You expend 1000 W of power in moving a piano 5 meters in 5 seconds. How much force did you exert?

Answer:

The third image

Explanation:

The one with the thumb pointing to the right

Answer:

3, correct on Edge 2020

Answer:

In electromagnetic waves, energy is transferred through vibrations of electric and magnetic fields. In sound waves, energy is transferred through vibration of air particles or particles of a solid through which the sound travels. In water waves, energy is transferred through the vibration of the water particles.

Answer:

All elements in a row have the same number of electron shells. Each next element in a period has one more proton and is less metallic than its predecessor. Arranged this way, groups of elements in the same column have similar chemical and physical properties, reflecting the periodic law.

Answer:

Option A, Sea Breeze

Explanation:

Ssea breeze is a wind that blows from the ocean or any water body to the nearby land mass. This breeze is cold as compared to the air on land. The water in water bodies has high specific heat capacity and hence takes longer time to cool as compared to the surrounding objects. The warmer air over the land rises upward thereby reducing the pressure on land and hence the sea breeze starts flowing from region of high pressure (i.e above the water body) towards the low pressure region that is the land.

Hence, option A is correct

Answer:

Time of flight=3.5 seconds

Speed at maximum height is 0

Explanation:

Φ=60°

initial velocity=u=20m/s

Acceleration due to gravity=g=9.8 m/s^2

Total time of flight=T

Final speed=v

question 1:

T=(2 x u x sinΦ)/g

T=(2 x 20 x sin60)/9.8

T=(2 x 20 x 0.8660)/9.8

T=34.64/9.8

T=3.5 seconds

Question 2

Speed at maximum height is 0

Answer:

t = 2.94 x 10⁶ years

Explanation:

The equation used in the case of constant speed is:

s = vt

t = s/v

where,

s = distance = 12 light years

s = (12 light years)(9.461 x 10¹² km/light year) =  113.532 x 10¹² km

v = speed = 440 km/hr

t = time passed = ?

Therefore,

t = (113.532 x 10¹² km)/(440 km/hr)

t = 2.58 x 10¹¹ hr

Now, converting it to years:

t = (2.58 x 10¹¹ hr)(1 year/8766 hr)

t = 2.94 x 10⁶ years

Answer:

coefficient of friction (μ) and normal force (N)

Answer: How hard the surfaces push together and the types of surfaces involved

Explanation:

Answer:

b) 20 kJ

Explanation:

Efficiency of carnot engine = (T₁ - T₂ ) / T₁  Where T₁ is temperature of hot source  and T₂ is temperature of sink .

T₁ = 270 + 273 = 543K

T₂ = 50 + 273 = 323 K

Putting the given values of temperatures

efficiency = (543 - 323) / 543

= .405

heat input = 50 KJ

efficiency = output work / input heat energy

.405 = output work / 50

output work = 20.25 KJ.

= 20 KJ .

Answer is Weaker. If it is talking about the objects' gravitational forces.

Answer:

Explanation:

Let v₁ and v₂ be velocities at lowest and topmost position . Let r be the radius of the circle .

Let N₁ and N₂ be the normal reaction force .

At the top position

centripetal force = N₂ + mg ;  so

N₂ + mg  = m v₂² / r

At the bottom  position

centripetal force = N₁ - mg ;  so

N₁ - mg  = m v₁² / r

subtracting these two equations

N₁ - mg - N₂ - mg = m v₁² / r  - m v₂² / r

N₁ - N₂ - 2mg = 1/r (m v₁²   - m v₂²  )

N₁ - N₂ - 2mg = 1/r x mg x 2r  ( loss of potential energy = gain of kinetic energy )

N₁ - N₂ =  2mg +  2mg

= 4 mg .

Answer:

a) 1504.8 J

b) 991.76 J

c) 0J

d) 0J

Explanation:

(a) The work done by the force P on the box is given by the following formula:

[tex]W_P=Px[/tex]

P: applied force = 171N

x: distance in which the for P is applied = 8.80m

you replace the values of P and x and obtain:

[tex]W_P=(171N)(8.80m)=1504.8J[/tex]

(b) The work don by the friction force is:

[tex]W_f=F_fx=\mu N x=\mu Mg x[/tex]

μ = coefficient of kinetic friction = 0.250

M: mass of the box = 46.0kg

g: gravitational constant = 9.8 m/s^2

[tex]W_f=(0.250)(46.0kg)(9.8m/s^2)(8.80m)=991.76J[/tex]

(c) The Normal force is

[tex]N=Mg=(46.0kg)(9,8m/s^2)=450.8N[/tex]

but this force does not do work on the box because the direction is perpendicular to the direction of the force P.

[tex]W_N=0J[/tex]

(d) the same as before:

[tex]W_g=0J[/tex]

Answer:

Explanation:

We shall use Ampere's circuital law to find magnetic field at required point.

The point is outside the circumference of two given wires so whole current will be accounted for .

Ampere's circuital law

B = ∫ Bdl = μ₀ I

line integral will be over circular path of radius r = 41 cm .

Total current  I  = 5A -3A = 2A .

∫ Bdl = μ₀ I

2π r B = μ₀ I

2π x .41  B = 4π x 10⁻⁷ x 2

B = 2 x 10⁻⁷ x 2 / .41

= 9.75 x 10⁻⁷ T . It will be along - ve Y - direction.

Answer:

-7.23 rad/s

Explanation:

Given that

Mass of the cockroach, m = 0.157 kg

Radius of the disk, r = 14.9 cm = 0.149 m

Rotational Inertia, I = 5.92*10^-3 kgm²

Speed of the cockroach, v = 2.92 m/s

Angular velocity of the rim, w = 3.89 rad/s

The initial angular momentum of rim is

Iw = 5.92*10^-3 * 3.89

Iw = 2.3*10^-2 kgm²/s

The initial angular momentum of cockroach about the axle of the disk is

L = -mvr

L = -0.157 * 2.92 * 0.149

L = -0.068 kgm²/s

This means that we can get the initial angular momentum of the system by summing both together

2.3*10^-2 + -0.068

L' = -0.045 kgm²/s

After the cockroach stops, the total inertia of the spinning disk is

I(f) = I + mr²

I(f) = 5.92*10^-3 + 0.157 * 0.149²

I(f) = 5.92*10^-3 + 3.49*10^-3

I(f) = 9.41*10^-3 kgm²

Final angular momentum of the disk is

L'' = I(f).w(f)

L''= 9.41*10^-3w(f)

Using the conservation of total angular momentum, we have

-0.068 = 9.41*10^-3w(f) + 0

w(f) = -0.068 / 9.41*10^-3

w(f) = -7.23 rad/s

Therefore, the speed of the lazy Susan after the cockroach stops is -7.23 and is directed in the opposite direction of the initial lazy Susan angular speed

b)

The mechanical energy of the cockroach is not converted as it stops

Answer:147 joules

Explanation:

Mass=m=5kg

Acceleration due to gravity=g=9.8m/s^2

Height=h=3 meter

Potential energy=m x g x h

Potential energy=5 x 9.8 x 3

Potential energy=147 joules

Answer:

the answers the correct one is cη

Explanation:

In this simple pendulum experiment the student observes a significant change in time between each period. This occurs since an approximation used is that the sine of the angle is small, so

              sin θ = θ

with this approach the equation will be surveyed

     d² θ / dt² = - g / L sin θ

It is reduced to

      d² θ / dt² = - g / L θ

in which the time for each oscillation is constant, for this approximation the angle must be less than 10º so that the difference between the sine and the angles is less than 1%

The angle is related to the height of the pendulum

         sin θ = h / L

         h = L sin θ.

Therefore the student must be careful that the height is small.

When reviewing the answers the correct one is cη

Considering the approximation of simple harmonic motion, the correct option is:

(c) Make sure that the difference in height between the pendulum's release position and rest position is not too large.

According to Newton's second law in case of rotational motion, we have;

[tex]\tau = I \alpha[/tex]

Applying this, in the case of a simple pendulum, we get;

[tex]-mg\,sin\,\theta =mL^2 \,\frac{d^2 \theta}{dt^2}[/tex]

On, rearranging the above equation, we get;

[tex]mL^2 \,\frac{d^2 \theta}{dt^2} + mg\,sin\,\theta=0\\\\\implies \frac{d^2 \theta}{dt^2} +\frac{g}{L} sin \,\theta=0[/tex]

Now, if angular displacement is very small, i.e.; the bob of the pendulum is only raised slightly.

Then, [tex]sin\, \theta \approx \theta[/tex]

[tex]\implies \frac{d^2 \theta}{dt^2} +\frac{g}{L} \,\theta=0[/tex]

This is now in the form of the equation of a simple harmonic motion.

[tex]\frac{d^2 \theta}{dt^2} +\omega^2 \,\theta=0[/tex]

Comparing both these equations, we can say that;

[tex]\omega = \sqrt{\frac{g}{L}}[/tex]

[tex]T=2\pi\sqrt{\frac{L}{g}}[/tex]

This relation for the time period can only be obtained if the angular displacement is very less.

So, the correct option is;

Option (c): Make sure that the difference in height between the pendulum's release position and rest position is not too large.

Learn more about simple harmonic motion here:

https://brainly.com/question/26114128

Answer:

A.  falling water

Explanation:

I got it right on Edgenuity. Good luck on your quiz.

In hydroelectric plants, water falls on turbine and makes it rotate. In generator, this mechanical energy transforms into electrical energy.

Hydroelectric power is generated by turbines that turn the potential energy of falling or swiftly flowing water into mechanical energy, which is then used to power generators. The most popular renewable energy source in the early 21st century was hydroelectricity, which in 2019 accounted for more than 18% of the world's total power producing capacity.

Water is gathered or stored at a higher elevation during the production of hydroelectric power and then transported through substantial pipes or tunnels (penstocks) to a lower elevation; the difference between these two elevations is referred to as the head. The falling water turns turbines as it nears the bottom of the pipelines. In turn, the turbines power generators, which transfer the mechanical energy of the turbines into electricity.

Learn more about  hydroelectric power here:

https://brainly.com/question/15228003

#SPJ2

Answer:

The center of mass move with the velocity of -3.487 m/s.

Explanation:

Given values of block A.

Mass of block A, (M1) = 4 kg

Speed of block A, (V1) = 2 m/s

Given values of block B.

 Mass of block B, (M2) = 8.4 kg

Speed of block B, (V2) = -6.1 m/s

Below is the formula to find the velocity of center of mass.

[tex]Velocity = \frac{M1V1 + M2V2}{M1 + M2} \\[/tex]

[tex]= \frac{4 \times 2 + 8.4 \times (-6.1) }{4 + 8.4} \\[/tex]

[tex]= \frac{- 43.24}{12.4}\\[/tex]

[tex]= - 3.487 m/s[/tex]

Answer:

Total internal reflection (TIR) is the phenomenon that involves the reflection of all the incident light off the boundary. TIR only takes place when both of the following two conditions are met: the light is in the more dense medium and approaching the less dense medium.

Explanation: Hope i helped!!!

Given Information:

Power = P = 100 Watts

Voltage = V = 220 Volts

Required Information:

a) Current = I = ?

b) Resistance = R = ?

Answer:

a) Current = I = 0.4545 A

b) Resistance = R = 484 Ω

Explanation:

According to the Ohm’s law, the power dissipated in the light bulb is given by

[tex]P = VI[/tex]

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and P is the power dissipated in the light bulb.

Re-arranging the above equation for current I yields,

[tex]I = \frac{P}{V} \\\\I = \frac{100}{220} \\\\I = 0.4545 \: A \\\\[/tex]

Therefore, 0.4545 A current is flowing through the light bulb.

According to the Ohm’s law, the voltage across the light bulb is given by

[tex]V = IR[/tex]

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and R is the resistance of the light bulb.

Re-arranging the above equation for resistance R yields,

[tex]R = \frac{V}{I} \\\\R = \frac{220}{0.4545} \\\\R = 484 \: \Omega[/tex]

Therefore, the resistance of the bulb is 484 Ω

Answer:

bulb will burn out!

Explanation:

Answer:

20cal/s

Explanation:

Question:

There are two questions. The first one has been answered:

From the formular, Power = Q/t = (kA∆T)/l

the amount heat depends on the duration of time interval, length of the iron rod, the thermal conductivity of iron and the temperature difference between the ends of the rod.

The amount of heat that flows through the rod by conduction during a given time interval does not depend upon the mass of the iron rod (D).

Second question:

The ends of a cylindrical steel rod are maintained at two different temperatures. The rod conducts heat from one end to the other at a rate of 10 cal/s. At what rate would a steel rod twice as long and twice the diameter conduct heat between the same two temperatures?

Solution:

Power = 10cal/s

Power = energy per unit time = Q/t

Where Q = energy

Power = (kA∆T)/l

k = thermal conductivity of iron

A = area

Area = πr^2

r = radius

Diameter = d = 2r

r = d/2

Area = (πd^2)/4

Length = l

∆T = change in temperature

10 = (kA∆T)/l

For a steel rod with length doubled and diameter doubled:

Let Length (L) = 2l

Diameter (D)= 2d

Area = π [(2d)^2]/4 = (π4d^2)/4

Area = 4(πd^2)/4

Using the formula Power = (kA∆T)/l, insert the new values for A and l

Power = [k × 4(πd^2)/4 × ∆T]/2l

Power = [4k((πd^2)/4) ∆T]/2l

Power = [(4/2)×k((πd^2)/4) ∆T]/l

Power = [2k(A) ×∆T]/l = 2(kA∆T)/l

Power of a steel that has its length doubled and diameter doubled = 2(kA∆T)/l

Recall initial Power = (kA∆T)/l = 10cal/s

And ∆T is the same

2[(kA∆T)/l] = 2 × 10

Power of a steel that has its length doubled and diameter doubled = 20cal/s

Answer:

Question 3: 2.25 watts

Question 4: 405 joules

Explanation:

question 3:

Current =0.5 amps

Voltage =4.5 volts

Power= current x voltage

Power=0.5 x 4.5

power=2.25 watts

Question 4

Current =0.5 amps

Voltage =4.5v

Time=3 minutes

Time =3x60

Time =180 seconds

Energy=current x voltage x time

Energy =0.5 x 4.5 x 180

Energy =405 joules

Answer:

The thickness of the door is 0.4230 m

Explanation:

Given;

mass of bullet, m = 0.009 kg

initial velocity of the bullet, u = 803 m/s

final velocity of the bullet, v = 617 m/s

average resistive force of the door on the bullet, F = 5620 N

Apply Newton's second law of motion;

Force exerted by the door on the bullet = Force of the moving bullet

F = ma

where;

F is applied force

m is mass

a is acceleration

Also, Force exerted by the door on the bullet = Force of the moving bullet

[tex]F =ma, \ But \ a =\frac{dv}{dt} = \frac{u-v}{t} \\\\F = \frac{m(u-v)}{t}[/tex]

where;

v is the final velocity of the bullet

u is initial velocity of the bullet

t is time

We need to calculate the time spent by the bullet before it passes through the door.

[tex]t = \frac{m(u-v)}{F} \\\\t = \frac{0.009(803-617)}{5620} = 0.0002979 \ s[/tex]

Distance traveled by the bullet within this time period = thickness of the door

This distance is equivalent to the product of average velocity and time

[tex]S = (\frac{u+v}{2}) t[/tex]

where;

s is the distance traveled

[tex]S = (\frac{u+v}{2}) t\\\\S = (\frac{803+617}{2}) 0.0002979\\\\S = 0.4230 \ m[/tex]

Therefore, the thickness of the door is 0.4230 m

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From the question we are told that

     The mass of the bat is [tex]m_b = 0.800 \ kg[/tex]

      The bat length is  [tex]L_b = 0.900 \ m[/tex]

      The distance of the bat's center of mass to the handle end is  [tex]z_c = 0.600 \ m[/tex]

      The moment of inertia of the bat is    [tex]I = 0.0530 \ kg \cdot m^2[/tex]

The objective of the solution is to find  x   which is the distance from the handle of the bat to the point where the baseball hit the bat

Generally the velocity change at the end of the bat is mathematically represented as

         [tex]\Delta v_e = \Delta v_c - \Delta w* z_c[/tex]

         Where  [tex]\Delta v_c[/tex] is the velocity change at the center of the bat  which is mathematically represented as

                [tex]\Delta v_c = \frac{Impulse}{m_b }[/tex]

We are told that the impulse is  J so

              [tex]\Delta v_c = \frac{J}{m_b }[/tex]

And   [tex]\Delta w[/tex] is the change in angular velocity which is mathematically represented as

         [tex]\Delta w = \frac{J (z -z_c)}{I}[/tex]

Now we have that

           [tex]\Delta v_e = \frac{J}{m_b } - [\frac{J (x- z_c)}{I} ] * z_c[/tex]

Before a swing when the bat is at rest the velocity change a the end of the bat handle is zero  and the impulse will be  1

   So  

            [tex]0 = \frac{1}{m_b } - [\frac{J (x- z_c)}{I} ] * z_c[/tex]

=>           [tex]x = \frac{I}{m_b z_c} + m_b[/tex]

substituting values

            [tex]x = \frac{0.530}{0.800 * 0.600} + 0.600[/tex]

           [tex]x = 0.710 \ m[/tex]

Answer:

A) equal to the battery's terminal voltage.

Explanation:

When the capacitor is fully charged after long hours of charging , its  potential becomes equal to the emf of the battery and its polarity is opposite to that of battery . Hence net emf becomes equal . The capacitor itself becomes a battery which is connected in the circuit with opposite polarity . This results in the net emf and  current becoming zero . There is no charging current when the capacitor is fully charged .

Answer:

The angular acceleration of CD player is [tex]13.93\ rad/s^2[/tex].

Explanation:

Initial angular speed of a CD player is 0 and final angular speed is 41.8 rad/s. Time to change the angular speed is 3 s.

It is required to find the angular acceleration. The change in angular speed of the CD player divided by time taken is called its angular acceleration. It can be given by :

[tex]a=\dfrac{\omega_f-\omega_i}{t}\\\\a=\dfrac{41.8-0}{3}\\\\a=13.94\ rad/s^2[/tex]

So, the angular acceleration of CD player is [tex]13.93\ rad/s^2[/tex].

Answer:

A.

Explanation:

Refraction light rays are bent as they pass into a different medium where its speed is different. As refracted light rays pass from a fast medium to a slow medium, the light ray bends toward the normal to the boundary between the two medium. Light refracts as it travels at an angle into a medium with a different refractive index.

Answer:

the answer is a

Explanation:

Answer:

v_max = (1/6)e^-1 a

Explanation:

You have the following equation for the instantaneous speed of a particle:

[tex]v(t)=ate^{-6t}[/tex]   (1)

To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:

[tex]\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))][/tex]  (2)

where you have use the derivative of a product.

Next, you equal the expression (2) to zero in order to calculate t:

[tex]a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}[/tex]

For t = 1/6 you obtain the maximum speed.

Then, you replace that value of t in the expression (1):

[tex]v_{max}=a(\frac{1}{6})e^{-6(\frac{1}{6})}=\frac{e^{-1}}{6}a[/tex]

hence, the maximum speed is v_max = ((1/6)e^-1)a

Answer:

26.92 N

Explanation:

The normal reaction of the ball is due to two force component acting on it.

However ; the acceleration is in polar coordinate which is given by the relation:

[tex]a^ { ^ \to} = (r- r \omega^2) \hat {e_r} + ( r \theta + 2 r \omega ) \hat {e_ \theta}[/tex]

[tex]a_{\theta} = r \theta + 2 r \omega[/tex]

Given that :

ω = 5 rad/s

mass m = 6 kg

θ = 30 ◦

r = 0.9 m

speed v =  0.4 m/s

[tex]a_{\theta} = 0 + 2(-0.4)*5[/tex]

[tex]a_{\theta}= -4 \ m/s[/tex]

The normal force reaction (N) that the arm applies to the ball at this instant is :

N = mg cos θ  + [tex]ma_{\theta}[/tex]

N = (6 × 9.8× cos 30) + (6 ×(-4))

N = 26.92 N