2}}

<u><em>t = 1.05s</em></u>
4 0
3 years ago
Read 2 more answers
A glass flask whose volume is 1000 cm^3 at a temperature of 1.00°C is completely filled with mercury at the same temperature. W
Marat540 [252]

Answer:

the coefficient of volume expansion of the glass is \mathbf{  ( \beta_{glass} )= 1.333 *10^{-5} / K}

Explanation:

Given that:

Initial volume of the glass flask = 1000 cm³ = 10⁻³ m³

temperature of the glass flask and mercury= 1.00° C

After heat is applied ; the final temperature = 52.00° C

Temperature change ΔT = 52.00° C - 1.00° C = 51.00° C

Volume of the mercury overflow = 8.50 cm^3 = 8.50 ×  10⁻⁶ m³

the coefficient of volume expansion of mercury is 1.80 × 10⁻⁴ / K

The increase in the volume of the mercury =  10⁻³ m³ ×  51.00 × 1.80 × 10⁻⁴

The increase in the volume of the mercury = 9.18*10^{-6} \ m^3

Increase in volume of the glass =  10⁻³ × 51.00 × \beta _{glass}

Now; the mercury overflow = Increase in volume of the mercury - increase in the volume of the flask

the mercury overflow = (9.18*10^{-6}  -  51.00* \beta_{glass}*10^{-3})\ m^3

8.50*10^{-6} = (9.18*10^{-6}  -51.00* \beta_{glass}* 10^{-3} )\ m^3

8.50*10^{-6} - 9.18*10^{-6} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3

-6.8*10^{-7} =  ( -51.00* \beta_{glass}* 10^{-3} )\ m^3

6.8*10^{-7} =  ( 51.00* \beta_{glass}* 10^{-3} )\ m^3

\dfrac{6.8*10^{-7}}{51.00 * 10^{-3}}=  ( \beta_{glass} )

\mathbf{  ( \beta_{glass} )= 1.333 *10^{-5} / K}

Thus; the coefficient of volume expansion of the glass is \mathbf{  ( \beta_{glass} )= 1.333 *10^{-5} / K}

5 0
3 years ago
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