Micro-bats use a form of radar called echolocation to navigate and find their prey such as flying insects. They locate the surrounding objects by bouncing sound wave pulses off these objects and detecting the time delay between the emitted pulses and the reflected pulses. Determine the time delay between the pulse emitted by the micro-bat and the detected pulse reflected from an insect located 10 m away from the micro-bat. Assume the approximate speed of sound waves to be 340 m-s-1

Answer:

ionized particles from the sun.

* interactions in radiation belts.

* the friction of the planet in the solar wind

q = +9 10⁵ C

Explanation:

Due to being made up of matter, the planet Earth has a series of positive and negative charges, in general these charges should be balanced and the net charge of the planet should be zero, but there are several phenomena that introduce unbalanced charges, for example:

* ionized particles from the sun.

* interactions in radiation belts.

* the friction of the planet in the solar wind

This creates that the planet has a net electrical load

We can roughly calculate the charge of the planet

             E = k q / r²

             q = E r² / k

let's calculate

             q = 200 (6.37 10⁶)²/9 10⁹

              q = +9 10⁵ C

Answer:

- time t taken for car to travel is 64.57 s

- distance travelled between A and B is 1.4887 km

Explanation:

Given the data in the question;

[tex]U_{BC}[/tex] =  83 km/h = ( 83×1000 / 60×60) =  23.0555 m/s

[tex]U_{CD}[/tex] = 41 km/h = ( 41×1000 / 60×60) =  11.3888 m/s

now, we calculate the acceleration;

a =  (  [tex]U_{BC}[/tex] -  [tex]U_{CD}[/tex] ) / t

we substitute

a =  ( 23.0555 -  11.3888 ) / 4.4

a = 11.6667 / 4.4

a = 2.6515 m/s²

Now equation for displacement from BC

[tex]S_{BC}[/tex] =  [tex]U_{BC}[/tex]t + 1/2.at²

we substitute

[tex]S_{BC}[/tex] =  23.0555×4.4 + 1/2×a×(4.4)²

we substitute -2.6515m/s² for a

[tex]S_{BC}[/tex] =  23.0555×4.4 + 1/2×(-2.6515)×(4.4)²

= 101.4442 - 25.6665

[tex]S_{BC}[/tex] = 75.7792 m

Now, for total distance covered = 2.3km = ( 2.3×1000) = 2300m

so

[tex]S_{AB}[/tex] +  [tex]S_{BC}[/tex] +  [tex]S_{CD}[/tex]  =  2300 m

we substitute substitute

[tex]S_{AB}[/tex] +  75.7792 m +  [tex]S_{CD}[/tex]  =  2300 m

[tex]S_{AB}[/tex] +  [tex]S_{CD}[/tex] = 2300 - 75.7792

[tex]S_{AB}[/tex] +  [tex]S_{CD}[/tex]  = 2224.2208 m

so we substitute 23.0555t for [tex]S_{AB}[/tex]  and 11.3888t for  [tex]S_{CD}[/tex]  

23.0555t + 11.3888t  = 2224.2208

34.4443t = 2224.2208

t = 2224.2208 / 34.4443

t = 64.57 s

Therefore, time t taken for car to travel is 64.57 s

Distance Between A to B

[tex]S_{AB}[/tex]  = t ×  [tex]U_{AB}[/tex]

we substitute

[tex]S_{AB}[/tex]  = 64.57 s × 23.0555

[tex]S_{AB}[/tex]  = 1488.69 m

[tex]S_{AB}[/tex]  = 1.4887 km

Therefore, distance travelled between A and B is 1.4887 km

The potential energy of the football with mass 2.5 kg which is lifted up to the top of a cliff is 4410 Joules.

Potential energy is the stored energy which depends upon the relative position of the various parts of a system of objects. Potential energy is the product of mass of the object, acceleration due to gravity, and the height. The SI unit of potential energy is Joule (J).

PE = m × g × h

PE = Potential energy,

m = mass of the object,

g = acceleration due to gravity,

h = height

PE = 2.5 × 9.8 × 180

PE = 4410 Joules

Therefore, the potential energy of the football is 4410 Joules.

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Answer:

The second option - the papers are acted upon by an unbalanced force from the fan.

Explanation:

Answer:

16

Explanation:

[tex] \frac{1}{2} \times 2 \times {4}^{2} = 16[/tex]

Answer:

Explanation:

We shall first calculate the velocity at height h = 575 m .

acceleration a = 2.2 m /s²

v² = u² + 2 a s

u is initial velocity , v is final velocity , s is height achieved

v² = 0 + 2 x 2.2 x 575

v = 50.3 m /s

After 575 m , rocket moves under free fall so g will act on it downwards

If it travels further by height H

from the relation

v² = u² - 2 g H

v = 0 , u = 50.3 m /s

H = ?

0 = 50.3² - 2 x 9.8 H

H = 129.08 m

Total height attained by rocket

= 575 + 129.08

= 704.08 m .

Answer:

a. 2500 cm³.

b. 2.5 litres.

Explanation:

Given the following data:

Density = 0.8g/cm³

Mass = 2000g

To find the volume of the petrol;

Density can be defined as mass all over the volume of an object.

Simply stated, density is mass per unit volume of an object.

Mathematically, density is given by the equation;

[tex]Density = \frac{mass}{volume}[/tex]

Making volume the subject of formula, we have;

[tex]Volume = \frac{mass}{density}[/tex]

Substituting into the equation, we have:

[tex]Volume = \frac{2000}{0.8}[/tex]

Volume = 2500 cm³

a. The volume of the petrol in the can in cubic centimeters (cm³) is 2000.

b. The volume of the petrol in the can in litres;

1000 cm³ = 1 litre

2500 cm³ = x litres

Cross-multiplying, we have;

1000x = 2500

x = 2500/1000

x = 2.5 litres.

Therefore, the volume of the petrol in the can in litres is 2.5.

Answer:

this is about momentum p=mv

A, the elephant has more mass

Explanation:

C. meter per second squared

B. meter per second

A. meter

Answer:

b. meters per second

c.meters per second squared

c.meters

Explanation:

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Answer:

A ball hits the ground and the ground pushes up on it

Explanation:

Newton's third law basically states that for every action, there's a reaction.

a ball hitting the ground would be the action. the ground pushing up on it with the same force is the reaction.

Hope this Helps!!! :)

Answer:

a=1m/s^2

Explanation:

1÷1÷1=1m/s^2

Answer:

Please find attached pdf

Explanation:

Answer:  a) weight on Earth = mass of the object and gravity n the Earth. = 65*10 = 650 kg.

Explanation:

An astronaut has a mass of 65 kg on Earth where the gravitational field strength is 10 N kg A calculate the astronaut's weight on Earth

          hope this helps :)

Answer:

650

Explanation:

use the equation

weight = gm

Answer:

Acceleration due to gravity, g = 2.68m/s²

Explanation:

Given the following data;

Period = 5.14s

Length = 0.25m

To find acceleration due to gravity, g;

[tex] Period, T = 2 \pi \sqrt {lg} [/tex]

Substituting into the equation, we have;

[tex] 5.14 = 2*3.142 \sqrt {0.25g} [/tex]

[tex] 5.14 = 6.284 \sqrt {0.25g} [/tex]

[tex] \frac {5.14}{6.284} = \sqrt {0.25g} [/tex]

[tex] 0.8180 = \sqrt {0.25g} [/tex]

Taking the square of both sides

[tex] 0.8180^{2} = 0.25g [/tex]

[tex] 0.6691 = 0.25*g[/tex]

[tex] g = \frac {0.6691}{0.25} [/tex]

Acceleration due to gravity, g = 2.68m/s²

so we will use Newton's gravitational law :

gravitational acceleration = G*m/r^2  

G is the gravitational constant = G = 6.673×10^-11 N m^2 kg^-2  

after substitution :  

6.673×10^-11 * m / (1.15 x 10^6)^2 = 0.61

= 5.04575*10^-23 * m = 0.61

dividing over 5.04575*10^-23 :

m = 1.20894*10^22 kg

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Answer:

it has a higher amplitude than that of the original waves

Explanation:

trust me its right

The two waves combined with constructive interference have a higher amplitude than the original wave.

High energy waves are characterized by high amplitude. Low energy waves are characterized by their small amplitude. As explained in Lesson 2, wave amplitude is the maximum amount of particle movement on a medium from a resting position.

Superposition is a combination of two waves in the same place. Constructive interference occurs when two identical waves interfere in phase. Destructive interference occurs when two identical waves are exactly out of phase and overlap.

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Answer: 4

Explanation: because my pet monkey said it was

Answer:

θ = 225 rad

Explanation:

given data

angle = 25 rad

to find out

angular velocity after 3t?

solution

let angular acceleration α in t

θ = ω × t + 0.5 × α × t²        ........................1

here ω  = 0 (initial velocity )

so put this value here

25 = 0 + 0.5 × α × t²             ..........................2

α = 25 ÷ (0.5 t²)

α = 50 ÷ t²                            .........................3

now here we take in 3t

θ = ω × 3t + 0.5 × α × (3t)²

for ω  = 0

θ = 0 + 0.5 × α × 9t²

now put value in eq 2

so

θ = (0.5) ×  (50 ÷ t²) ×  (3t)²

θ = 25 × 9

θ = 225 rad

Answer:

38.09 m

Explanation:

We'll begin by calculating the distance travelled by the car during the reaction time. This can be obtained as follow:

Reaction time (tᵣ) = 0.404 s

Initial velocity (u) = 22.8 m/s,

Distance travelled during the reaction time (sᵣ) =?

sᵣ = utᵣ

sᵣ = 22.8 × 0.404

sᵣ = 9.21 m

Next, we shall determine the distance travelled by the car when the brake was applied. This can be obtained as follow:

Initial velocity (u) = 22.8 m/s

Acceleration (a) = –9 m/s² (since the car is decelerating)

Final velocity (v) = 0 m/s

Distance travelled when the brake was applied (s₆) =?

v² = u² + 2as₆

0² = 22.8² + (2 × –9 × s₆)

0 = 519.84 – 18s₆

Collect like terms

0 – 519.84 = –18s₆

–519.84 = –18s₆

Divide both side by –18

s₆ = –519.84 / –18

s₆ = 28.88 m

Finally, we shall determine the stopping distance of the car, as measured from the point where the driver first notices the red light. This can be obtained as follow:

Distance travelled during the reaction time (sᵣ) = 9.21 m

Distance travelled when the brake was applied (s₆) = 28.88 m

Stopping distance =?

Stopping distance = sᵣ + s₆

Stopping distance = 9.21 + 28.88

Stopping distance = 38.09 m

Answer: light from the object

Explanation:

When light is reflected off an object like a lamp or a door is travels in a straight line but in a new direction so if the light enters our eyes we will see the object because our eyes can detect light

Answer:

84.1 m

Explanation:

Given :

The distance from the ship to submarine :

20 + y

Using Pythagoras :

Tan θ = opposite / Adjacent

Tan θ = 20 / 12

12 tan θ = 20

θ = tan^-1(20/12)

20

θ = 59.036°

The angle phi;

n1sinθ1 = n2sin θ

Sin 59.036 = 1.33 * sin phi

Sin phi = sinsin(59.04) ÷1.33

0.8574907 = 1.33 * sin phi

Sin phi = 0.8574907 / 1.33

Sin phi = 0.6447298

phi = sin(0.6447298

Phi = 40.15°

From Pythagoras :

y = 76tan40.15°

y = 76 * 0.8435707

y = 64.11

20 + y

20 + 64.11 = 84.11

Answer:

C. zero for both

Explanation:

In case of solid metal sphere , when it is given any charge , all the charges are transferred on the surface and within surface no charge exists . In case of hollow metal sphere , all charges reside on surface . In this way , in both solid and hollow sphere , all charge reside on the surface and no charge resides inside it . Hence due to absence of any charge inside  , there is no electric field inside the sphere in both the cases .  

Hence in both the case electric field is zero .

option C is correct .

Answer:

here

Explanation:

Examples of insulators include plastics, Styrofoam, paper, rubber, glass and dry air.

Examples of conductors include metals, aqueous solutions of salts

Answer:

c

Explanation:

Answer:

2m/s

Explanation:

v=f×wavelength

v=2×1

=2m/s

Answer:

yes

Explanation:

Answer:yeeeeeeeeeeeeeeeeeeeeeees

Explanation:

Answer:

it 1.5 meters

Explanation:

if u could put the option number it will be cool and hope it help and if it doesnt am really sorry ;)

Answer:

     l = 0.548 m

Explanation:

For this exercise we compensate by finding the speed of the car

         p = m v

         v = p / m

         v = 0.58 / 0.2

         v = 2.9 m / s

this is how fast you get to the ramp, let's use conservation of energy

starting point. Lowest point

         Em₀ = K = ½ m v²

final point. Point where it stops on the ramp

         [tex]Em_{f}[/tex] = U = m g h

  mechanical energy is conserved

          Em₀ = Em_{f}

          ½ m v² = m g h

           h = [tex]\frac{m v^2}{2 g}[/tex]

let's calculate

          h = [tex]\frac{0.2 \ 2.9^2}{2 \ 9.8}[/tex]

          h = 0.0858 m

to find the distance that e travels on the ramp let's use trigonometry, we look for the angle

          tan θ = y / x

          tan θ = 12/75 = 0.16

          θ = tan⁻¹ 0.16

          θ = 9º

therefore

           sin 9 = h / l

           l = h / sin 9

           l = 0.0858 / sin 9

           l = 0.548 m