Question 1 of 5
In which way are electromagnetic waves different from mechanical waves?

Answer:

14 m/s

Explanation:

The following data were obtained from the question:

Mass = 50 kg

Initial velocity (u) = 0 m/s

Height (h) = 10 m

Acceleration due to gravity (g) = 9.8 m/s²

Final velocity (v) =?

The velocity (v) with which the person hit the water can be obtained as shown below:

v² = u² + 2gh

v² = 0² + (2 × 9.8 × 10)

v² = 0 + 196

v² = 196

Take the square root of both side

v = √196

v = 14 m/s

Therefore, he will hit the water with a speed of 14 m/s

Answer:

Mutualism - Bee to flower. Bee eats - flower reproduces

Commensalism - Tree Frog to plant or tree. Frog uses plant for protection.

Parasitism - Flea or tick to host. Parasite feeds off host.

Explanation:

Competition - relationship between organisms that strive for same resources. intraspecific and interspecific. ex) two males competing for mates.

predation - one organism kills and consumes another. wolf hunting moose, cat hunting mouse. venus fly trap killing insect

Answer:

1,439,283.10Watts

Explanation:

The question looks incomplete. We need the workdone to get the power generated.

Power = Workdone/Time

Power = Force * Distance/Time

Let

Force = 24340 Newtons

Distance covered = 259meters

Given

Time = 4.38secs

Required

Power generated

Substitute the given parameters into the formula;

Power generated = 24340*259/4.38

Power generated = 6,304,060/4.38

Power generated = 1,439,283.10

Hence the amount of power generated is 1,439,283.10Watts

Answer:

(a) the change in length of the silk is 0.001585 cm

(b) the maximum weight that a single thread can support is 17.67 N

Explanation:

Given;

mass of the spider, m = 0.50 g = 0.5 x 10⁻³ kg

length of the silk, L = 12 mm = 0.012 m

diameter of the silk, d = 0.15 mm

radius of the silk, r = d / 2 = 0.075 mm = 0.075 x 10⁻³ m

The cross sectional area of the silk;

A = πr² = π(0.075 x 10⁻³)²

A = 1.767 x 10⁻⁸ m²

The Young's modulus of elasticity of spider-silk is given by;

2.1 Gpa = 2.1 x 10⁹ N/m²

(a)

Apply  Young's modulus of elasticity equation to determine the change in length of the silk;

[tex]E = \frac{FL}{Ax} = \frac{F_gL}{Ax}\\\\x = \frac{F_gL}{AE}\\\\x = \frac{(0.5*10^{-3}*9.8)(0.12)}{(2.1*10^9)(1.767*10^{-8})}\\\\x = 1.585*10^{-5} \ m\\\\x = 0.01585 \ mm[/tex]

[tex]x = 0.001585 \ cm[/tex]

(b)

the maximum weight that a single thread can support is given by;

[tex]T_{tensile \ strength} = \frac{F_{max}}{A}[/tex]

The tensile strength of spider-silk is given by 1 Gpa = 1 x 10⁹ N/m²

[tex]F_{max} = T_{tensile \ strength}*A\\\\F_{max} = (1*10^9)(1.767*10^{-8})\\\\F_{max} = 17.67 \ N[/tex]

Answer:

Explanation:

Step one:

given data

initial velocity u= 0m/s since the ball is at rest

time of contact t= 0.3s

final velocity v=15m/s

Required

acceleration a

from the first law of motion

v=u+at

substitute our given data

15=0+a*0.3

15=0.3a

divide both sides by 0.3

a=15/0.3

a=50m/s

The average acceleration is 50m/s^2

Answer:

9.75 m/s²

Explanation:

Given that,

v= 14.5 m/s

u = 27.5 m/s

s = 28 m

a = ?

v² = u² -2as [ minus sign due to deceleration]

14.5² = 27.5² - 2 × a × 28

210.25 - 756.25 = -56a

-56a = -546

a = 9.75 m/s²

Answer:

Mercury is the only one in liquid state at room temperature. It's used in thermometers because it has high coefficient of expansion.

Explanation:

please mark me brainlist

Mercury is the only one in liquid state at room temperature. It's used in thermometers because it has high coefficient of expansion. they still use mercury even though it is the poorest conductor of heat.

Answer:

Explanation:

No da esta aplicación y si para que usted vella

Answer:Highest frequency  =618.89Hz

Lowest frequency=582.22Hz

Explanation:

 The linear velocity of a sound generator  is related to angular velocity and is given as

Vs = rω  where

r = the radius of circular path = 1.0 m

ω is the angular velocity of the sound generator. = 100 rpm

1 rev/min = 0.10472 rad/s

100rpm =10.472 rad/ s

Vs = rω

= 1m x 10.472rad/ s=  10.472m/s

A) Highest frequency  heard by a student in the classroom = Maximum frequency. Using the Doppler effect formulae,

f max = (v/ v-vs) fs

Where , v is the speed of the sound in air at 20 degrees celcius =

343 metres per second

vs is the linear velocity of the sound generator=10.472m/s

fs is the frequency of the sound generator= 600 Hz

f max = (343/ 343 - 10.472) x 600

=343/332.528) x600

=618.89Hz

B) Lowest frequency  heard by a student in the classroom = Minimum frequency

f min = (v/ v+vs) fs

(343/ 343 + 10.472) x 600

=343/353.472) x 600

=582.22hz

Answer:

Satellites may move north to south, or south to north, or west to east, but never from east to west. When satellites are launched, they always head eastward to take advantage of the Earth's rotation, going more than 1,000 miles per hour near the equator. This saves a lot of fuel.

Explanation:

yooooooooo

Answer:

(a) (i) The orientation of the coil which gives maximum torque on the coil is 90⁰

(a)(ii) The maximum torque is 0.132 Nm

(b) The orientation of the coil is 45⁰

Explanation:

Given;

diameter of the circular wire, d = 8.6 cm = 0.086 m

radius of the wire, r = d /2 = 0.043 m

number of turns, N = 15 turns

magnetic field, B = 0.56 T

The torque on the wire is given by;

τ = NIABsinθ

where;

θ is the orientation of the wire

(a) maximum torque occurs when the orientation of the wire is at 90⁰

The maximum torque is given by;

τ = NIABsin(90⁰)

τ = NIAB

τ = (15)(2.7)(π x 0.043²)(0.56)

τ = 0.132 Nm

(b)

71% of 0.132 = 0.71 x 0.132 = 0.0937 Nm

[tex]\tau = NIAB sin\theta\\\\sin\theta = \frac{\tau}{NIAB }\\\\ sin\theta = \frac{0.0937}{(15)(2.7)(\pi *0.043^2)(0.56)} \\\\sin\theta = 0.7111\\\\\theta = sin^{-1}(0.7111)\\\\\theta =45.32\\\\\theta = 45^0[/tex]

Answer:

ok tbh i didnt cry when neji died but now when i think about it i just cry thinking about how himawari could have met her uncle

Explanation:

Answer:

a

[tex]\lambda = 3.68 *10^{-36} \ m[/tex]

b

[tex]\lambda_p = 1.28*10^{-14} \ m[/tex]

Explanation:

From the question we are told that

   The mass of the person is  [tex]m = 180 \ kg[/tex]

    The speed of the person is  [tex]v = 1 \ m/s[/tex]

    The energy of the proton is  [tex]E_ p = 5 MeV = 5 *10^{6} eV = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \ J[/tex]

Generally the de Broglie wavelength is mathematically represented as

      [tex]\lambda = \frac{h}{m * v }[/tex]

Here  h is the Planck constant with the value

      [tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]

So  

     [tex]\lambda = \frac{6.62607015 * 10^{-34}}{ 180 * 1 }[/tex]

=> [tex]\lambda = 3.68 *10^{-36} \ m[/tex]

Generally the energy of the proton is mathematically represented as

         [tex]E_p = \frac{1}{2} * m_p * v^2_p[/tex]

Here [tex]m_p[/tex]  is the mass of proton with value  [tex]m_p = 1.67 *10^{-27} \ kg[/tex]

=>     [tex]8.0*10^{-13} = \frac{1}{2} * 1.67 *10^{-27} * v^2[/tex]

=>   [tex]v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }[/tex]

=>   [tex]v = 3.09529 *10^{7} \ m/s[/tex]

So

        [tex]\lambda_p = \frac{h}{m_p * v_p }[/tex]

so    [tex]\lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }[/tex]

=>     [tex]\lambda_p = 1.28*10^{-14} \ m[/tex]

Answer:

106.24 kJ.

Explanation:

Given that,

Mass of sample of sand, m = 8 kg

Specific heat of sand, c = 664 J/kg-°C

The temperature changes from 20° C to 40° C. We need to find the change in thermal energy. It is given by :

[tex]Q=mc\Delta T\\\\Q=8\times 664(40-20)\\\\=106240\ J\\\\=106.24\ kJ[/tex]

So, the change in thermal energy is 106.24 kJ.

Answer:

true

Explanation:

Im in k12 and I got an 100%

Answer:

B. 1275 kg*m/s

Explanation:

I = F(deltaT) = (deltaP) = mv2- mv1

Therefore,

I = mv2-mv1

m = 850 kg

v2 = 5 m/s

v1 = 3.5 m/s

I = (850)(5)-(850)(3.5)

I = 1275 kg* m/s

Answer:

Human activity has fundamentally changed our planet. We live on every continent and have directly affected at least 83% of the planet’s viable land surface. Our influence has impacted everything from the makeup of ecosystems to the geochemistry of Earth, from the atmosphere to the ocean. Many scientists define this time in the planet’s history by the scale of human influence, and label it as a new geological epoch called the Anthropocene.

Explanation:

Answer:

Explanation:

The centripetal acceleration is expressed as;

a = v²/r

a is the acceleration = 50m/s²

v is the velocity = 10m/s

r is the radius

To get the radius

r = v²/a

r = 10²/50

r = 100/50

r = 2m

Hence its radius is 2m

Answer:

B 0.4 have a nice day and hiiii

Answer:

F = - 3.56*10⁵ N

Explanation:

To attempt this question, we use the formula for the relationship between momentum and the amount of movement.

I = F t = Δp

Next, we try to find the time that the average speed in the contact is constant (v = 600m / s), so we say

v = d / t

t = d / v

Given that

m = 26 g = 26 10⁻³ kg

d = 50 mm = 50 10⁻³ m

t = d/v

t = 50 10⁻³ / 600

t = 8.33 10⁻⁵ s

F t = m v - m v₀

This is so, because the bullet bounces the speed sign after the crash is negative

F = m (v-vo) / t

F = 26*10⁻³ (-500 - 640) / 8.33*10⁻⁵

F = - 3.56*10⁵ N

The negative sign is as a result of the force exerted against the bullet

Answer: it causes some parts of the earth to get more radiation than others.

Explanation: earth rotates around the sun on a tilted axis so the Rays of the sun cause earth to have more radiation than it needs.

Answer:

0.546 [tex]\hat k[/tex]

Explanation:

From the given information:

The force on a given current-carrying conductor is:

[tex]F = I ( \L \limits ^ {\to } \times B ^{\to})\\ \\ dF = I(dL\limits ^ {\to } \times B ^{\to})[/tex]

where the length usually in negative (x) direction can be computed as

[tex]\L ^ {\to } = -x\hat i \\dL\limits ^ {\to }- dx\hat i[/tex]

Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:

[tex]\int dF = \int ^3_1 I ( dL^{\to} \times B ^{\to})[/tex]

[tex]F = I \int^3_1 ( -dx \hat i ) \times ( 4.0 \hat i + 9.0 \ x^2 \hat j)[/tex]

[tex]F = I \int^3_1 - 9.0x^2 \ dx \hat k[/tex]

[tex]F = I (9.0) \bigg [\dfrac{x^3}{3} \bigg ] ^3_1 \hat k[/tex]

[tex]F = I (9.0) \bigg [\dfrac{3^3}{3} - \dfrac{1^3}{3} \bigg ] \hat k[/tex]

where;

current I = 7.0 A

[tex]F = (7.0 \ A) (9.0) \bigg [\dfrac{27}{3} - \dfrac{1}{3} \bigg ] \hat k[/tex]

[tex]F = (7.0 \ A) (9.0) \bigg [\dfrac{26}{3} \bigg ] \hat k[/tex]

F = 546 × 10⁻³ T/mT [tex]\hat k[/tex]

F = 0.546 [tex]\hat k[/tex]

Answer:

7.62

Explanation:

because you have to divide 32/4.2

and can you do a friend request so i can accept it

Answer:

a

 [tex]d = 0.0223 \ m[/tex]

b

[tex]d = 0.0481 \ m[/tex]  

Explanation:

From the question we are told that

   The maximum allowable shear stress is  [tex]\sigma = 70 MPa = 70 *10^{6} \ Pa[/tex]

    The power is  [tex]P = 40 \ kW = 40 *10^{3} \ W[/tex]

considering question a

   The shaft speed is given as [tex]v = 2500\ rev/min[/tex]

Generally the torque experienced by the shaft is mathematically represented as

       [tex]\tau = \frac{ 9.55 * P}{v}[/tex]

=>    [tex]\tau = \frac{ 9.55 * 40 *10^{3}}{ 2500}[/tex]

=>    [tex]\tau = 152.8 \ N \cdot m[/tex]

Generally the maximum torque experienced by the shaft is mathematically represented as

          [tex]\tau_m = \frac{2 \tau }{ \pi r^2 }[/tex]

Generally diameter  =  2 *  radius (r)

So

        [tex]\tau_m = \frac{2 \tau }{ \pi 4 d^2 }[/tex]

Generally the maximum allowable shear stress is mathematically represented as

                [tex]\sigma = \frac{2 \tau }{ \pi 4 d^2 } * \frac{32}{d}[/tex]

=>             [tex]\sigma = \frac{16 \tau }{ \pi d^3}[/tex]

=>            [tex]d = \sqrt[3]{\frac{16 * \tau }{ \pi \sigma } }[/tex]

=>          [tex]d = \sqrt[3]{\frac{16 * 152.8 }{ \pi * 70 *10^{6} } }[/tex]

=>          [tex]d = 0.0223 \ m[/tex]

considering question b

   The shaft speed is given as [tex]v = 250\ rev/min[/tex]

Generally the torque experienced by the shaft is mathematically represented as

       [tex]\tau = \frac{ 9.55 * 40 *10^{3}}{250 }[/tex]

=>    [tex]\tau = 1528 \ N \cdot m[/tex]

Generally the shaft diameter is mathematically represented as

   [tex]d = \sqrt[3]{\frac{16 * \tau }{ \pi \sigma } }[/tex]

=>[tex]d = \sqrt[3]{\frac{16 * 1528 }{ 3.142 * 70 *10^{6} } }[/tex]  

=>[tex]d = 0.0481 \ m[/tex]  

Answer:

Nope

Explanation:

Answer:

answer is 61.8 m/s; 32.9

l am not sure