Match the scienntist with their accomplishment / discoveries

Answer:

Knowing that these metals are infact good conductors of electricity we can infer that metals are able to hold and conduct certain temperatures. Another thing we can infer is that these good conductors can be used in connection to transferring energy or electricity.

Answer:

a) 0 V

b) 10 turns

c) 4000 turns

d) 12.5 A

e) 400 W

f) 0.5 A

g) 95.4%

Explanation:

A

0

B

To solve this, we would be using the simple relationship between voltage and number of turns

V1/V2 = N1/N2

500/25 = 200/N2

20 = 200/N2

N2 = 200/20

N2 = 10 turns

C

Here also, we would be using the relationship between current and the number of turns

I1/I2 = N2/N1

500/25 = N2/20

20 = N2/20

N2 = 20 * 20

N2 = 4000 turns

D

Like in the previous question, current and the number of turn relationship is used

N1/N2 = I2/I1

400/80 = I2/2.5

5 = I2/2.5

I2 = 5 * 2.5

I2 = 12.5 A

E

The power remains unchanged at 400 W

F

Power = Voltage * Current

P = VI

I = P/V

I = 60/120

I = 0.5 A

G

95.4%

The transformer is a device used to step up or step down voltage.

Part A;

Given that;

Es/Ep = Ns/Np

Es = voltage in the secondary coil

Ep = voltage in primary coil

Ns = Number of turns in secondary coil

Np = Number of coils in primary coil

Es = Ns/Np ×  Ep

Es = 200/100  × 1.5 V

Es = 3 V

Part B

Ns = Es/Ep × Np

Ns = 25/500 × 200

Ns = 10 turns

Part C

Ns/Np = Ip/Is

Ns = Ip/Is × Np

Ns = 500/25 × 200

Ns = 4000 turns

Part D

Ns/Np = Ip/Is

NsIs = NpIp

Is = NpIp/Ns

Is = 400 × 2.5/80

Is =12.5 A

Part E

The power in the primary coil is the same as the power in the secondary coil. The power in the secondary coil is 400 watts.

Part F

Power supplied = 60 watts

Voltage of primary coil = 120 V

Since;

P = IV

I = P/V = 60/120 = 0.5 A

Part G

Since;

E = 100Pout/Pin

Pin = 120 V × 2 A = 240 W

Pout =  19.4 V ×  11.8 A = 228.92 W

E = 100(228.92/240)

E = 95.4%

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Answer:

Angular speed = 27.78 rad/s (Approx)

Explanation:

Given:

Diameter = 21.6 cm

Speed = 3 m/s

Find:

Angular speed

Computation:

Radius = 21.6 / 2 = 10.8 cm = 0.108 m

Angular speed = v / r

Angular speed = 3 / 0.108

Angular speed = 27.78 rad/s (Approx)

Answer:

Nose has mucous glands with hairs which helps the body in trapping pollutants and infectants from entering inside the body. On the other hand,our feet is composed of millions of sweat pores when dirt and other things accumulate,it smells because of sweat mixed with the dirt and other dirty things of the ground.

Explanation:

Hope this helps

By conservation of momentum :

[tex]m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2\\\\96v + 0 = 96v_1 + 135\times 1.03\\\\96v = 96v_1 +139.05 \ \ \ \ \ \ .....1[/tex]

By conservation of energy :

[tex]\dfrac{m_1u_1^2}{2}+\dfrac{m_1u_2^2}{2} =\dfrac{m_2v_1^2}{2} +\dfrac{m_2v_2^2}{2}\\\\96v^2 + 0 = 96v_1^2 + 135(1.03)^2\\\\96v^2 = 96v_1^2 +143.22 \ \ \ \ \ ....2)[/tex]

Solving equation 1 and 2, we get :

v = 1.23 m/s

Therefore, the initial speed of 96 kg car is 1.23 m/s .

its A or D but im not sure which one ik it moves fast

Answer:

radation because the heat gets too intense and is forced to rise

Given :

Brandon hits a golf ball with an initial velocity of 30 m/s at an angle of 30 above the horizontal.

To Find :

How long is it in the air.

Solution :

We know, the formula of time of flight is :

[tex]T = \dfrac{2usin\ \theta}{g}\\\\T = \dfrac{2\times 30\times sin\ 30^o}{9.8}\\\\T = 3.06\ seconds[/tex]

Therefore, the ball is in air for 3.06 seconds.

Answer:

The value is  [tex]\mu_k = 0.102[/tex]

Explanation:

From the question we are told that

   The initial speed of the pluck is  [tex]u = 10 \ m/s[/tex]

    The  distance it slides on the horizontal ice is  [tex]s = 50 \ m[/tex]

Generally from kinematic equation we have that

       [tex]v^2 = u^2 + 2as[/tex]      

Here v is  is the final velocity and the value is 0 m/s given that the pluck came to rest, so

      [tex]0^2 = 10 ^2 + 2* a * 50[/tex]      

=>   [tex]a = - 1 \ m/s^2[/tex]

Here the negative sign show that the pluck is decelerating  

 Generally the force applied on the pluck is  equal to the frictional force experienced by the pluck

      So  

                [tex]F = F_f[/tex]

=>            [tex]m * a = m* g * \mu_k[/tex]

=>             [tex]1 = 9.8 * \mu_k[/tex]

=>              [tex]\mu_k = 0.102[/tex]

Answer:ratio of the radii of their orbits = 1.3 --- C

Explanation:

1- eV = to the kinetic energy of the electrons

and kinetic energy is given as

K.E= 1/2mv2

v = √(2E/m)----- equation 1

The force on the particles  relating to the magnetic and circular motion ( centripetal force is given as

F = magnetic force = centripetal force

F= qvB = mv2/r

qvB = mv2/r

  r = mv/qB ------ equation 2

We know from equation 1 that v = √(2E/m)

Therefore,

r = √(2mE)/qB------ equation 3

We can now say that the  ratio of the two radii of their orbits can be calculated as

r1/r2 =(√(2mE1)/qB) /(√(2mE2)/qB

Where E1 = 500-eV  and E2 = 300-eV (1- eV = to the kinetic energy of the electrons)

r1/r2 = (√(2m x500)/qB) /(√(2mx 300)/qB

Cancelling out common variables, we are left with

r1/r2 =[tex]\sqrt{500/300}[/tex]

r1/r2= 1.29 ≈ 1.3

Answer:

12445 J

Explanation:

Given that

Power output, P = 5 kW

efficiency of the engine, e = 27% = 0.27

Thermal energy expelled, Q(c) = 9085 J

Heat absorbed, Q(h) = ?

Using the formula

e = W/Q(h)

e = [Q(h) - Q(c)] / Q(h)

e = 1 - Q(c)/Q(h)

Now, substituting the values into the formula, we have

0.27 = 1 - 9085/Q(h)

9085/Q(h) = 1 - 0.27

9085/Q(h) = 0.73

Q(h) = 9085 / 0.73

Q(h) = 12445 J

Thus, the heat absorbed is 12445 J

The exercise involves filling in the gaps with the possible wave

properties that can be obtained from a spring.

The correctly completed exercise is presented as follows;

A wave is a disturbance that moves through a medium. There are two

types of waves. A mechanical wave requires matter to travel. List some

examples of this type: sound wave, water wave, spring waves.

A electromagnetic wave does not require a medium. Examples include: Light waves

In order to start and transmit a mechanical wave, a source of

disturbance and a physical medium are required. A single disturbance is

referred to as a pulse, and a series of disturbance is a wave train.

This type of wave is called transverse wave. Its pulses are called crest

and troughs.

Now send a pulse by quickly pushing the spring forward and pulling it

back, as shown. This type of wave is called longitudinal wave. Watch the

motion of the ribbon. In this type, the particles move parallel to the

direction the wave travels. Its pulses are called compression and

rarefactions. Note that all waves transfer energy without transferring

matter. In mechanical waves, particle of the medium vibrate back and

forth in simple harmonic motion while the disturbance (or energy)

moves from one place to another.

B. Wave speed

Does the pulse catch up to the other? yes. The size of the pulse is called

the amplitude of the wave.

Did the size of the pulse affect the speed? No.

The average time wave it takes the wave to travel

Without changing your positions therefore keeping the distance the

pulse travels the same), pull the slinky tighter using only about 3/4 of

coils. This makes a completely different medium through which the

pulse will travel. Time the journey as before time record. Does the kind

of medium affect the speed of the pulse? Yes

C. Wavelength and Frequency

High frequency waves have short wavelengths and low frequency waves

have long wavelengths.

The speed of a wave in any medium is equal to the frequency of the wave × the wavelength. This wave equation [tex]\underline{f = \dfrac{v}{\lambda } }[/tex] shows that f and λ are

inversely proportional. The units of the variables are;

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Answer:

bus momentum

p_bus= m_bus x v_bus

=18,200 x 16.5

basball momentum

pball=mball x vball

=0.142 x v

p_bus = pball

18200 x 16.5 = 0.142 x v

v=(18200 x 16.5)/0.142

v is the answer for baseball

Explanation:

⚠️not my answer tryna be honest here⚠️

The momentum of the bus of 18200 kg and velocity of 13.5 m/s is 245700 Kg m/s. To have equal momentum the base ball with 145 g have to throw in a speed of 1.7 × 10 ⁶  m/s.

Momentum of a moving body is the product of mass and velocity. Thus it have the unit of g m/s or Kg m/s. Momentum is a vector quantity and thus having magnitude and direction.

Given that one bus is having a mass of  18200 Kg and 13.2  m/s speed. The momentum is:

p = mv

 =18200 kg × 13.5  m/s

 = 245700 Kg m/s

To have a momentum of 245700 Kg m/s   the base ball with 0. 142 g have to have a velocity  = 245700 Kg m/s   / 0.142 g

                          =1.7 × 10 ⁶  m/s

Hence, the baseball weighs0. 142 g have to move in 1.7 × 10 ⁶  m/s

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Answer:

The maximum angular velocity is 20 rad/s

Explanation:

Given;

torque, τ = 10 N

maximum mechanical power, P = 200 J/s

The output power of the pmdc is given as;

P = τω

where;

P is the maximum mechanical power

ω is the maximum angular velocity

ω = P / τ

ω = (200) / (10)

ω = 20 rad/s

Therefore, the maximum angular velocity is 20 rad/s

Answer:

The value is [tex]T_2 =416.9 \ K[/tex]

Explanation:

From the question we are told that

     The mass of the aluminum baking sheet is  [tex]m = 225 \ g = 0.225 \ kg[/tex]

      The energy absorbed is [tex]E = 2.4 *10^{4} \ J[/tex]

       The initial  temperature is  [tex]T_1 = 25 ^oC = 25 + 273 = 298 \ K[/tex]

Generally the heat absorbed is mathematically represented as

         [tex]Q = m * c_a * [T_2 - T_1][/tex]

Here  [tex]c_a[/tex] is the specific heat capacity of  aluminum with value  [tex]c_a = 897 \ J / kg \cdot K[/tex]

So

           [tex]2.4 *10^{4 } =0.225 * 897 * [ T_ 2- 298][/tex]

=>         [tex]T_2 - 298 = 118.915[/tex]

=>          [tex]T_2 =416.9 \ K[/tex]

Answer:

600 KPa.

Explanation:

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

Initial volume (V1) = 0.075 m³

Final volume (V2) = 0.45 m³

Final pressure (P2) = 100 KPa

Initial pressure (P1) =?

Temperature = constant

The initial pressure can be obtained by using the Boyle's law equation as shown below:

P1V1 = P2V2

P1 × 0.075 = 100 × 0.45

P1 × 0.075 = 45

Divide both side by 0.075

P1 = 45 / 0.075

P1 = 600 KPa.

Thus, the initial pressure in the balloon is 600 KPa.

Answer:

Your average speed is 60 kilometers per hour

Explanation:

240 divided by 4 is 60

240/4=60

Answer:

As the ski jumper starts moving downhill, some of his potential energy changes into kinetic energy (KE). Kinetic energy moves him down the slope to the ramp. When the ski jumper takes off from the ramp, some of his kinetic energy is changed back into potential energy as he rises in the air.

Explanation: hope this helps

Answer:

at the top of the slope

Explanation:

Answer:

roughly 8-10 hours for the average human being

Explanation:

Answer:

the placement of the ammeter in the circuit

Explanation:

Answer:

-240

Explanation:

A motorcycle skids for a distance of 2.0 m on an icy road, then the change in kinetic energy for the motorcycle will be equal to -240 J.

The force which a moving object has is referred to as kinetic energy in physics. It is defined as the number of effort required to propel a person of a specific mass from still to a specific velocity.

Aside from slight fluctuations in speed, your body holds onto the kinetic energy it obtains during acceleration.

When the body slows down from its present level to a condition of rest, the same quantity of energy is used.

Formally, kinetic energy is any quantity that has a gradient concerning time in the Lagrangian of a system.

As per the given information in the question,

Distance, d = 2.0 m

Friction, f = 120 N

The angle between displacement and friction force, θ = 180°

Now, the change in kinetic energy for the motorcycle = Work done by the friction.

K.E = f × d(cos θ)

= 120 (2.0 m)(cos 180°)

Δ K.E = -240 J

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Answer:

A

Explanation:

Galaxy clusters are basically very large (>50 galaxies) groups

Answer:

The correct answer would be:

C.

Clusters contain a hot intracluster medium, whereas groups do not.

#PLATOFAM

Have a nice day!

Answer:

1.) 7.3 m/s 2.) 6.1 m/s

Explanation:

To calculate the initial horizontal velocity with just degrees and velocity alone is pretty simple. The formula is     Velocity*cos(degrees)

eg 9.5*cos(40)

2.  To calculate the initial vertical velocity with just degrees and velocity alone is pretty simple. The formula Velocity*sin(degrees)

eg 9.5*sin(40)

Answer:

Vs/c = 20 [m/s]

Explanation:

This is a problem of relative velocities, as velocity is a vector we can use addition or subtraction of vectors for the solution.

We are asked for the velocity of the sailboat with respect to an observer located in the canoe.

[tex]V_{s/c}=V_{sailboat}-V_{canoe}\\V_{s/c}=15-(-5)\\V_{s/c}=20[m/s][/tex]

Why is the speed of the canoe negative?, it is negative because the canoe moves in the opposite direction to the sailboat.

Answer: 54.88 meters/s

Explanation:

The final velocity will be calculated by using the formula:

v = u + at

where,

v = final velocity

u = initial velocity = 0

a = 9.8

t = 5.6

Therefore, we slot the value back into the formula. This will be:

v = u + at

v = 0 + (9.8 × 5.6)

v = 0 + 54.88

v = 54.88 meters per second

Therefore, the final velocity is 54.88m/s

Answer:

[tex]MA = 3[/tex]

Explanation:

Given

[tex]Box = 10kg[/tex]

[tex]Ramp\ Height = 5ft[/tex]

[tex]Ramp\ Length = 15ft[/tex]

Required

Determine the mechanical advantage

This is calculated as follows:

[tex]MA = \frac{Ramp\ Length}{Ramp\ Height}[/tex]

[tex]MA = \frac{15ft}{5ft}[/tex]

[tex]MA = 3[/tex]

Hence, the mechanical advantage is 3

Answer:

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 9.8 m/s²

PE = 10 × 9.8 × 20

We have the final answer as

Hope this helps you

[tex]{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{First \: penetrating \: length\:(s_{1}) = 3 \: cm}[/tex]

[tex]\\[/tex]

[tex]{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{Left \: Penetration \: length \: before \: it \: comes \: to \: rest \:( s_{2} )}[/tex]

[tex]\\[/tex]

[tex]{\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{Let \: Initial \: velocity = v\:m/s} \\\\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{Left \: velocity \: after \: s_{1} \: penetration = \dfrac{v}{2} \:m/s} \\\\ [/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{s_{1} = \dfrac{3}{100} = 0.03 \: m}[/tex]

[tex]\\[/tex]

☯ As we know that,

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ \bigg(\dfrac{v}{2} \bigg)^{2} = {v}^{2} + 2a s_{1}}[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} = {v}^{2} + 2 \times a \times 0.03 }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} - {v}^{2} = 0.06 \times a }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{\dfrac{ - 3{v}^{2} }{4} = 0.06 \times a }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{a = \dfrac{ - 3 {v}^{2} }{4 \times 0.06} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ a = \dfrac{ - 25 {v}^{2} }{2}\:m/s^{2} ......(1) }[/tex]

[tex]\\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{ Initial\:velocity=v\:m/s} \\\\ [/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{ Final \: velocity = 0 \: m/s }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as}[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{{0}^{2} = {v}^{2} + 2 \times \dfrac{ - 25 {v}^{2} }{2} \times s }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ - {v}^{2} = - 25 {v}^{2} \times s }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ s = \dfrac{ - {v}^{2} }{ - 25 {v}^{2} }}[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ s = \dfrac{1}{25} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ s = 0.04 \: m }[/tex]

[tex]\\[/tex]

☯ For left penetration (s₂)

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{s = s_{1} + s_{2} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ 0.04 = 0.03 + s_{2}}[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ s_{2} = 0.04 - 0.03 }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{s_{2} = 0.01 \: m = {\boxed{\sf{\purple{1 \: cm }}} }}[/tex]

[tex]\\[/tex]

[tex]\star\:\sf{Left \: penetration \: before \: it \: come \: to \: rest \: is \:{\bf{ 1 \: cm}}} \\ [/tex]