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wel
3 years ago
13

PLEASE HELP ME!!!! I'LL MARK YOU BRAINLIEST IF YOU ANSWER THIS CORRECTLY!!!!! I NEED THESE DONE TODAY!!!!

Physics
1 answer:
Reptile [31]3 years ago
4 0

Answer:

<h2>1)</h2><h2>TRUE</h2><h2>2)</h2><h2>TRUE</h2><h2>3)</h2><h2>TRUE</h2><h2>4)</h2><h2>\lambda = 0.773 m</h2><h2>5)</h2><h2>v = 100 m/s</h2>

Explanation:

1)

As we know by the formula of speed of sound we have

v = 332 + 0.6 t

t = 20 deg C

v = 332 + (0.6 \times 20)

v = 332 + 12 = 344 m/s

So approx speed of the sound is nearly 340 m/s

so its TRUE

2)

Wave speed is given as

v = \frac{\lambda}{T}

so it is given as

v = f \lambda

TRUE

3)

f = \frac{1}{T}

so from above equation we have

v = f \times \lambda

v = \frac{\lambda}{T}

4)

Wavelength is given as

\lambda = \frac{v}{f}

\lambda = \frac{340}{440}

\lambda = 0.773 m

5)

As we know that

f = 50 Hz

\lambda = 2.0 m

so we will have

v = \lambda f

v = 50 \times 2

v = 100 m/s

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2 years ago
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A drowsy cat spots a flower pot that sails first up and then down past an open window. The pot was in view for a total of 0.56 s
liberstina [14]

Answer:

h = 0.028 m

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As we know that

d = \frac{v_2 + v_1}{2} t

here we have

1.95 = \frac{v_2 + v_1}{2}(0.56)

v_2 + v_1 = 6.96

also we know

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v_2 - v_1 = 5.49

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4 0
4 years ago
A car is stopped at a traffic light. When the light turns green at t=0, a truck with a constant speed passes the car with a 20m/
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Answer:

At t = (70 / 3) \; {\rm s} (approximately 23.3 \; {\rm s}.)

Explanation:

Note that the acceleration of the car between t = 0\; {\rm s} and t = 20\; {\rm s} (\Delta t = 20\; {\rm s}) is constant. Initial velocity of the car was v_{0} = 0\; {\rm m\cdot s^{-1}}, whereas v_{1} = 35\; {\rm m\cdot s^{-1}} at t = 20\; {\rm s}\!. Hence, at t = 20\; {\rm s}\!\!, this car would have travelled a distance of:

\begin{aligned}x &= \frac{(v_{1} - v_{0})\, \Delta t}{2} \\ &= \frac{(35\; {\rm m\cdot s^{-1}} - 0\; {\rm m\cdot s^{-1}}) \times (20\; {\rm s})}{2} \\ &= 350\; {\rm m}\end{aligned}.

At t = 20\; {\rm s}, the truck would have travelled a distance of x = v\, t = 20\; {\rm m\cdot s^{-1}} \times 20\; {\rm s} = 400\; {\rm m}.

In other words, at t = 20\; {\rm s}, the truck was 400\; {\rm m} - 350\; {\rm m} = 50\; {\rm m} ahead of the car. The velocity of the car is greater than that of the truck by 35\; {\rm m\cdot s^{-1}} - 20\; {\rm m\cdot s^{-1}} = 15 \; {\rm m\cdot s^{-1}}. It would take another (50\; {\rm m}) / (15\; {\rm m\cdot s^{-1}}) = (10/3)\; {\rm s} before the car catches up with the truck.

Hence, the car would catch up with the truck at t = (20 + (10/3))\; {\rm s} = (70 / 3)\; {\rm s}.

3 0
2 years ago
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