Answer meeeeeeeeeeeeeee

2.     Voltage(V) in the circuit is 1.5 V and The current (I) in the circuit is 0.31 A.

So we have to find out power (p) in the circuit.

So  P = VI

       P= 1.5(0.31)

       P = 0.465

        P= 0.47 W

the total power output of the cell is approximately 0.47 W.

3.   We have to obtain  the energy dissipated per second in resistor R.

P = 0.465 - I²R

P = 0.465 - 0.31²(0.65)

P  = 0.402535 J/s

P =  0.40 J

Total 0.40 Joule per second energy has been dissipated from the resistor R.

4. The cell stores 14 kJ of energy when it is fully charged. The cell's emf and internal resistance are constant as the cell f discharged.

The energy dissipated per second(e) in resistor R is .40

Cell have total (E) 14 kJ (14000 J) of energy.

We have to calculate the maximum time (t) during which the fully-charged cell can deliver energy to resistor R.

[tex]\(E=\into{{e}_}}{t}\\\)[/tex]

14000 = 0.40 × t

t =14000  / 0.40  

t =  35 000 s or about 9.75 hrs.

Fully-charged cell can deliver energy to resistor R up to 9.75 hrs.

What is Electromotive Force?

Electromotive force, abbreviation E or emf, energy per unit electric charge that is imparted by an energy source, such as an electric generator or a battery. Energy is converted from one form to another in the generator or battery as the device does work on the electric charge being transferred within itself.

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Answer:

W

Explanation:

The closer you get to the center of the positive point of the electrical feild the stronger the pull gets. Therfore W is the best option.

Answer:

W

Explanation:

The closer you get to the center of the positive point of the electrical feild the stronger the pull gets. Therfore W is the best option.

Answer:

The correct option is D) Fission

Explanation:

There are several methods through which Radioactive isotopes are created.

Cheers

Answer:

a) false

b) True

c) True

d) False

e) False

Explanation:

a) False

For a diverging lens, the focal length is negative while it is positive for a converging lens

b) True

Image distances  for virtual images are always negative and it also forms on the the same side of the lens as the object and is enlarged

c) True

d) False

For a diverging lens, the focal length is negative while it is positive for a converging lens

e) False

Image distances  for virtual images are always negative and it also forms on the the same side of the lens as the object and is enlarged

Answer: [tex]0.0353\ s^{-1}[/tex]

Explanation:

Given

Radioactive material is found to decrease 40% of its original value in [tex]2.59\times 10\ s[/tex]

Sample at any time is given by

[tex]N=N_oe^{-\lambda t}[/tex]

where, [tex]\lambda=\text{decay constant}[/tex]

Put values

[tex]\Rightarrow 0.4N_o=N_oe^{-\lambda\cdot 2.59\times 10}\\\Rightarrow 0.4=e^{-\lambda\cdot 2.59\times 10[/tex]

Taking natural logarithm both side

[tex]\Rightarrow \lambda=\dfrac{\ln 2.5}{25.9}\\\\\Rightarrow \lambda =0.0353\ s^{-1}[/tex]

120 works for acellus

Answer: 119.9

Explanation:

F = (mv^2)/r

Here we know m (0.309kg), v (12.9m/s) and r (0.429m)

So F = (0.309*12.9^2)/0.429 = 119.861748

Answer:

Jill will move down first

Explanation:

Answer:

The tension will be "713 N".

Explanation:

The given values are:

Frequency,

= 890 Hz

Mass,

= 90 gm

or,

= 0.09 kg

Length of wire,

= 1.0 m

Wavelength,

= 0.10 m

The velocity will be:

=  [tex]Frequency\times Wavelength[/tex]

=  [tex]890\times 0.10[/tex]

=  [tex]89 \ m/s[/tex]

Now,

The tension will be:

⇒  [tex]T=v^2\times \frac{M}{L}[/tex]

On substituting the values, we get

⇒      [tex]=(89)^2\times \frac{0.09}{1.0}[/tex]

⇒      [tex]=7921\times \frac{0.09}{1.0}[/tex]

⇒      [tex]=7921\times 0.09[/tex]

⇒      [tex]=713 \ N[/tex]

Answer:

[tex]0.0084575\ \text{T}[/tex]

[tex]0.272\ \text{m}[/tex]

2.2 cm easily observable

Explanation:

[tex]m_1[/tex] = Mass of 12 C = [tex]1.99\times 10^{-26}\ \text{kg}[/tex]

[tex]m_2[/tex] = Mass of 13 C = [tex]2.16\times 10^{-26}\ \text{kg}[/tex]

[tex]r_1[/tex] = Radius of 12 C = [tex]\dfrac{25}{2}=12.5\ \text{cm}[/tex]

B = Magnetic field

v = Velocity of atom = 8.5 km/s

[tex]r_2[/tex] = Radius of 13 C

The force balance of the system is

[tex]qvB=\dfrac{m_1v^2}{r}\\\Rightarrow B=\dfrac{m_1v}{rq}\\\Rightarrow B=\dfrac{1.99\times 10^{-26}\times 8500}{12.5\times 10^{-2}\times 1.6\times 10^{-19}}\\\Rightarrow B=0.0084575\ \text{T}[/tex]

The required magnetic field is [tex]0.0084575\ \text{T}[/tex]

Radius is given by

[tex]r=\dfrac{mv}{qB}[/tex]

[tex]r\propto m[/tex]

So

[tex]\dfrac{r_2}{r_1}=\dfrac{m_2}{m_1}\\\Rightarrow r_2=\dfrac{m_2}{m_1}r_1\\\Rightarrow r_2=\dfrac{2.16\times 10^{-26}}{1.99\times 10^{-26}}\times 12.5\times 10^{-2}\\\Rightarrow r_2=0.136\ \text{m}[/tex]

The required diameter is [tex]2\times 0.136=0.272\ \text{m}[/tex]

Separation is given by

[tex]2(r_2-r_1)=2(0.136-0.125)=0.022\ \text{m}[/tex]

The distance of separation is 2.2 cm which is easily observable.

Answer:

the Earth's crust is broken into about 12 plates that float on hotter, softer rocks in the underlying mantle

Explanation:

Answer:

E

A

C

B

D

Explanation:

Answer:

Energy May be measured in joule

Answer:

hello your question is poorly written and I have tried to understand it hence I will give a general diagram as related to your question

answer : attached below

Explanation:

Attached below is the required diagram of the situation you are trying to describe

let the wide receiver be ; W

Corner back = C

westward velocity = Vw

eastward velocity = Vc

at initial position  X = Xw

x = 0 ( initial position )

The specific heat of the solid phase is 0.333 joules per gram-degree Celsius.

The specific heat of the liquid phase is 0.593 joules per gram-degree Celsius.

In this case, we need to determine that specific heat for solid and liquid states of matter. By Heat Physics, we understand that specific heat is contained in the slopes of the two sensible phases in the following form:

[tex]\frac{\Delta T}{\Delta Q} = \frac{1}{m\cdot c}[/tex] (1)

Where:

Solid phase

If we know that [tex]m = 9.80\,g[/tex], [tex]T_{1} = 40\,^{\circ}C[/tex], [tex]T_{2} = 235\,^{\circ}C[/tex], [tex]Q_{1} = 183\,J[/tex] and [tex]Q_{2} = 819\,J[/tex], then the specific heat of the solid phase is:

[tex]c = \frac{\Delta Q}{m\cdot \Delta T}[/tex]

[tex]c = \frac{819\,J-183\,J}{(9.80\,g)\cdot (235\,^{\circ}C - 40\,^{\circ}C)}[/tex]

[tex]c = 0.333\,\frac{J}{g\cdot ^{\circ}C}[/tex]

The specific heat of the solid phase is 0.333 joules per gram-degree Celsius.

Liquid phase

If we know that [tex]m = 9.80\,g[/tex], [tex]T_{3} = 230\,^{\circ}C[/tex], [tex]T_{4} = 471\,^{\circ}C[/tex], [tex]Q_{3} = 1470\,J[/tex] and [tex]Q_{4} = 2870\,J[/tex], then the specific heat of the liquid phase is:

[tex]c = \frac{\Delta Q}{m\cdot \Delta T}[/tex]

[tex]c = \frac{2870\,J - 1470\,J}{(9.80\,g)\cdot (471\,^{\circ}C - 230\,^{\circ}C)}[/tex]

[tex]c = 0.593\,\frac{J}{g\cdot ^{\circ}C}[/tex]

The specific heat of the liquid phase is 0.593 joules per gram-degree Celsius.

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Answer:

Voltage and resistance

Answer:

Part A:

Distance=864000 m=864 km

Part B:

Energy Used=ΔE=8638000 Joules

Part C:

[tex]\frac{\triangle m}{m}=0.004998=0.49985\%[/tex]

Explanation:

Given Data:

v=20m/s

Time =t=12 hours

In Secs:

Time=12*60*60=43200 secs

Solution:

Part A:

Distance = Speed**Time

Distance=v*t

Distance= 20*43200

Distance=864000 m=864 km

Part B:

Energy Used=ΔE= Energy Required-Kinetic Energy of swans

Energy Required to move= Power Required*time

Energy Required to move=200*43200=8640000 Joules

Kinetic Energy=[tex]\frac{1}{2}mv^2[/tex]

[tex]K.E\ of\ Swans=\frac{1}{2} *10*(20)^2=2000\ Joules[/tex]

Energy Used=ΔE=8640000 -2000

Energy Used=ΔE=8638000 Joules

Part C:

Fraction of Mass used=Δm/m

For This first calculate fraction of energy used:

Fraction of energy=ΔE/Energy required to move

ΔE is calculated in part B

Fraction of energy=8638000/8640000

Fraction of energy=0.99977

Kinetic Energy=[tex]\frac{1}{2}mv^2[/tex]

Now, the relation between energies ratio and masses is:

[tex]\frac{\triangle E}{E}=\frac{\triangle m}{2m}v^2[/tex]

[tex]\frac{\triangle m}{m}=\frac{2}{v^2} *\frac{\triangle E}{E}\\\frac{\triangle m}{m}=\frac{2}{20^2} *0.99977[/tex]

[tex]\frac{\triangle m}{m}=0.004998=0.49985\%[/tex]

The nonmechanical energy or the loss of energy after the collision will be equal to  [tex]E=6\ J[/tex]

The conservation of the momentum is defined as when two bodies collide with each other then the total energy of the masses will remain constant.

It is given in the question that

Mass of the body X is  [tex]M_X=4\ kg[/tex]

The velocity of the body X is  =[tex]V_x=3 \ \frac{m}{s}[/tex]

Mass of the body Y is   [tex]M_Y= 2\ kg[/tex]

The velocity of body Y is [tex]V_y= 0[/tex]

Now to find out the energy converted after the collision we will first find the final velocity of both the bodies.

Now from the conservation of the momentum

[tex]M_X V_X +M_YV_Y=(M_X+M_Y)V_F[/tex]

[tex](4\times 3)+(2\times0)=(4+2)\times V_F[/tex]

[tex]V_F= \dfrac{12}{6}[/tex]

[tex]V_F = 2 \frac{m}{s}[/tex]

Now to find the Change in the energy of the body

[tex]\Delta E=E_i-E_f[/tex]

[tex]\Delta E=\dfrac{1}{2} M_XV_X^2+\dfrac{1}{2} M_YV_Y^2-\dfrac{1}{2} (M_X+M_Y)V_F^2[/tex]

[tex]\Delta E=\dfrac{1}{2} (4)(3^2)+\dfrac{1}{2} (2)(0^2)-\dfrac{1}{2} ( 4+2) 2^2[/tex]

[tex]\Delta E= 6\ J[/tex]

Thus the nonmechanical energy or the loss of energy after the collision will be equal to  [tex]E=6\ J[/tex]

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Answer:

No.

Explanation:

The scale actually measures the force that the object does against it, and that force is called the weight.

Such that if we have an object with mass M and we are on Earth, where the gravitational acceleration is g, the weight is:

W = M*g

Now, there is a unit called "kilogram-force"

Such that on Earth, an object that has a mass of 10 kilograms, weighs 10 kilograms-force.

Then from the weight measure, we can instantly know the mass of the object, but the thing that is being measured is the weight, not the mass.

Answer:

[tex]B_{int}=-0.015T[/tex]

Explanation:

From the question we are told that:

RF source oscillation speed [tex]\sigma= 34 MHz[/tex]

The external field [tex]Bext =0.78 T[/tex].

Pro- ton magnetic moment component [tex]\mu=1.41 X 10-26 J/T[/tex]

Generally the equation for magnitude of [tex]B_{int}[/tex] is mathematically given by

[tex]B_{int}=B_{ext}-\frac{h\triangle \sigma}{2 \mu}[/tex]

[tex]B_{int}=0.78-\frac{6.6*10^{-34}*34*10^6}{2*1.41*10^{26}}[/tex]

[tex]B_{int}=0.78-0.7957[/tex]

[tex]B_{int}=-0.015T[/tex]

Answer:

89 N downward.

Explanation:

Both forces are working in the same direction -- down.

Therefore you can add them together.

F = F1 + F2

F is the total force

F1 is the gravitational force

F2 is the force the person provides

F1 = 19

F2 = 70

F = ?

F = 19 + 70

F = 89 Newtons downward.

Answer:

wE dONt taLk aBt brunooooOoooOoOo ;')

Explanation:

Recall that

P = F v

where

P = power produced by the engine

F = magnitude of the component force supplied by the engine in the direction of the car's motion

v = speed of the car

Draw a free-body diagram for the car, and decompose all the forces acting on it into components that act parallel and perpendicular to the slope. The car moves parallel to the slope, so we only care about 2 forces: the parallel component of the car's weight, and the force provided by the engine.

By Newton's second law, since the car is moving at a constant speed,

∑ F = F - m g sin(12°) = 0

where m = 950 kg and g = 9.80 m/s². Solve for F :

F = m g sin(12°) ≈ 1935.658 N

The engine provides P = 6.5 × 10⁴ W, so the car's speed v is

v = P / F = (6.5 × 10⁴ W) / (1935.658 N) ≈ 33.6 m/s

which makes C : 34 ms⁻¹ the closest answer.

The Steady speed of a car moving up a hill, with a slope of 12° to the horizontal,  whose output power is 6.5×10⁴ W and whose mass is 950 kg is A. 7.0 m/s

Speed: This can be defined as the ratio of distance to the time travelled by a body. The S.I unit of speed is m/s

[tex]P = (mgcosx)V[/tex].................... Equation 1

Where P = output Power of the engine, m = mass of the car, g = acceleration due to gravity, x = angle of the hill with the horizontal, V = steady speed of the car.

[tex]V = P/mgcosx[/tex]............... Equation 2

From the question,

[tex]V = (6.5*10^{4} )/(950*9.8*cos12)[/tex]

[tex]V = 65000/9106.55[/tex]

[tex]V = 7.13 m/s[/tex]

From the question, the right option is A. 7.0 m/s

Hence the steady speed of the car is 7.0 m/s

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Answer:

E = 1.38 x 10⁸ eV = 138 MeV

Explanation:

The energy associated with the given wavelength can be found from the following formula:

[tex]E = \frac{hc}{\lambda}[/tex]

where,

E = Energy of electron = ?

h = Plank's Constant = 6.625 x 10⁻³⁴ J.s

c = Speed of Light = 3 x 10⁸ m/s

λ = wavelength = 9 fm = 9 x 10⁻¹⁵ m

Therefore,

[tex]E = \frac{(6.625\ x\ 10^{-34}\ J.s)(3\ x\ 10^8\ m/s)}{9\ x\ 10^{-15}\ m}\\\\E = (2.21\ x\ 10^{-11}\ J)(\frac{1\ eV}{1.6\ x\ 10^{-19}\ J})[/tex]

E = 1.38 x 10⁸ eV = 138 MeV

Answer:

1482.35J

Explanation:

Efficency=(output energy/input energy) x100

17.0 =(252/input energy) x100

input energy =100x252/17=1482.35J

wasted energy=1482.35-252=1230.35J

Answer: See explanation

Explanation:

The reason why astronomers think that the rates of impact for the Moon must have been higher earlier than 3.8 billion years ago is because on the older highlands, there are ten times more craters than on the younger maria.

It is believed that the impact rate was higher earlier and thus can be seen when the numbers of the craters that can be seen on the lunar highlands is being compared to that on the maria. It should be noted that there are about 10 times more craters that can be found on the highlands than those on the maria.

If there was a constant rate of impact throughout the history of the Moon, then the highlands be about 10 times older and therefore will have been formed about 38 billion years ago.

Answer:

moving

Explanation:

hope it helped!!!

Answer:

B

Explanation: