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Karo-lina-s [1.5K]
4 years ago
14

Which of the following actions would make a pulse travel faster along a stretched string? More than one answer may be correct. I

f so, give all that are correct. a. Move your hand up and down more quickly as you generate the pulse. b. Move your hand up and down a larger distance as you generate the pulse. c. Use a heavier string of the same length, under the same tension. d. Use a lighter string of the same length, under the same tension. e. Stretch the string tighter to increase the tension. f. Loosen the string to decrease the tension. g. Put more force into the wave.
Physics
1 answer:
MaRussiya [10]4 years ago
5 0

Answer:

Option d and e are correct.

Explanation:

The expression for velocity of pulse in a stretched string can be given as follows

v = \sqrt{\frac{T}{m} }

where T is tension in the string , m is mass of string per unit length.

Use of  lighter string of the same length, under the same tension  amounts to higher m so velocity will decrease. Hence option d is correct.

Similarly, v is directly  proportional to square root of tension. So if we increase tension , velocity also increases. So option e ) is correct.

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An object is thrown straight up with an initial velocity of 10 m/s, and there is an air resistance force causing an acceleration
lana [24]

Answer:

Vf= 7.29 m/s

Explanation:

Two force act on the object:

1) Gravity

2) Air resistance

Upward motion:

Initial velocity = Vi= 10 m/s

Final velocity = Vf= 0 m/s

Gravity acting downward =  g = -9.8 m/s²

Air resistance acting downward = a₁ = - 3 m/s²

Net acceleration = a = -(g + a₁ ) = - ( 9.8 + 3 ) = - 12.8 m/s²

( Acceleration is consider negative if it is in opposite direction of velocity )

Now

2as = Vf² - Vi²

⇒ 2 * (-12.8) *s = 0 - 10²

⇒-25.6 *s = -100

⇒ s = 100/ 25.6

⇒ s = 3.9 m

Downward motion:

Vi= 0 m/s

s = 3.9 m

Gravity acting downward =  g = 9.8 m/s²

Air resistance acting upward = a₁ = - 3 m/s²

Net acceleration = a = g - a₁  =  9.8 - 3  = 6.8 m/s²

Now

2as = Vf² - Vi²

⇒ 2 * 6.8 * 3.9 = Vf² - 0

⇒ Vf² = 53. 125

⇒ Vf= 7.29 m/s

8 0
4 years ago
If you're driving towards the sun late in the afternoon, you can reduce the glare from the road by wearing sunglasses that permi
AveGali [126]

Correct answer choice is:

C. Polarized in a vertical plane.

Explanation:

Polarized sunglasses give excellent glare shield, particularly on the water. Polarized lenses include a specific filter that prevents this type of strong reflected light, diminishing glare.

This is because when you angle one polarized lens to different perpendicularly, they prevent glare both horizontally and vertically. The polarized lenses are enduringly tinted sunglasses that exceedingly decrease glare.

7 0
4 years ago
Why do organs have different types of tissues?<br><br><br> PLZ HELP
Setler [38]

Explanation:

its hard to explain its very complex but its so they can function properly

6 0
3 years ago
What is the magnitude of δv12 if the bulb is removed from the socket (i.e. the circuit is not closed)?
andreev551 [17]

Since the circuit is incomplete or not closed, no current flows in the circuit. as per ohm's law , Voltage is directly proportional to current and is given as

V = Voltage = i R where i = current , R = resistance

as no current flows in the circuit, i = 0

the resistance R can not be zero. hence

V = 0 (R)

V = 0 Volts

so the magnitude of the Voltage is zero Volts

4 0
4 years ago
Read 2 more answers
In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1
asambeis [7]

Explanation:

It is given that,

Frequency of monochromatic light, f=5\times 10^{14}\ Hz

Separation between slits, d=2.2\times 10^{-5}\ m

(a) The condition for maxima is given by :

d\ sin\theta=n\lambda

For third maxima,

\theta=sin^{-1}(\dfrac{n\lambda}{d})

\theta=sin^{-1}(\dfrac{n\lambda}{d})

\theta=sin^{-1}(\dfrac{nc}{fd})  

\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})  

\theta=4.69^{\circ}

(b) For second dark fringe, n = 2

d\ sin\theta=(n+1/2)\lambda

\theta=sin^{-1}(\dfrac{5\lambda}{2d})

\theta=sin^{-1}(\dfrac{5c}{2df})

\theta=sin^{-1}(\dfrac{5\times 3\times 10^8}{2\times 2.2\times 10^{-5}\times 5\times 10^{14}})

\theta=3.90^{\circ}

Hence, this is the required solution.

8 0
3 years ago
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