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2 years ago

A bat emitts a sonar sound wave (343 m/s) that bounces off a mosquito 8.42 m away

Physics

1 answer:

Varvara68 [4.7K]2 years ago

8 0

The time lapse between when the bat emits the sound and when it hears the echo is 0.05 s.

From the question given above, the following data were obtained:

Velocity of sound (v) = 343 m/s

Distance (x) = 8.42 m

Time (t) =?

We can obtain obtained the time as illustrated below:

v = 2x / t

343 = 2 × 8.42 / t

343 = 16.84 / t

Cross multiply

343 × t = 16.84

Divide both side by 343

t = 16.84/343

t = 0.05 s

Thus, the time between when the bat emits the sound and when it hears the echo is 0.05 s.

<h3>How does a bat know how far away something is?</h3>

A bat emits a sound wave and carefully listens to the echoes that return to it. The returning information is processed by the bat's brain in the same way that we processed our shouting sound with a stopwatch and calculator. The bat's brain determines the distance of an object by measuring how long it takes for a noise to return.

Learn more about time elapses between when the bat emits the sound :

<u>brainly.com/question/16931690</u>

#SPJ4

Correction question:

A bat emits a sonar sound wave (343 m/s) that bounces off a mosquito 8.42 m away. How much time elapses between when the bat emits the sound and when it hears the echo? (Unit = s)

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Answer:

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3 years ago

A mass MM uniform solid cylinder of radius RR and a mass MM thin uniform spherical shell of radius RR roll without slipping. If

vampirchik [111]

Answer:

vcyl / vsph = 1.05

Explanation:

The kinetic energy of a rolling object can be expressed as the sum of a translational kinetic energy plus a rotational kinetic energy.
The traslational part can be written as follows:

$K_{trans} = \frac{1}{2}* M* v_{cm} ^{2} (1)$

The rotational part can be expressed as follows:

$K_{rot} = \frac{1}{2}* I* \omega ^{2} (2)$

where I = moment of Inertia regarding the axis of rotation.
ω = angular speed of the rotating object.
If the object has a radius R, and it rolls without slipping, there is a fixed relationship between the linear and angular speed, as follows:

$v = \omega * R (3)$

For a solid cylinder, I = M*R²/2 (4)
Replacing (3) and (4) in (2), we get:

$K_{rot} = \frac{1}{2}* \frac{1}{2} M*R^{2} * \frac{v_{cmc} ^{2}}{R^{2}} = \frac{1}{4}* M* v_{cmc}^{2} (5)$

Adding (5) and (1), we get the total kinetic energy for the solid cylinder, as follows:

$K_{cyl} = \frac{1}{2}* M* v_{cmc} ^{2} +\frac{1}{4}* M* v_{cmc}^{2} = \frac{3}{4}* M* v_{cmc} ^{2} (6)$

Repeating the same steps for the spherical shell:

$I_{sph} = \frac{2}{3} * M* R^{2} (7)$

$K_{rot} = \frac{1}{2}* \frac{2}{3} M*R^{2} * \frac{v_{cms} ^{2}}{R^{2}} = \frac{1}{3}* M* v_{cms}^{2} (8)$

$K_{sph} = \frac{1}{2}* M* v_{cms} ^{2} +\frac{1}{3}* M* v_{cms}^{2} = \frac{5}{6}* M* v_{cms} ^{2} (9)$

Since we know that both masses are equal each other, we can simplify (6) and (9), cancelling both masses out.
And since we also know that both objects have the same kinetic energy, this means that (6) are (9) are equal each other.
Rearranging, and taking square roots on both sides, we get:

$\frac{v_{cmc}}{v_{cms}} =\sqrt{\frac{10}{9} } = 1.05 (10)$

This means that the solid cylinder is 5% faster than the spherical shell, which is due to the larger moment of inertia for the shell.

3 0

3 years ago

In 1994, Leroy Burrell of the United States set what was then a new world record for the men’s 100 m run. He ran the 1.00  102

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Answer:

61.33 Kg

Explanation:

From the question given above, the following data were obtained:

Distance = 1×10² m

Time = 9.5 s

Kinetic energy (KE) = 3.40×10³ J

Mass (m) =?

Next, we shall determine the velocity Leroy Burrell. This can be obtained as follow:

Distance = 1×10² m

Time = 9.5 s

Velocity =?

Velocity = Distance / time

Velocity = 1×10² / 9.5

Velocity = 10.53 m/s

Finally, we shall determine the mass of Leroy Burrell. This can be obtained as follow:

Kinetic energy (KE) = 3.40×10³ J

Velocity (v) = 10.53 m/s

Mass (m) =?

KE = ½mv²

3.40×10³ = ½ × m × 10.53²

3.40×10³ = ½ × m × 110.8809

3.40×10³ = m × 55.44045

Divide both side by 55.44045

m = 3.40×10³ / 55.44045

m = 61.33 Kg

Thus, the mass of Leroy Burrell is 61.33 Kg

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2 years ago

A physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor in 0.350s. ignore air resistance.

Nookie1986 [14]

Answer:

a. 0.6 m b. 0.385 m c. 3.6 m/s at 287.78° to the horizontal

Explanation:

a. Using s = ut - 1/2gt² for motion under gravity where s = vertical distance = height of table, u = initial vertical velocity of book = 0 m/s, t = time of flight = 0.350 s and g = acceleration due to gravity = 9.8 m/s².

Substituting these these values into s and taking the top of the table as position 0 m, we have.

0 - s = 0t - 1/2gt²

-s = -1/2gt²

s = 1/2gt²

s = 1/2 × 9.8 m/s² × (0.350 s)²

s = 0.6 m

b. Using d = v't where d = horizontal distance from table, v' = horizontal velocity of book = 1.10 m/s and t = time of flight = 0.350 s

d = v't = 1.10 m/s × 0.350 s = 0.385 m

c. Using v² = u² - 2gs where u = initial vertical velocity of book = 0 m/s and g = 9.8 m/s², s = -0.6 m (negative since we are at the bottom and 0 m is at the top)and v = final vertical velocity of book

v² = u² - 2gs

= 0 - 2 × 9.8 m/s² × (-0.6 m)

= 11.76 m²/s²

v = √11.76 m/s

= 3.43 m/s

So, the magnitude of the resultant velocity is V = √(v² + v'²)

= √((3.43 m/s)² + (1.10 m/s)'²)

= √(11.76 m²/s² + 1.21 m²/s²)

= √12.97 m²/s²

= 3.6 m/s

Its direction Ф = tan⁻¹(-v/v') since v is in the negative y direction

= tan⁻¹(-3.43 m/s/1.10 m/s)

= tan⁻¹(-3.1182)

= -72.22°

Ф = -72.22°+ 360 = 287.78° since it is in the third quadrant

7 0

3 years ago