A scientist who studies fossils of ancient life forms .O ornithologist O Paleontologist O Ichthyologist O Marine Biologist .

Hurry !! First one to answer gets points !

Answer:

B. Smaller than

Explanation:

This question is from the Doppler effect. As the object which is in motion goes off from the other, there's a reduction in the frequency. This is due to the fact that successive soundwave get to be longer. So that the pitch will then be lowered. When the person observing moves towards what is making the sound, each soundwave that follows gets faster than the previous.

Answer:

22.1 years

Explanation:

Since the current in the wire is I = nevA where n = electron density = 8.50 × 10²⁸ electrons/cm³ × 10⁶ cm³/m³= 8.50 × 10³⁴ electrons/m³, e = electron charge = 1.602 × 10⁻¹⁹ C, v = drift velocity of electrons and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.00 cm = 2 × 10⁻² m

Making v subject of the formula, we have

v = I/neA

So, v = I/neπd²/4

v = 4I/neπd²

Since I = 1,015 A, substituting the values of the other variables into the equation, we have

v = 4I/neπd²

v = 4(1,015 A)/[8.50 × 10³⁴ electrons/m³ × 1.602 × 10⁻¹⁹ C × π ×(2 × 10⁻² m)²]

v = 4(1,015 A)/[8.50 × 10³⁴ electrons/m³ × 1.602 × 10⁻¹⁹ C × π × 4 × 10⁻⁴ m²]

v = (1,015 A)/[42.779 × 10¹¹ electronsC/m]

v = 23.73 × 10⁻¹¹ m/s

v = 2.373 × 10⁻¹⁰ m/s

Since distance d = speed, v × time, t

d = vt

So, the time it takes one electron to travel the full length of the cable is t = d/v

Since d = distance moved by free charge = length of transmission line = 165 km = 165 × 10³ m and v = drift velocity of charge = 2.373 × 10⁻¹⁰ m/s

t = 165 × 10³ m/2.373 × 10⁻¹⁰ m/s

t = 69.54 × 10⁷ s

t = 6.954 × 10⁸ s

Since we have 365 × 24 hr/day × 60 min/hr × 60 s/min = 31536000 s in a year = 3.1536 × 10⁷ s

So,  6.954 × 10⁸ s =  6.954 × 10⁸ s × 1yr/3.1536 × 10⁷ s = 2.21 × 10 yrs = 22.1 years

It will take one electron 22.1 years to travel the full length of the cable

Answer:

(a) 8.57 x 10^8 Hz

(b) 23.3 cm

Explanation:

Wavelength = 35 cm = 0.35 m

speed =3 x10^8 m/s

Let the frequency is f.

(a) The relation is

speed  = frequency x wavelength

3 x 10^8 = 0.35 x f

f = 8.57 x 10^8 Hz

(b) refractive index of glass  is 1.5

The relation for the refractive index and the wavelength is

wavelength in glass= wavelength in air/ refractive index.

Wavelength in glass= 35/1.5 = 23.3 cm

Answer:

300 meters

Explanation:

a= 2m/s^2

t= 5 min

Convert into seconds, 5*60= 300seconds

v0= 0

x0=0

use x-x0= v0t + 1/2at^2

plug in values

x= 1/2*2*(300)

Solve

x= 300 meters

Answer: a rippling fountain

Explanation: diffuse reflection happens on rough surfaces, so using the process of elimination, that leaves us with b, a rippling fountain (I also just took this test I'm pretty sure I'm right)

Explanation:

Given that,

Amplitude, A = 2.5 nm

Wavelength,[tex]\lambda=1.8\ m[/tex]

The speed of the wave, v = 36 m/s

At time t = 0 the left end of the string has its maximum upward displacement.

(a) Let f is the frequency. So,

[tex]f=\dfrac{v}{\lambda}\\\\f=\dfrac{36}{1.8}\\\\f=20\ Hz[/tex]

(b) Angular frequency of the wave,

[tex]\omega=2\pi f\\\\=2\pi \times 20\\\\=125.7\ rad/s[/tex]

(c) The wave number of the wave[tex]=\dfrac{1}{\lambda}[/tex]

[tex]=\dfrac{1}{1.8}\\\\=0.56\ m^{-1}[/tex]

Answer:

I DON'T UNDERSTAND

Explanation:

GUESS A MISUNDERSTANDING PLZ PUT A UNDERSTANDABLE QUESTION.

Answer:

3.44 rad

Explanation:

The rotational kinetic energy change of the disk is given by ΔK = 1/2I(ω² - ω₀²) where I = rotational inertia of solid sphere = MR²/2 where m = mass of solid disk = 4 kg and R = radius of solid disk = 4 m, ω₀ = initial angular speed of disk = 0 rad/s (since it starts from rest) and ω = final angular speed of disk

Since the kinetic energy is increasing at a rate of 21 J/s, the increase in kinetic energy in 3.3 s is  ΔK = 21 J/s × 3.3 s = 69.3 J

So, ΔK = 1/2I(ω² - ω₀²)

Since ω₀ = 0 rad/s

ΔK = 1/2I(ω² - 0)

ΔK = 1/2Iω²

ΔK = 1/2(MR²/2)ω²

ΔK = MR²ω²/4

ω² = (4ΔK/MR²)

ω = √(4ΔK/MR²)

ω = 2√(ΔK/MR²)

Substituting the values of the variables into the equation, we have

ω = 2√(ΔK/MR²)

ω = 2√(69.3 J/( 4 kg × (4 m)²))

ω = 2√(69.3 J/[ 4 kg × 16 m²])

ω = 2√(69.3 J/64 kgm²)

ω = 2√(1.083 J/kgm²)

ω = 2 × 1.041 rad/s

ω = 2.082 rad/s

The angular displacement θ is gotten from

θ = ω₀t + 1/2αt² where ω₀ = initial angular speed = 0 rad/s (since it starts from rest), t = time of rotation = 3.3 s and α = angular acceleration = (ω - ω₀)/t = (2.082 rad/s - 0 rad/s)/3.3 s = 2.082 rad/s ÷ 3.3 s = 0.631 rad/s²

Substituting the values of the variables into the equation, we have

θ = ω₀t + 1/2αt²

θ = 0 rad/s × 3.3 s + 1/2 × 0.631 rad/s² (3.3 s)²

θ = 0 rad + 1/2 × 0.631 rad/s² × 10.89 s²

θ = 1/2 × 6.87159 rad

θ = 3.436 rad

θ ≅ 3.44 rad

Answer: 2.4 km/hr

Explanation:

Distance = 600m

Time= 15 minutes = 15 x 60 second/minute = 900 seconds

Speed =  [tex]\frac{Distance}{Time}[/tex]  =  [tex]\frac{600}{900}[/tex]  = [tex]\frac{2}{3}[/tex] m/sec

⇒ [tex]\frac{2}{3}[/tex] x [tex]\frac{18}{5}[/tex] = 2.4 km/hr (1 m/sec = 3.6 km/hr)

Answer:

A will attract

B will repare

Answer:

answer here

Explanation:

the unit of work is fundamental unit because it doesn't depend on other units.

__________________

Thx

Answer:

The SI unit of work is joule (J)

Explanation:

Joule is a derived unit. ∴ unit of work is a derived unit

Answer:

a=10m/s^2

Explanation:

acceleration= final velocity+ initial velocity/time taken

v-u/t=a

100-0/5=a

100/5=a

a=20m/s^2

case2

100-0/10=a

100/10=a

a=10m/s^2

Don't forget to write the units.

Hope this helps

please mark me as brainliest.

Hi there!

[tex]\large\boxed{\approx 35.29 C}[/tex]

Use the following formula:

E = F / C, where:

E = electric field (N/C)

F = force (N)

C = Charge (C)

Thus:

6.8 × 10⁻⁵ = 2.4 × 10⁻³ / C

Isolate for C:

C = 2.4 × 10⁻³  / 6.8 × 10⁻⁵

Solve:

≈ 35.29 C

Explanation:

Forces on Block A:

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass [tex]m_A[/tex] as

[tex]x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)[/tex]

[tex]y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)[/tex]

Substituting (2) into (1), we get

[tex]\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)[/tex]

where [tex]f_N= \mu_kN[/tex], the frictional force on [tex]m_A.[/tex] Set this aside for now and let's look at the forces on [tex]m_B[/tex]

Forces on Block B:

Let the x-axis be (+) up along the inclined plane. We can write the forces on [tex]m_B[/tex] as

[tex]x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)[/tex]

[tex]y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)[/tex]

From (5), we can solve for N as

[tex]N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)[/tex]

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension T is given by

[tex]T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)[/tex]

Substituting (7) into (4) we get

[tex]m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba[/tex]

Collecting similar terms together, we get

[tex](m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag[/tex]

or

[tex]a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)[/tex]

Putting in the numbers, we find that [tex]a = 1.4\:\text{m/s}[/tex]. To find the tension T, put the value for the acceleration into (7) and we'll get [tex]T = 21.3\:\text{N}[/tex]. To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get [tex]N = 50.9\:\text{N}[/tex]

efficiency=work output/work input×100

since it exhausts(use up)3000j of heat that's the work input and the 1500j is the work input

efficiency=1500/3000×100

=50%

Answer:

D

Explanation:

This is a great way to manage health.

A would be avoiding everything which isnt good.

B. would be emotionally draining and damaging to bottle feelings and ignore them.

C. is unhealthy to not exercise and eat food while doing nothing.

Answer:

a)  fr = 266.92 N,   fy = 1300 N,  b)    μ = 0.36

Explanation:

a) This is a balancing act.

Let's write the rotational equilibrium relations, where the turning point is the bottom of the ladder and the counterclockwise rotations are positive

             -w x - W x₂ + R y = 0         (1)

usemso trigonometry to find distances

            cos 60.08 = x / 7.5

            x = 7.5 cos 60.08

            x = 3.74 m

fireman

           cos 60.08 = x₂ / 4

           x2 = 4 cos 60

           x2 = 2 m

wall support

           sin 60.08 = y / 15

           y = 15 are 60.08

           y = 13 m

we substitute in equation 1

           R y = w x + W x2

            R = (w x + W x2) / y

            R = (500 3.74 +800 2) / 13

            R = 266.92 N

now let's write the expressions for the translational equilibrium

X axis

           R -fr = 0

           R = fr

           fr = 266.92 N

Y Axis  

           Fy - w-W = 0

           fy = 500 + 800

           fy = 1300 N

b) ask the friction coefficient

the firefighter's distance is

          cos 60.08 = x₃ / 9.00

          x₃ = 9 cos 60

          x₃ = 5.28 m

from equation 1

          R = (w x + W x₃) / y

          R = 500 3.74 + 800 5.28) / 13

          R = 468.769 N

we saw that

          fr = R = 468.769

The expression for the friction force is

          fr = μ N

in this case the normal is the ratio to pesos

        N = Fy

       N = 1300 N

        μ N = fr

        μ = fr / N

        μ = 468,769 / 1300

         μ = 0.36

Answer:

[tex]K.E_{max}=0.8973J[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=1.64kg[/tex]

Equation of Mass

[tex]X=0.33cos(3.17t)[/tex]...1

Generally equation for distance X is

[tex]X=Acos(\omega t)[/tex]...2

Therefore comparing equation

Angular Velocity [tex]\omega=3.17rad/s[/tex]

Amplitude A=0.33

Generally the equation for Max speed is mathematically given by

[tex]V_{max}=A\omega[/tex]

[tex]V_{max}=0.33*3.17[/tex]

[tex]V_{max}=1.0461m/s[/tex]

Therefore

[tex]K.E_{max}=0.5mv^2[/tex]

[tex]K.E_{max}=0.5*1.64*(1.0461)^2[/tex]

[tex]K.E_{max}=0.8973J[/tex]

The question is incomplete. The complete question is :

A high-speed bullet train accelerates and decelerates at the rate of 4 ft/s^2. Its maximum cruising speed is 90 mi/h. What is the maximum distance the train can travel if it accelerates from rest until it reaches its cruising speed and then runs at that speed for 15 minutes?

Solution :

Given :

Speed of the bullet train, v = 90 mi/h

                                            = [tex]$90 \times \frac{5280}{3600}$[/tex]

                                            = 132 ft/s

Time = 15 minutes

        = 15 x 60

        = 900 s

Acceleration from rest,

[tex]$a(t) = 4 \ ft/s^2$[/tex]

[tex]$v(t) = 4t + C$[/tex]

Since, v(0) = 0, then C = 0, so velocity is

v(t) = 4t ft/s

Then find the position function,

[tex]$s(t) = \frac{4}{2}t^2 + C$[/tex]

      [tex]$=2t^2+C$[/tex]

It is at position 0 when t = 0, so C = 0, and the final position function for only the time it is accelerating is :

[tex]$s(t) = 2t^2$[/tex]

Time to get maximum cruising speed is :

4t = 132

t = 33 s

Distance travelled (at cruising speed) by speed to get the remaining distance travelled.

[tex]$900 \ s \times 132 \ \frac{ft}{s} = 118800 \ ft$[/tex]

Total distance travelled, converting back to miles,

[tex]$2178 + 118800 = 120978\ ft . \ \frac{mi}{5280 \ ft}$[/tex]

                      = 22.9125 mi

Therefore, the distance travelled is 22.9125 miles

Answer:

a)[tex]|\Delta E|=4.58\: J[/tex]  

b)[tex]F=61.90\: N[/tex]

Explanation:

a)

We can use conservation of energy between these heights.

[tex]\Delta E=mgh_{2}-mgh_{1}=mg(h_{2}-h_{1})[/tex]  

[tex]\Delta E=0.608*9.81(0.6026-1.37)[/tex]

Therefore, the lost energy is:

[tex]|\Delta E|=4.58\: J[/tex]  

b)

The force acting along the distance create a work, these work is equal to the potential energy.

[tex]W=\Delta E[/tex]

[tex]F*d=mgh[/tex]

Let's solve it for F.

[tex]F=\frac{mgh}{d}[/tex]

[tex]F=\frac{0.608*9.81*1.37}{0.132}[/tex]

Therefore, the force is:

[tex]F=61.90\: N[/tex]

I hope is helps you!

Answer:

answer is 18.58 because

Answer:

My answer is d.25.1 because

Solids cannot diffuse.

Answer:

converts elastic energy to mechanical energy

Explanation:

Answer:

The correct answer is "[tex]4.49\times 10^{10} \ joules[/tex]".

Explanation:

According to the question,

The work will be:

⇒ [tex]Work=-\frac{kQq}{R}[/tex]

              [tex]=-\frac{1}{4 \pi \varepsilon \times (18-30)\times 3\times 0.2}[/tex]

              [tex]=-\frac{1}{4 \pi \varepsilon \times (-12)\times 3\times 0.2}[/tex]

              [tex]=\frac{0.3978}{\varepsilon }[/tex]

              [tex]=4.49\times 10^{10} \ joules[/tex]

Thus the above is the correct answer.    

We have that the workdone  is mathematically given as

W=4.49*10e10 J

From the question we are told

Generally the equation for the workdone   is mathematically given as

W=-kQq/R

Therefore

0.3978/ε0 =-1/(4πε0*(18-30)*3*0.2

Hence

For more information on Charge visit

https://brainly.com/question/9383604

Answer:

The answer is C. Isolated System

Answer:

C. Isolated system

Explanation :

∵According to law of  conservation of momentum ,In an isolated system ,the total momentum remains conserved.

Answer:

Answer:

because it look more impressive than empty dark sky .

Answer:

Distance = velocity x time, so 10 m/s X 10 s = 100 m

Explanation:

If you accelerate at 2 m/s^2 for 10 seconds, at the end of the 10 seconds you are moving at a rate of 20 m/s.

V(f) = V(i) + a*t

Final velocity = initial velocity + acceleration x time

Your average velocity will be half of your final, because you accelerated at a constant rate. So your average velocity is 10 m/s.

Distance = velocity x time, so 10 m/s X 10 s = 100 m

Answer:

100 m

Explanation:

Given,

Initial velocity ( u ) = 0 m/s

Acceleration ( a ) = 2 m/s^2

Time ( t ) = 10 sec s

To find : Displacement ( s ) = ?

By 2nd equation of motion,

s = ut + at^2 / 2

= ( 0 ) ( 10 ) + ( 2 ) ( 10 )^2 / 2

= 0 + ( 2 ) ( 100 ) / 2

= 200 / 2

s = 100 m