A seesaw made of a plank of mass 10.0 kg and length 3.00 m is balanced on a fulcrum 1.00 m from one end of the plank. A 20.0-kg mass rests on the end of the plank nearest the fulcrum. What mass must be on the other end if the plank remains balanced?

Answer:

converts elastic energy to mechanical energy

Explanation:

Answer:

4.86 m

Explanation:

Given that,

The frequency produced by a humming bird, f = 70 Hz

The speed of sound, v = 340 m/s

We need to find how far does the sound  travel between wing flaps. Let the distance is equal to its wavelength. So,

[tex]v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{340}{70}\\\\\lambda=4.86\ m[/tex]

So, the sound travel 4.86 m between wings flaps.

Answer:

The electromotive force is the maximum potential difference between the terminals of a battery.

The electromotive force is the maximum potential difference between the terminals of a battery.  The correct option is b.

The electromotive force also called as EMF, is the force which causes current to flow from  the positive to negative terminal of the battery.

The electromotive force is the maximum potential difference between the terminals of a battery.

Thus, the correct option is b.

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Answer:

Option (d).

Explanation:

According to the Boyle's law, for a given mass of a gas, the pressure of the gas is inversely proportional to the volume of the gas keeping the temperature of the gas is constant.

So,

Let the pressure is P, volume is V and T is the absolute temperature of the gas.

Pressure proportional to the reciprocal of the volume.

[tex]P \alpha \frac{1}{V}\\\\P V = constant[/tex]

The correct option is (d).

Answer:

a) the required angle in both radian and degree is  1.25 rad and 71.6°

b) the plane's angular speed relative to the earth is 3.86 × 10⁻⁵ rad/sec

Explanation:

Given the data in the question;

a)

we know that The expression for the angle subtended by an arc of circle at the center of the circle is,

θ = Length / radius

given that Length is 5000 miles and radius is 4000 miles

we substitute

θ = 5000 miles / 4000 miles

θ = 1.25 rad

Radian to Degree

θ = 1.25 rad × ( 180° / π rad )

θ =  71.6°

Therefore, required angle in both radian and degree is  1.25 rad and 71.6°

b)

The flight from Kampala to Singapore take 9 hours.

the plane's angular speed relative to the earth = ?

we know that, the relation between angular velocity and angular displacement is;

ω = θ / t

given that θ is 1.25 rads and time t is 9 hours or ( 9 × 3600 sec ) = 32400 sec

we substitute

ω = 1.25 rad / 32400 sec

ω = 3.86 × 10⁻⁵ rad/sec

Therefore, the plane's angular speed relative to the earth is 3.86 × 10⁻⁵ rad/sec

Answer:

By the principle of corona discharge.

Explanation:

The charge on each ball will decreases over time due to the electrical discharge in air.

According to the principle of corona discharge, when the curvature is small, the discharge of the charge takes placed form the pointed ends.

Answer:

The pulse speed depends on the properties of the medium and not on the amplitude or pulse length of the pulse.

Explanation:

Hope this helps.

Answer:

[tex]d_2=1.5*10^-3m[/tex]

Explanation:

From the question we are told that:

Initial Wavelength [tex]\lambda_1=550nm=550*10^{-9}[/tex]

Space 1 [tex]d_1=2.3*10^{-3}[/tex]

Final wavelength [tex]\lambda_2=360*10^{-9}[/tex]

Generally the equation for Fringe space at [tex]\lambda _2[/tex] is mathematically given by

 [tex]d_2=\frac{d_1}{\lambdaI_1}*\lambda_2[/tex]

 [tex]d_2=\frac{2.3*10^{-3}}{550*10^{-9}}*360*10^{-9}[/tex]

 [tex]d_2=1.5*10^-3m[/tex]

Answer:

The angle between the electric field lines and the equipotential surface is 90 degree.

Explanation:

The equipotential surfaces are the surface on which the electric potential is same. The work done in moving a charge from one point to another on an equipotential surface is always zero.

The electric field lines are always perpendicular to the equipotential surface.

As

[tex]dV = \overrightarrow{E} . d\overrightarrow{r}\\\\[/tex]

For equipotential surface, dV = 0 so

[tex]0 = \overrightarrow{E} . d\overrightarrow{r}\\\\[/tex]

The dot product of two non zero vectors is zero, if they are perpendicular to each other.

Answer:

b.have frequencies that are integer multiples of the frequency of the complex waveform

Explanation:

Please correct me if I am wrong

Answer:

219 ft

Explanation:

Here we can define the value t = 0s as the moment when the car starts decelerating.

At this point, the acceleration of the car is given by the equation:

A(t) = -10 ft/s^2

Where the negative sign is because the car is decelerating.

To get the velocity equation of the car, we integrate over time, to get:

V(t) = (-10 ft/s^2)*t + V0

Where V0 is the initial velocity of the car, we know that this is 88 ft/s

Then the velocity equation is:

V(t) = (-10 ft/s^2)*t + 88ft/s

To get the position equation we need to integrate again, this time we get:

P(t) = (1/2)*(-10 ft/s^2)*t^2 + (88ft/s)*t + P0

Where P0 is the initial position of the car, we do not know this, but it does not matter for now.

We want to find the total distance that the car traveled in a 3 seconds interval.

This will be equal to the difference in the position at t = 3s and the position at t = 0s

distance = P(3s) - P(0s)

 = ( (1/2)*(-10 ft/s^2)*(3s)^2+ (88ft/s)*3s + P0) - ( (1/2)*(-10 ft/s^2)*(0s)^2 + (88ft/s)*0s + P0)

=  ( (1/2)*(-10 ft/s^2)*9s^2+ (88ft/s)*3s + P0) - ( P0)

=  (1/2)*(-10 ft/s^2)*9s^2+ (88ft/s)*3s = 219ft

The car advanced a distance of 219 ft in the 3 seconds interval.

Answer:

Explanation:

The energy for an isothermal expansion can be computed as:

[tex]\mathsf{Q_H =nRTIn (\dfrac{V_b}{V_a})}[/tex] --- (1)

However, we are being told that the volume of the gas is twice itself when undergoing adiabatic expansion. This implies that:

[tex]V_b = 2V_a[/tex]

Equation (1) can be written as:

[tex]\mathtt{Q_H = nRT_H In (2)}[/tex]

Also, in a Carnot engine, the efficiency can be computed as:

[tex]\mathtt{e = 1 - \dfrac{T_L}{T_H}}[/tex]

[tex]e = \dfrac{T_H-T_L}{T_H}[/tex]

In addition to that, for any heat engine, the efficiency e =[tex]\dfrac{W}{Q_H}[/tex]

relating the above two equations together, we have:

[tex]\dfrac{T_H-T_L}{T_H} = \dfrac{W}{Q_H}[/tex]

Making the work done (W) the subject:

[tex]W = Q_H \Big(\dfrac{T_H-T_L}{T_H} \Big)[/tex]

From equation (1):

[tex]\mathsf{W = nRT_HIn(2) \Big(\dfrac{T_H-T_L}{T_H} \Big)}[/tex]

[tex]\mathsf{W = nRIn(2) \Big(T_H-T_L} \Big)}[/tex]

If we consider the adiabatic expansion as well:

[tex]PV^y[/tex] = constant

i.e.

[tex]P_bV_b^y = P_cV_c^y[/tex]

From ideal gas PV = nRT

we can have:

[tex]\dfrac{nRT_H}{V_b}(V_b^y)= \dfrac{nRT_L}{V_c}(V_c^y)[/tex]

[tex]T_H = T_L \Big(\dfrac{V_c}{V_b}\Big)^{y-1}[/tex]

From the question, let us recall  aw we are being informed that:

If the volumes changes by a factor = 5.7

Then, it implies that:

[tex]\Big(\dfrac{V_c}{V_b}\Big) = 5.7[/tex]

∴

[tex]T_H = T_L (5.7)^{y-1}[/tex]

In an ideal monoatomic gas [tex]\gamma = 1.6[/tex]

As such:

[tex]T_H = T_L (5.7)^{1.6-1}[/tex]

[tex]T_H = T_L (5.7)^{0.67}[/tex]

Replacing the value of [tex]T_H = T_L (5.7)^{0.67}[/tex] into equation [tex]\mathsf{W = nRIn(2) \Big(T_H-T_L} \Big)}[/tex]

[tex]\mathsf{W = nRT_L In(2) (5.7 ^{0.67 }-1}})[/tex]

From in the question:

W = 930 J and the moles = 1.90

using 8.314 as constant

Then:

[tex]\mathsf{930 = (1.90)(8.314)T_L In(2) (5.7 ^{0.67 }-1}})[/tex]

[tex]\mathsf{930 = 15.7966\times 1.5315 (T_L )})[/tex]

[tex]\mathsf{T_L= \dfrac{930 }{15.7966\times 1.5315}}[/tex]

[tex]\mathbf{T_L \simeq = 39 \ K}[/tex]

From [tex]T_H = T_L (5.7)^{0.67}[/tex]

[tex]\mathsf{T_H = 39 (5.7)^{0.67}}[/tex]

[tex]\mathbf{T_H \simeq 125K}[/tex]

Answer:

Radio waves are used to carry satellite signals. These waves travel at 300,000 km/s (the speed of light). This means that a signal sent to a satellite 38,000 km away takes 0.13 s to reach the satellite and another 0.13 s for the return signal to be received back on Earth.

Explanation:

hope it help

(D)

Explanation:

The more massive an object is, the greater is the curvature that they produce on the space-time around it.

The theory of relativity explain the gravity exerted by massive objects is

more massive objects create larger curves of space-time (option-d).

The term "gravitational force" refers to the attraction between masses. The gravitational force increases in size as the masses get bigger (also called the gravity force). As the distance between masses grows, the gravitational force progressively lessens.

Greater gravitational forces will be used to attract heavier things since the gravitational force is directly proportional to the mass of both interacting objects. Therefore, when two things' respective masses increase, so does their gravitational pull to one another.

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Answer:

The right solution is "[tex]4.5\times 10^{-10} \ Cm[/tex]".

Explanation:

Given that,

q = 0.50 nC

d = 900 mm

As we know,

⇒ [tex]P=qd[/tex]

By putting the values, we get

⇒     [tex]=0.50\times 900[/tex]

⇒     [tex]=(0.50\times 10^{-9})\times 0.9[/tex]

⇒     [tex]=4.5\times 10^{-10} \ Cm[/tex]  

Answer:

The dipole moment is 4.5 x 10^-10 Cm.

Explanation:

Charge on each ball, q = 0.5 nC

Length, L = 900 mm = 0.9 m

The dipole moment is defined as the product of either charge and the distance between them.

It is a vector quantity and the direction is from negative charge to the positive charge.

The dipole moment is

[tex]p = q L\\\\p = 0.5 \times 10^{-9}\times 0.9\\\\p = 4.5\times 10^{-10} Cm[/tex]

Answer:

[tex]\frac{dV}{dt}=21.21cm^3/min[/tex]

Explanation:

We are given that

[tex]PV^{1.4}=C[/tex]

Where C=Constant

[tex]\frac{dP}{dt}=-7KPa/minute[/tex]

V=420 cubic cm and P=99KPa

We have to find the rate at which the  volume increasing at this instant.

Differentiate w.r.t t

[tex]V^{1.4}\frac{dP}{dt}+1.4V^{0.4}P\frac{dV}{dt}=0[/tex]

Substitute the values

[tex](420)^{1.4}\times (-7)+1.4(420)^{0.4}(99)\frac{dV}{dt}=0[/tex]

[tex]1.4(420)^{0.4}(99)\frac{dV}{dt}=(420)^{1.4}\times (7)[/tex]

[tex]\frac{dV}{dt}=\frac{(420)^{1.4}\times (7)}{1.4(420)^{0.4}(99)}[/tex]

[tex]\frac{dV}{dt}=21.21cm^3/min[/tex]

Answer:

[tex]\dot V=2786.52~cm^3/min[/tex]

Explanation:

Given:

initial pressure during adiabatic expansion of air, [tex]P_1=99~kPa[/tex]

initial volume during the process, [tex]V_1=420~cm^3[/tex]

The adiabatic process is governed by the relation [tex]PV^{1.4}=C[/tex] ; where C is a constant.

Rate of decrease in pressure, [tex]\dot P=7~kPa/min[/tex]

Then the rate of change in volume, [tex]\dot V[/tex] can be determined as:

[tex]P_1.V_1^{1.4}=\dot P.\dot V^{1.4}[/tex]

[tex]99\times 420^{1.4}=7\times V^{1.4}[/tex]

[tex]\dot V=2786.52~cm^3/min[/tex]

[tex]\because P\propto\frac{1}{V}[/tex]

[tex]\therefore[/tex] The rate of change in volume will be increasing.

Answer:

v = 804.23 m/s

Explanation:

Given that,

The maximum distance covered by a cannon, d = 33 km = 33000 m

We need to find the initial velocity of the shell. Let it is v. It can be calculated using the conservation of energy such that,

[tex]v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 33000} \\\\v=804.23\ m/s[/tex]

So, the initial velocity of the shell is 804.23 m/s.

Answer:

Scalar quantity is the Physical quantity expresed only by their magnitude.

Answer:

w = 1,066 rad / s

Explanation:

For this exercise we use Newton's second law

         F = m a

the centripetal acceleration is

         a = w² r

indicate that the force is the mass of the body times the acceleration

        F = m 0.58g = m 0.58 9.8

        F = 5.684 m

we substitute

       5.684 m = m w² r

       w = [tex]\sqrt{5.684/r}[/tex]

To finish the calculation we must suppose a cylinder radius, suppose it has r = 5 m

       w = [tex]\sqrt{ 5.684/5}[/tex]

       w = 1,066 rad / s

Answer:

 ΔL = 3.82 10⁻⁴ m

Explanation:

This is a thermal expansion exercise

          ΔL = α L₀ ΔT

          ΔT = T_f - T₀

where ΔL is the change in length and ΔT is the change in temperature

Let's reduce the length to SI units

          L₀ = 90.5 mm (1m / 1000 mm) = 0.0905 m

let's calculate

          ΔL = 25.10⁻⁶ 0.0905 (154.6 - (14.4))

          ΔL = 3.8236 10⁻⁴ m

using the criterion of three significant figures

          ΔL = 3.82 10⁻⁴ m

Answer:

b)

Explanation:

Answer:

a) [tex]a_{avg}=1.25ft/s^2[/tex]

b) [tex]v_b=13.35ft/s[/tex]

Explanation:

From the question we are told that:

Speed at point A [tex]v_A=10.6ft/s[/tex]

Speed at point C [tex]v_C=15.6ft/s[/tex]

Time from Point A to C [tex]T_{ac}=4.00s[/tex]

Time from Point B to C [tex]T_{bc}=1.80s[/tex]

Generally the equation for acceleration From A to B is mathematically given by

 [tex]a_{avg}=\frac{v_c-v_a}{\triangle t}[/tex]

 [tex]a_{avg}=\frac{15.6-10.6}{4.0 }[/tex]

 [tex]a_{avg}=1.25ft/s^2[/tex]

Generally the equation for cart speed at B is mathematically given by

 [tex]a_{avg}=\frac{v_c-v_a}{T_{bc}}[/tex]

 [tex]v_b=v_c-a_{avg}*T_{bc}[/tex]

 [tex]v_b=15.6ft/s-(1.25ft/s^2)(1.80)[/tex]

  [tex]v_b=13.35ft/s[/tex]

Answer:

paleontologist

Explanation:

Paleontologists are scientists that investigate the fossils of extinct life forms. Thus, the correct option is B.

A fossil is defined as the preserved trace, imprint, or proof of a once-living entity from a past geological era. Exoskeletons, bones, shells, impressions of animals or microbes in stone, objects preserved in amber, hair, petrified wood, and genetic traces are a few examples. The collection of all the fossils is called as the fossil record.

An organism from a past geologic era that has been preserved in the Earth's crust is referred to as a fossil. Paleontologists are scientists that investigate the fossils of extinct life forms. The intricate system of fossil records is the main source of information about the evolution of life on Earth.

Therefore, the correct option is B.

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Answer:

v_f = 17.4 m / s

Explanation:

For this exercise we can use conservation of energy

starting point. On the hill when running out of gas

          Em₀ = K + U = ½ m v₀² + m g y₁

final point. Arriving at the gas station

         Em_f = K + U = ½ m v_f ² + m g y₂

energy is conserved

         Em₀ = Em_f

         ½ m v₀ ² + m g y₁ = ½ m v_f ² + m g y₂

        v_f ² = v₀² + 2g (y₁ -y₂)

we calculate

        v_f ² = 20² + 2 9.8  (10 -15)

        v_f = √302

         v_f = 17.4 m / s

Answer:

The correct answer is c

Explanation:

Flow is defined by

        Ф =  B . A

bold letters indicate vectors.

The magnetic field is directed to the y axis, The area of ​​the coil is represented by a vector normal to the plane of the coil, so to have a flux

                i.i = j.j = k.k = 1

and the tori scalar products are zero

a) If the coil must be in the xy plane so that its normal vector is in the Z axis, so there is no flux

b) if the coil is in the plane yz the normal veto is in the x axis, so the flux is zero

C) If the coil is in XZ, the normal vector points in the y direction, usually the scalar product is one and there is a flux in this configuration

The correct answer is c

Answer:

fair palybtgshsisuehdh

Answer:

proton have higher energy than electron

Explanation:

tag me brainliest

Answer:

proton

Explanation:

proton is higher energy than the electron

Answer: hello your question lacks some vital information below is the complete question

Suppose a power plant uses a Carnot engine to generate electricity, using the atmosphere at 300 K as the low-temperature reservoir. Suppose the power plant produces 1 × 106 J of electricity with the hot reservoir at 500 K during Day One and then produces 1 × 106 J of electricity with the hot reservoir at 600 K during Day Two. The thermal pollution was

answer:

Total thermal pollution = 2.5 * 10^6 J

Explanation:

Low temperature reservoir = 300 K

hot reservoir temperature = 500 K

Electrical energy produced by plant ( W ) = 1 * 10^6 J

lets assume ; Q1 = energy absorbed , Q2 = energy emitted

W = Q1 - Q2  or  Q2 = Q1 - W  ( we will apply this as the formula for determining thermal pollution )

For day 1

T1 = 500k , T2 = 300k

applying Carnot engine formula

W / Q1 = 1 - T2/T1

∴ Q1 = 10^6 / ( 1 - (300/500)) = 2.5 * 10^6 J

thermal pollution ; Q2 = Q1 - W = ( 2.5 * 10^6 - 1 * 10^6 ) = 1.5 * 10^6 J

for Day 2

T1 = 600k,  T2 = 300k

Q1 = 10^6 / ( 1 - (300/600)) = 2 * 10^6 J

Thermal pollution; Q2 = Q1 - W  = 1 * 10^6 J

Therefore the Total thermal pollution =  1 * 10^6  + 1.5 * 10^6  = 2.5 * 10^6 J