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svlad2 [7]
2 years ago
9

Lacie kicks a football from ground level at a velocity of 13.9 m/s and at an angle of 25.0° to the ground. How long will the bal

l be in the air before it lands? Round your answer to the nearest tenth. s How far will the football travel before it lands? Round your answer to the nearest tenth. m
Physics
2 answers:
RUDIKE [14]2 years ago
6 0

Answer:

1.2s for the first one

15.1m for the second one on time4learning

Explanation:

Step2247 [10]2 years ago
4 0

Answer:

T = 1.2 s

T = 15.1 m = 15 m

Explanation:

This is a case of projectile motion:

TOTAL TIME OF FLIGHT:

The formula for total time of flight in projectile motion is:

T = 2 V₀ Sinθ/g

where,

T = Total Time of Flight = ?

V₀ = Launch Speed = 13.9 m/s

θ = Launch Angle = 25°

g = 9.8 m/s²

Therefore,

T = (2)(13.9 m/s)(Sin 25°)/(9.8 m/s²)

<u>T = 1.2 s</u>

<u></u>

RANGE OF BALL:

The formula for range in projectile motion is:

R = V₀² Sin2θ/g

where,

R = Horizontal Distance Covered by ball = ?

Therefore,

T = (13.9 m/s)²(Sin 2*25°)/(9.8 m/s²)

<u>T = 15.1 m = 15 m</u>

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Two speakers face each other, and they each emit a sound of wavelength λ. One speaker is 180∘ out of phase with respect to the o
mel-nik [20]

Answer:

a. 3/4λ

d. 1/4λ

Explanation:

When the wavelength of the sound waves is λ and the two waves are having same frequency the waves are said to be out of phase if their phase difference is in the multiples of \frac{\lambda}{2} or 180°.

When the two waves are out of phase then their opposite maxima coincide at the same time resulting in the minimum amplitude of the resulting wave throughout.

  • As we observe from the schematic that the a wave has sinusoidal pattern of variation and we get a maxima after each \frac{\lambda}{4} of the distance.
  • Here we have two speakers out of phase therefore on shifting one of the speakers by the odd multiples of \frac{\lambda}{2} we have the maxima or the extreme amplitudes.

So, we must place the microphone at  3/4λ and 1/4λ to pickup the loudest sound.

4 0
2 years ago
Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to t
yarga [219]

Answer

The rate at which the magnetic field is changing is  [\frac{dB}{dt} ] =  0.000467 T/s

Explanation

From the question we are told that

   The electric field strength is E =  3.5mV/m =  3.5 *10^{-3} \ V/m

    The radius is  r =  1.5 \ m

The rate of change of the  magnetic  field  is mathematically represented as

        \frac{d \phi }{dt}  =  \int\limits^{} {E \cdot dl}

Where dl is change of a unit length

     \frac{d \phi}{dt}  =  A *  \frac{dB}{dt}

Where A is the area which is mathematically represented as

     A = \pi r^2

    So

    E \int\limits^{} {  dl} =  ( \pi r^2) (\frac{dB}{dt} )  

  E L  =  ( \pi r^2) (\frac{dB}{dt} )  

where L is the circumference of the circle which is mathematically represented as

     L = 2 \pi r

So

     E (2 \pi r ) =  (\pi r^2 ) [\frac{dB}{dt} ]

      E  =   \frac{r}{2}  [\frac{dB}{dt} ]

       [\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }

substituting values

      [\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }

      [\frac{dB}{dt} ] =  0.000467 T/s    

8 0
2 years ago
Tammy leaves the office, drives 26 km due
fredd [130]

Answer:

72.98 km

Explanation:

Her displacement is simply the distance from her final position to her initial position.

Now, I've drawn and attached a triangle diagram to depict this her movement.

Point O is her initial starting point.

Point A is the first point she gets to after travelling north while point B is the final point after travelling north east.

From the triangle, the displacement will be the distance OB which is denoted by x and can be solved from cosine rule.

Thus;

x² = 62² + 26² - 2(62 × 26)cos 120

x² = 4520 + 806

x² = 5326

x = √5326

x = 72.98 km

4 0
2 years ago
In a supermarket, you place a 22.3-N (around 5 lb) bag of oranges on a scale, and the scale starts to oscillate at 2.7 Hz. What
allsm [11]

Answer:

Force constant, k = 653.3 N/m

Explanation:

It is given that,

Weight of the bag of oranges on a scale, W = 22.3 N

Let m is the mass of the bag of oranges,

m=\dfrac{W}{g}

m=\dfrac{22.3}{9.8}

m = 2.27 kg

Frequency of the oscillation of the scale, f = 2.7 Hz

We need to find the force constant (spring constant) of the spring of the scale. We know that the formula of the frequency of oscillation of the spring is given by :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

k=4\pi^2 f^2m

k=4\pi^2 \times (2.7)^2\times 2.27

k = 653.3 N/m

So, the force constant of the spring of the scale is 653.3 N/m. Hence, this is the required solution.

7 0
3 years ago
At what condition does a body become weightless at equator
salantis [7]
Hope this is helpful <span>Weightlessness

</span>
3 0
2 years ago
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