A motorist travels due North at 90 km/h for 2 hours. She changes direction and travels West at 60 km/for 1 hour.
a) Calculate the average speed of the motorist [4]
b) Calculate the average velocity of the motorist.

Answer:

A) vectors: veloicty, force

scalar:  speed, work

B)  t = 1.75 s,  C)   v = - 17 2 m / s

Explanation:

We answer each part separately

A) A vector magnitude has magnitude and direction instead a scalar magnitude has only magnitude

vector quantities: the speed of a car number is the magnitude and direction is where it goes

Force, the number is the magnitude and above that applies gives direction

Scalar magnitude: how quickly the number of the speedometer of the car

Temperature, work

B) I = 15 m height to the soil and get to calculate time = 0

        y = y₀ + v₀ t - ½ g t²

as the ball is loose its initial velocity is zero

       0 = 0 +0  - ½ g t²

       t = [tex]\sqrt{2y_o/g}[/tex]

       t = [tex]\sqrt{2 \ 15/ 9.8}[/tex]

       t = 1.75 s

C) the velocity to the reach the floor

      v = vo - g t

      v = 0 - g t

      v = - 9.8 1.74

       v = - 17 2 m / s

The negative signt iindicates that the speed goes down

Answer:

B) Blackbody radiation

MARK THIS AS BRAINLIEST PLEAWESSS :)

Answer:

Black body radiation

Explanation:

Answer:

0.94 m³/s

Explanation:

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

Air flow (in ft³/min) = 2×10³ ft³/min

Air flow (in m³/s) =.?

Next, we shall convert 2×10³ ft³/min to m³/min. This can be obtained as follow:

35.315 ft³/min = 1 m³/min

Therefore,

2×10³ ft³/min = 2×10³ ft³/min × 1 m³/min / 35.315 ft³/min

2×10³ ft³/min = 56.63 m³/min

Finally, we shall convert 56.63 m³/min to m³/s. This can be obtained as follow:

1 m³/min = 1/60 m³/s

Therefore,

56.63 m³/min = 56.63 m³/min × 1/60 m³/s ÷ 1 m³/min

56.63 m³/min = 0.94 m³/s

Thus, 2×10³ ft³/minis equivalent to 0.94 m³/s.

Answer:

The free electrons tend to Negatively Charged body to Positively Charged body

Explanation:

As they have different charges, Electrons are more attracted to the Positive or it's opposite charge.

Answer:

λ  = 4.023 10⁻⁷ m

Explanation:

The double-slit interference phenomenon is described by

          d sin θ = (m + ½) λ          destructive interference

          d sin θ = m λ                  constructive interference

we can use trigonometry

         tan θ = y / L

how these experiments occur for small angles

         tan θ = sin θ/cos θ = sin θ

         sin θ = y / L

we substitute

        d y / L = (m + ½) λ        destructive interference

        d y / L = m λ                 constructive interference

with the expression for constructive interference we look for the separation of the slits

        d = m λ L / y

        d = 1 603 10⁻⁹ 3 /4.84 10⁻³

        d = 3.738 10⁻⁴ m

Now let's analyze the case where the distance for constructive and destructive interference occurs at the same point y = 4.84 mm = 4.84 10⁻³m

          d y / L = (m + ½)  λ

          λ = [tex]\frac{ d \ y}{L\ (m+ 1/2) }[/tex]

the first strip is for m = 1  

let's calculate

         λ  = [tex]\frac{3.738 \ 10^{-4} 4.84 \ 10^{-3} }{ 3 \ ( 1 + 0.5) }[/tex]

         λ  = 4.023 10⁻⁷ m

Answer:

a)   μ = 0.0136, b)   F = 22.8 N

Explanation:

This exercise must be solved in parts. Let's start by using conservation of moment.

a) We define a system formed by the downward and the box, therefore the forces during the collision are internal and the momentum is conserved

initial instant. Before the crash

        p₀ = m v₀

final instant. After inelastic shock

        p_f = (m + M) v

the moment is preserved

        p₀ = p_f

        m v₀ = (m + M) v

        v = [tex]\frac{m}{m + M} \ v_o[/tex]

We look for the speed of the block with the bullet inside

        v = [tex]\frac{0.00325}{0.00325 + 2.50 } \ 345[/tex]

        v = 0.448 m / s

Now we use the relationship between work and kinetic energy for the block with the bullet

in this journey the force that acts is the friction

         W = ΔK

          W = ½ (m + M) [tex]v_f^2[/tex]  - ½ (m + M) v₀²

the final speed of the block is zero

the work between the friction force and the displacement is negative, because the friction always opposes the displacement

         W = - fr x

we substitute

           - fr x = 0 - ½ (m + M) vo²

           fr = ½ (m + M) v₀² / x

the friction force is

          fr = μ N

          μ = fr / N

equilibrium condition

          N - W = 0

          N = W

          N = (m + M) g

we substitute

         μ = ½ v₀² / x g

we calculate

          μ = ½ 0.448 ^ 2 / 0.75 9.8

          μ = 0.0136

b) Let's use the relationship between work and the variation of the kinetic energy of the block

          W = ΔK

initial block velocity is zero vo = 0

         F x₁ = ½ M v² - 0

         F = [tex]\frac{1}{2} M \frac{x}{y} \frac{v^2}{x1}[/tex]

         F = ½ 2.50 0.448² / 0.0110

         F = 22.8 N

Answer:

kinetic energy

Explanation:

hop it is helpful

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Answer:

b = e > a = c = f > d  

Explanation:

Since the satellites complete one circular orbit in the same amount of time, their speed is the same.

The force needed to maintain the orbit is the centripetal force given by F = mv²/L where m = mass of artificial satellite and L = radius of orbit.

So, for artificial satellite a

a. m=200 kg and L= 5000m

F =  mv²/L

F =  200 kgv²/5000 m

F =  0.04v² N

So, for artificial satellite b

b. m=400 kg and L= 2500m

F =  mv²/L

F =  400 kgv²/2500 m

F =  0.16v² N

So, for artificial satellite c

c. m=100 kg and L= 2500m

F =  mv²/L

F =  100 kgv²/2500 m

F =  0.04v² N

So, for artificial satellite d

d. m=100 kg and L= 10000m

F =  mv²/L

F =  100 kgv²/10000 m

F =  0.01v² N

So, for artificial satellite e

e. m=800 kg and L= 5000m

F =  mv²/L

F =  800 kgv²/5000 m

F =  0.16v² N

So, for artificial satellite f

f. m=300 kg and L= 7500m

F =  mv²/L

F =  300 kgv²/7500 m

F =  0.04v² N

So, the net force are in the order b = e > a = c = f > d  

Answer:

Copper would experience induced magnestism mostly easily.

Explanation:

Permalloy would experience induced magnetism most easily.

I don't know the options but that is correct too!!

Answer:

1 cal/s =4.184w

p=50 cal/s =2093w

v=12v

P = V*I

I =P/V

I = 17.43 A

P =1²*R

R = P/I²

Answer:

The friction coefficient's minimum value will be "0.173".

Explanation:

The given query seems to be incomplete. Below is the attached file of the complete question.

According to the question,

(a)

The net friction force's magnitude will be:

⇒ [tex]F_{net}=ma[/tex]

           [tex]=5\times 1.7[/tex]

           [tex]=8.5 \ N[/tex]

(b)

For m₃,

⇒ [tex]ma=\mu m_3 g[/tex]

Or,

⇒    [tex]\mu=\frac{a}{g}[/tex]

          [tex]=\frac{1.7}{9.8}[/tex]

          [tex]=0.173[/tex]

Answer:

the rate of heat transfer after the system achieves steady state is -3.36 kW

Explanation:

Given the data in the question;

mass of water m = 50 kg

N = 300 rpm

Torque T = 0.1 kNm

V = 110 V

I = 2 A

Electric work supplied W₁ = PV = 2 × 110 = 220 W = 0.22 kW

Now, work supplied by paddle wheel W₂ is;

W₂ = 2πNT/60

W₂ = (2π × 0.1 × 300) / 60

W₂ = 188.495559 / 60

W₂ = 3.14 kW

So the total work will be;

W = 0.22 + 3.14

W = 3.36 kW

Hence total work done on the system is 3.36 kW.

At steady state, the properties of the system does not change so the heat transfer will be 3.36 KW.

The heat will be rejected by the system so the sign of heat will be negative.

i.e Q = -3.36 kW

Therefore,  the rate of heat transfer after the system achieves steady state is -3.36 kW

Answer:

the man has returned from his trip

Answer:

just did by typing this lol

Answer:

     A = 2,8333  s

Explanation:

El periodo es definido como el tiene que toma de dar una oscilación.

En este caso realiza varias osicilacion por lo cual debemos encontrar el promedio del perdono.

              T = t/n

calculemos

              A = 34,0/ 12,0

              A = 2,8333  s

Answer:

crystallography.......

Answer:

The velocity of the fish when it hits the water is:

v = 10.93 m/s and 71.88 ° below the x-direction.

Explanation:

Let's find the velocity of the fish in the y-direction.

[tex]v_{fy}^{2}=v_{iy}^{2}-2gh[/tex]

Here, v(iy) of the fish is zero, and the heigh h = 5.50 m, then the velocity will be:

[tex]v_{fy}^{2}=0-2(9.81)(5.50)[/tex]

[tex]v_{fy}^{2}=-2(9.81)(5.50)[/tex]

[tex]v_{fy}=-10.39 \: m/s[/tex]

Now, we know that the velocity in the x-direction is constant, so we can calculate the velocity of the fish when it hits the water.

[tex]v=\sqrt{v_{x}^{2}+v_{y}^{2}}[/tex]

[tex]v=\sqrt{3.40^{2}+(-10.39)^{2}}[/tex]

[tex]v=10.93 \: m/s[/tex]

And the direction will be:

[tex]\theta=tan^{-1}(\frac{10.39}{3.40})[/tex]

[tex]\theta=71.88^{\circ}[/tex]

The angle is 71.88 ° belox the x-direction.

I hope it helps you!

Answer:

Things you should do for your family

things you shouldn't

Answer:

a. None

b. Both

Explanation:

a. Which rider is traveling faster at the bottom?

Since both riders fall from the same height, h, their potential energy, U at the top equals their kinetic energy, K at the bottom.

U = mgh and K = 1/2mv²

Since U is he same for both water-slide riders, then K will be the same and thus their speed at the bottom will be the same. This is shown below.

K = U

1/2mv² = mgh

v² = 2gh

v =√(2gh) where v = speed of rider at the bottom, g = acceleration due to gravity and h = height of slide.

Since the height is the same, so their speed at the bottom is the same. So, none of the riders travels faster than the other since they have the same speed at the bottom.

b. Which rider makes it to the bottom first? Ignore friction and assume both slides have the same path length.

Since the path length of the water slides are the same and friction is neglected, both water-slide rider get to the bottom at the same time since the distance moved is the same and they both start from rest.

So, both riders make it to the bottom at the same time.

The acceleration due to gravity acts vertically downwards, and the component of gravity acceleration is larger when the slope is steeper.

Reasons:

The acceleration of the riders are due to gravity

The component acceleration due to gravity acting on a slope is a = g·sin(θ)

As the steepness of the slope increase, the angle, θ, and sin(θ) increases, therefore, the acceleration increases.

Rider A is on a slide with gentle slope, such that if the slide is flat, rider A will be stationary.

Therefore;

a. The rider that is travelling faster at the bottom is rider B

b. Given that friction is ignored, and the path have the same length to the

bottom, the rider that makes to the bottom first is the rider that is moving

faster, which is rider B

Learn more here:

https://brainly.com/question/13218675

Answer:

Units named after scientists are written in lowercase but their symbols are written as capital

Unit of power is watt, since it is named after a scientist, then symbol will be written as W

Farad, symbol = F

Henry, symbol = H

Units whose names aren't culled from that of scientists are written in lower case and their symbols are also in lower case.

Units such as meter, kilogram should be represented with symbols in small letters as: m and kg respectively.

Explanation:

Kindly check answer

Answer:

45 m

Explanation:

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

Initial velocity (u) = 30 m/s

Deceleration (a) = –10 m/s²

Final velocity (v) = 0 m/s

Distance (s) =?

The distance of the dog from the car before the driver hits the brake can be obtained as follow:

v² = u² + 2as

0² = 30² + (2 × –10 × s)

0 = 900 + (–20s)

0 = 900 – 20s

Collect like terms

0 – 900 = –20s

–900 = –20s

Divide both side by –20

s = –900 / –20

s = 45 m

Therefore, the dog was 50 m away when the driver first hit the brake

Answer:

[tex]\boxed {\boxed {\sf 2 \ Newtons}}[/tex]

Explanation:

According to Newton's Second Law of Motion, force is the product of mass and acceleration.

[tex]F= m \times a[/tex]

The mass of the stone is 0.2 kilograms and the acceleration is 10.0 meters per square second.

Substitute the values into the formula.

[tex]F= 0.2 \ kg * 10.0 \ m/s^2[/tex]

Multiply.

[tex]F=2 \ kg*m/s^2[/tex]

Convert the units.

[tex]F= 2 \ N[/tex]

The force is 2 Newtons.

Answer:

The period the field must be reduced to zero is 9.81 x 10⁻⁵ s

Explanation:

Given;

initial value of the magnetic field, B₁ = 0.276 T

number of turns of the solenoid, N = 517 turns

diameter of the solenoid, d = 10.5 cm = 0.105 m

induced emf, = 12.6 kV = 12,600 V

when the field becomes zero, then the final magnetic field value, B₂ = 0

The induced emf is given by Faraday's law;

[tex]emf = -\frac{NA\Delta B}{t} \\\\emf = -\frac{NA (B_2 -B_1)}{t} \\\\t = -\frac{NA (B_2 -B_1)}{emf}\\\\t = \frac{NA (B_1 -B_2)}{emf}\\\\where;\\\\t \ is \ the \ time \ when \ B = 0 \ \ (i.e\ B_2 = 0)\\\\A \ is \ the \ area \ of \ the \ coil\\\\A = \frac{\pi d^2}{4} = \frac{\pi (0.105)^2}{4} = 0.00866 \ m^2\\\\t= \frac{(517) \times (0.00866)\times (0.276 -0)}{12,600}\\\\t = 9.81 \times 10^{-5} \ s[/tex]

Therefore, the period the field must be reduced to zero is 9.81 x 10⁻⁵ s

OPTION D is the correct answer.

Refer to the attachment for complete calculation...

the answer to the question is D

Answer:

time of fall is 1.75 s and the velocity with which it strikes the water is 17.15 m/s.

Explanation:

Height, h =  15 m

Newton's second law

Force = mass x acceleration

The unit of gravitational force is Newton and the value is m x g.

where, m is the mas and g is the acceleration due to gravity.  

Let the time of fall is t.

Use second equation of motion

[tex]s= u t +0.5 at^2\\\\15 = 0 +0.5\times 9.8\times t^{2}\\\\t = 1.75 s[/tex]

Let the final speed is v.

Use third equation of motion

[tex]v^2 = u^2 + 2 a s\\\\v^2 = 0 + 2 \times 9.8\times 15\\\\v =17.15 m/s[/tex]

Answer:

A. Archaea

Explanation:

Living organisms have been taxonomically classified into a highest ranking taxa called DOMAIN. The domains are as follows: Eukarya, Prokarya (bacteria) and Archaea. Although both Domains Bacteria and Archaea are unicellular, the Archaea is specifically characterized by their ability to survive in harsh weather conditions like the thermal vents tiny organisms were collected from.

Also, members of the domain Archaea have a unique cell wall different from other organisms like bacteria. Hence, based on the matching characteristics of the collected tiny organisms, they would be found in the Domain ARCHAEA.