(20) A rocket is launched vertically. At time t = 0 seconds, the rocket’s engine shuts down. At the time, the rocket has reached an altitude of 500m and is rising at a velocity of 125 m/s. Gravity then takes over. The height of the rocket as a function of time is h(t)=-9.8/2 t^2+125t+500,t>0. Using your function file from HW2A: Generate a plot of height (vertical axis) vs. time (horizontal axis) from 0 to 30 seconds. Include proper axis labels. Find the maximum height and the time at which it occurs: Analytically, showing your steps and equations. This part should be done entirely in the write-up: no coding Using the data cursor on the plot. Using the MAX function on your data from part (a) Using FMINSEARCH on your m file Comment on the differences between the methods. How closely does each method match the "true" (analytical) value? Find the time when the rocket hits the ground: Analytically, showing your equations. This part should be done entirely in the write-up: no coding Using the data cursor on the plot. Using FZERO on your m file Comment on the differences between the methods in each of part (B) and (C). How closely does each method match the "true" (analytical) value? Use a quantitative comparison to make your argument.

Answer:

Explanation:

Emf e generated in a coil with no of turn n and area A rotating in a magnetic field B  with angular speed of ω is given by the expression

e = e₀ sinωt

where e₀ = nωAB which is the maximum emf generated

Putting the given values

15.7 = 1000xω x 20 x 10⁻² x 5

ω = .0157

frequency of rotation

= ω / 2π

= .0157 / 2 x 3.14

= .0025 /s

9 rotation / hour .

Answer:

a)  μ = 0.475 , b)   μ = 0.433

Explanation:

a) For this exercise of Newton's second law, we create a reference system with the x-axis parallel to the plane and the y-axis perpendicular to it

X axis

     Wₓ - fr = m a

the friction force has the expression

     fr = μ N

y Axis

     N - [tex]W_{y}[/tex] = 0

let's use trigonometry for the components the weight

     sin 27 = Wₓ / W

     Wₓ = W sin 27

     cos 27 = W_{y} / W

     W_{y} = W cos 27

     N = W cos 27

     W sin 27 - μ W cos 27 = m a

     mg sin 27 - μ mg cos 27 = m a

      μ = (g sin 27 - a) / (g cos 27)

      very = tan 27 - a / g sec 27

      μ = 0.510 - 0.0344

      μ = 0.475

b) now the block starts with an initial speed of 3m / s. In Newton's second law velocity does not appear, so this term does not affect the result, the change in slope does affect the result

         μ = tan 25 - 0.3 / 9.8 sec 25

         μ = 0.466 -0.03378

         μ = 0.433

Answer:

(a) Ra = 9.25 kN; Rb = 5.75 kN

(b) 26.7 kN

Explanation:

(a) Draw a free-body diagram of the crane.  There are four forces:

Reaction Ra pushing up at A,

Reaction Rb pushing up at B,

Weight force 12.5 kN pulling down at G,

and weight force 2.5 kN pulling down at F.

Sum of moments about B in the counterclockwise direction:

∑τ = Iα

-Ra (0.66 m + 0.42 m + 2.52 m) + 12.5 kN (2.52 m + 0.42 m) − 2.5 kN ((3.6 m + 0.9 m) cos 30° − 2.52 m) = 0

-Ra (3.6 m) + 12.5 kN (2.94 m) − 2.5 kN (1.38 m) = 0

Ra = 9.25 kN

Sum of moments about A in the counterclockwise direction:

∑τ = Iα

Rb (0.66 m + 0.42 m + 2.52 m) − 12.5 kN (0.66 m) − 2.5 kN ((3.6 m + 0.9 m) cos 30° + 0.66 m + 0.42 m) = 0

Rb (3.6 m) − 12.5 kN (0.66 m) − 2.5 kN (4.98 m) = 0

Rb = 5.75 kN

Alternatively, you can use sum of the forces in the y direction as your second equation.

∑F = ma

Ra + Rb − 12.5 kN − 2.5 kN = 0

Ra + Rb = 15 kN

9.25 kN + Rb = 15 kN

Rb = 5.75 kN

However, you must be careful.  If you make a mistake in the first equation, it will carry over to this equation.

(b) At the maximum weight, Ra = 0.

Sum of the moments about B in the counterclockwise direction:

∑τ = Iα

12.5 kN (2.52 m + 0.42 m) − F ((3.6 m + 0.9 m) cos 30° − 2.52 m) = 0

12.5 kN (2.94 m) − F (1.38 m) = 0

F = 26.7 kN

Answer:Velocity

Explanation:

Velocity is directly proportional to the frequency of a wave.

Velocity=frequency x wavelength

Carbon dioxide is the most abundant greenhouse gas in the atmosphere.

Answer:C

Explanation:

Carbon dioxide is the most abundance greenhouse gas in The atmosphere.

Answer:

The deceleration is  [tex]a = - 76.27 m/s^2[/tex]

Explanation:

From the question we are told that

   The height above  firefighter safety net is [tex]H = 14 \ m[/tex]

   The length by which the net is stretched is [tex]s = 1.8 \ m[/tex]

From the law of energy conservation

    [tex]KE_T + PE_T = KE_B + PE_B[/tex]

 Where [tex]KE_T[/tex] is the kinetic energy of the person before jumping which equal to zero(because to kinetic energy at maximum height )

   and  [tex]PE_T[/tex] is the potential energy of the before jumping  which is mathematically represented at

          [tex]PE_T = mg H[/tex]

and  [tex]KE_B[/tex] is the kinetic energy of the person just before landing on the safety net  which is mathematically represented at

        [tex]KE_B = \frac{1}{2} m v^2[/tex]

and  [tex]PE_B[/tex] is the potential energy of the person as he lands on the safety net which has a value of zero (because it is converted to kinetic energy )

   So the above equation becomes

          [tex]mgH = \frac{1}{2} m v^2[/tex]

=>           [tex]v = \sqrt{2 gH }[/tex]

    substituting values

                [tex]v = 16.57 m/s[/tex]

Applying the equation o motion

             [tex]v_f = v + 2 a s[/tex]

Now the final velocity is zero because the person comes to rest

      So

         [tex]0 = 16.57 + 2 * a * 1.8[/tex]

            [tex]a = - \frac{16.57^2 }{2 * 1.8}[/tex]

            [tex]a = - 76.27 m/s^2[/tex]

Answer:Reflection

Explanation:

The throwing back of a sound wave without absorbing it is called reflection

In acoustic reflection of sound is termed as echo i.e. sound arrived at the listener after a particular delay depending upon the position of barrier to the observer.

The reflection of sound is used in many devices like megaphone, trumpets, etc. It is also used in auditorium such that the ceiling of the auditorium is curved for multiple reflections of sound so that sound can be reached at every corner of the auditorium.

Answer:

B. add more loops of wire between the magnets

Explanation:

this would increase the magnetic force acting on the rod therefore increasing

By adding more loops of wire between the magnets the motor is changed to be able to lift a heavier box.

An electric motor is a mechanism that turns electricity into mechanical energy.

The interaction between the motor's magnetic field and electric current in a wire winding generates force in the form of torque imparted to the motor's shaft in most electric motors.

An electric generator is physically equivalent to an electric motor, but it converts mechanical energy into electrical energy using a reversed flow of power.

The load capacity in the motor can be increased by increasing the no of loops. So by adding more loops of wire between the magnets the motor is changed to be able to lift a heavier box.

Hence option B is correct.

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The rate of heat loss, in watts, does this achieve is 37.66 W

It leads in cooling since water absorbs heat equivalent to mass times latent heat of evaporation to get converted into vapor .

So,

latent heat of evaporation of water = 2260 x 10³ J / kg or 2260 J / g

Now

in the evaporation of 1 g of water , heat lost = 2260 J

And,

heat lost per minute = 2260 J

So,

heat lost per second = 2260 / 60

= 37.66 J /s

= 37.66 W

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Answer:

3.4 rad/sec^2

Explanation:

rotational inertia = 2.5 kg-m^2   radius = 1.1 m   force = 7.7 N

t = rotational inertia * angular acceleration    equation 1

also t = force * radius

therefore to calculate angular acceleration equation 1 becomes

f * r = inertia * angular acceleration   hence

angular acceleration = f * r / inertia = [tex]\frac{7.7 * 1.1 }{2.5}[/tex]   8.47 / 2.5 = 3.388 ≈ 3.4 rad/sec^2

Answer:

The time period for this pendulum is 1.68 seconds

Explanation:

Solution

Given that:

The length of the pendulum is measured from the axis of rotation to the center of mass of the bob of the pendulum

Now,

In this case, the length becomes:

L= 80 - 15+5

L = 70 cm

The time period = T = 2π √L/g

T = 2* 3.14 *√0.7/9.8

= 1.68 seconds

Note: Kindly find an attached work to the part of the solution of the given question

Answer:

a)5.88J

b)-5.88J

c)0.78m

d)0.24m

Explanation:

a) W by the block on spring is given by

W= [tex]\frac{1}{2}[/tex]kx² = [tex]\frac{1}{2}[/tex](530)(0.149)² =  5.88 J

b)  Workdone by the spring = - Workdone by the block = -5.88J

c) Taking x = 0 at the contact point we have U top = U bottom

So, mg[tex]h_o[/tex] = [tex]\frac{1}{2}[/tex]kx² - mgx

And, [tex]h_o[/tex]= ( [tex]\frac{1}{2}[/tex]kx² - mgx )/(mg) = [tex][\frac{1}{2} (530)(0.149^2)-(0.645)(9.8)(0.149)[/tex]]/(0.645x9.8)    

   [tex]h_o[/tex]=   0.78m            

d) Now, if the initial initial height of block is 3[tex]h_o[/tex]

[tex]h_o[/tex] = 3 x 0.78 = 2.34m

then, [tex]\frac{1}{2}[/tex]kx² - mgx - mg[tex]h_o[/tex] =0

[tex]\frac{1}{2}[/tex](530)x²  - [(0.645)(9.8)x] - [(0.645)(9.8)(2.34) = 0

265x² - 6.321x - 14.8 = 0  

a=265

b=-6.321

c=-14.8

By using quadratic eq. formula, we'll have the roots

x= 0.24 or x=-0.225

Considering only positive root:

x= 0.24m (maximum  compression of the spring)

Answer:

C. The left wire attracts the right wire and exerts as much force as the right wire does.

Explanation:

To know what is the answer you first take into account the magnetic field generated by each current, for a distance of d:

[tex]B_1=\frac{\mu_oI_1}{2\pi d}=\frac{\mu_o}{2\pi d}(1A)\\\\B_2=\frac{\mu_oI_2}{2\pi d}=\frac{\mu_o}{2\pi d}(2A)=2B_1\\\\[/tex]

Next, you use the formula for the magnetic force produced by the wires:

[tex]\vec{F_B}=I\vec{L}\ X \vec{B}[/tex]

if the direction of the L vector is in +k direction, the first wire produced a magnetic field with direction +y, that is, +j and the second wire produced magnetic field with direction -y, that is, -j (this because the direction of the magnetic field is obtained by suing the right hand rule). Hence, the direction of the magnetic force on each wire, produced by the other one is:

[tex]\vec{F_{B1}}=I_1L\hat{k}\ X\ B_2(-\hat{j})=I_1LB_2\hat{i}=(2A^2)\frac{L\mu_o}{2\pi d}\hat{i}\\\\\vec{F_{B2}}=I_2L\hat{k}\ X\ B_2(\hat{j})=I_2LB_1\hat{i}=-(2A^2)\frac{L\mu_o}{2\pi d}\hat{i}[/tex]

Hence, due to this result you have that:

C. The left wire attracts the right wire and exerts as much force as the right wire does.

Answer:

C. is always perpendicular to the surface of the conductor

Explanation:

On a charged conductor , electric charge is uniformly distributed on its surface . The lines of forces are also uniformly  distributed on all directions . They repel each other so they emerge perpendicular to the surface so that they do nor cut each other and at the same time they remain at maximum distance from each other.

Answer:

a) 1.248 rad/s

b) 13.728 m/s

c) 0.52 rad/s^2

d) 17.132m/s^2

Explanation:

You have that the angles described by a astronaut is given by:

[tex]\theta=0.260t^2[/tex]

(a) To find the angular velocity of the astronaut you use the derivative og the angle respect to time:

[tex]\omega=\frac{d\theta}{dt}=\frac{d}{dt}[0.260t^2]=0.52t[/tex]

Then, you evaluate for t=2.40 s:

[tex]\omega=0.52(2.40)=1.248\frac{rad}{s}[/tex]

(b) The linear velocity is calculated by using the following formula:

[tex]v=\omega r[/tex]

r: radius if the trajectory of the astronaut = 11.0m

You replace r and w and obtain:

[tex]v=(1.248\frac{rad}{s})(11.0m)=13.728\frac{m}{s}[/tex]

(c) The tangential acceleration is:

[tex]a_T=\alpha r\\\\\alpha=\frac{\omega^2}{2\theta}=\frac{(1.248rad/s)^2}{2(0.260(2.40s)^2)}=0.52\frac{rad}{s^2}[/tex]

(d) The radial acceleration is:

[tex]a_r=\frac{v^2}{r}=\frac{(13.728m/s)^2}{11.0m}=17.132\frac{m}{s^2}[/tex]

Answer:

The horizontal distance between two adjacent crests or troughs is known as the wavelength.

Answer: Wavelength

Explanation:

From crest to crest, it is one full wavelength

Answer:

Here's your answer :

hope it helps!

Answer:

3) False. It is expensive since it requires sophisticated equipment and very low temperatures

Explanation:

Nuclear magnetic resonance imaging measurements consist of magnetic resonance imaging to analyze tissues by the transition of the unpaired electron at carbon 13, giving information on the structure and composition of tissues. This information is processed in computers and transformed into images.

So the physical measurement is the MRN

Now we can analyze the statements in the problem

1) True by itself a magnetic measurement is non-invasive

2) True. Measuring carbon transitions has information about the soft tissue of the body

3) False. It is expensive since it requires sophisticated equipment and very low temperatures

4) Right. The applied magnetic field is high to be able to induce carbon transaction

Answer:

true. b, c and d

Explanation:

Let's review each statement separately  

a) False. The kinematics studies the position, speed and acceleration of the bodies, but not what causes these changes

b) True. Velocity is the displacement between time, displacement is a vector, and time is a scalar, so the division between them gives a vector

c) True. speed is the displacement that is a length between time, so its unit is length / time

d) true  desaceleration = - aceleration

Answer:

An atom is the smallest constituent unit of ordinary matter that constitutes a chemical element. Every solid, liquid, gas, and plasma is composed of neutral or ionized atoms. Atoms are extremely small; typical sizes are around 100 picometers.Explanation:

An atom is the smallest constituent unit of ordinary matter that constitutes a chemical element.

Every solid, liquid, gas, and plasma is composed of neutral or ionized atoms. Atoms are extremely small; typical sizes are around 100 picometers.

Each atom is made up of a nucleus and one or more electrons that are linked to it. One or more protons and a significant number of neutrons make up the nucleus. Only the most prevalent type of hydrogen is neutron-free.

Atoms that are neutral or ionized make up every solid, liquid, gas, and form of plasma. Atoms are incredibly tiny, measuring typically 100 picometers across. The nucleus of an atom contains more than 99.94% of its mass.

Therefore, An atom is the smallest constituent unit of ordinary matter that constitutes a chemical element.

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#SPJ2

Answer:

F = (913.14 , 274.87 )

|F| = 953.61 direction 16.71°

Explanation:

To calculate the resultant force you take into account both x and y component of the implied forces:

[tex]\Sigma F_x=480N+513Ncos(32.4\°)=913.14N\\\\\Sigma F_y=513sin(32.4\°)=274.87N[/tex]

Thus, the net force over the body is:

[tex]F=(913.14N)\hat{i}+(274.87N)\hat{j}[/tex]

Next, you calculate the magnitude of the force:

[tex]F=\sqrt{(913.14N)+(274.87N)^2}=953.61N[/tex]

and the direction is:

[tex]\theta=tan^{-1}(\frac{274.14N}{913.14N})=16.71\°[/tex]

Answer:

option C is correct

................

Answer:

C- in an inelastic collision, both momentum & kinetic energy are conserved

Explanation:

Took the test

Answer:

The answer is diffraction

Explanation:

Answer:

The answer is diffraction

Explanation:

I did the test! HOPE THIS HELPS!

Answer:

......................

Answer:

Explanation:

Force between two charges of q₁ and q₂ at distance d is given by the expression

F = k q₁ q₂ / d₂

Here force between charge q₁ = - 15 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C when distance between them d = (1.66 - 1.24 ) = .42 mm

k = 1/ 4π x 8.85 x 10⁻¹²

putting the values in the expression

F = 1/ 4π x 8.85 x 10⁻¹²  x - 15 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 9 x 10⁹ x  - 15 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 35969.4 x 10⁻³ N .

force between charge q₂ =  34.5 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C when distance between them d = ( 1.24 - 0 ) = 1.24 mm .

putting the values in the expression

F = 1/ 4π x 8.85 x 10⁻¹²  x  34.5 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 9 x 10⁹ x  - 34.5 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 82729.6  x 10⁻³ N

Both these forces will act in the same direction towards the left (away from the origin towards - ve x axis)

Total force = 118699 x 10⁻³

= 118.7 N.

Answer:

A) 0.188

B) 53.1 ⁰

Explanation:

taking moment about 0

∑ Mo = Lo∝

mg 1/2 sin∅ = 1/3 m L^2∝

note ∝ = w[tex]\frac{dw}{d}[/tex]∅

forces acting along t-direction ( ASSUMED t direction)

∑ Ft = Ma(t) = mr∝

mg sin ∅ - F = m* 1/2 * 3g/2l sin∅

therefore F = mg/4 sin∅

forces acting along n - direction ( ASSUMED n direction)

∑ Fn = ma(n) = mr([tex]w^{2}[/tex])

= mg cos∅ - N = m*1/2*3g/1 ( 1 - cos∅ )

hence N = mg/2 ( 5cos∅ -3 )

A ) Angle given = 30⁰c find coefficient of static friction

∪ = F/N

  = [tex]\frac{\frac{mg}{4}sin30 }{\frac{mg}{2}(5cos30 -3) }[/tex]  = 0.188

B) when there is no slip

N = O

   = 5 cos ∅ -3 =0

   therefore cos ∅ = 3/5  hence ∅ = 53.1⁰

Answer:

A i. E = 9.62 × 10⁻⁷ J/s

ii. The absorbed dose is 4.81 × 10⁻⁶ Gy

iii. The equivalent dose is  3.37 × 10⁻⁴ rem/s

iv.  t = 593471.81 seconds

B. i. 4.025 × 10¹⁵/s

ii. 0.512 mW

C. 7218092.2 seconds

D. i. 6.3 × 10⁻¹ J

ii. 1.4 × 10⁻² W

iii. 1.57 × 10³ Curie

E. 0.129 Ω

Explanation:

The given parameters are;

Mass of tumor = 0.20 kg

Activity of Cobalt-60 = 2.60 × 10⁻⁴ Ci

Photon energy = 1.25 MeV

(i) The energy, E, delivered to the tumor is given by the relation;

[tex]E = \frac{1}{2}\left (Number \, of \, decay / seconds \right )\times \left (Energy \, of \, photon \right )[/tex]

[tex]E = \frac{1}{2}\left (2.6\times 10^{-4}Ci )\times \left (\frac{3.70\times 10^{10}decays/s}{1 Ci} \right )\times 1.25\times 10^{6}eV\times \frac{1.6\times 10^{-19}J}{1eV}[/tex]

E = 9.62 × 10⁻⁷ J/s

(ii) The equation for absorbed dose is given as follows;

Absorbed dose, D, in Grays Gy = (Energy Absorbed Joules J)/Mass kg

Therefore, absorbed dose = (9.62 × 10⁻⁷ J/s)/( kg) = 4.81 × 10⁻⁶ Gy

1 Gray = 100 rad

4.81 × 10⁻⁷ Gy = 100 × 4.81 × 10⁻⁶ = 4.81 × 10⁻⁴ rad/s

(iii) Equivalent dose, H, is  given by the relation;

H = D × Radiation factor, [tex]w_R[/tex]

∴ H = 0.7 × 4.81 × 10⁻⁴ rad/s = 3.37 × 10⁻⁴ Sv = 3.37 × 10⁻⁴ rem/s

(iv) The exposure time required for an equivalent dose of 200 rem is given as follows;

[tex]\dot{H} = \dfrac{H}{t}[/tex]

Therefore;

[tex]t= \dfrac{200}{{3.37 \times 10^{-4}} } = 593471.81 \, s[/tex]

∴ t = 6.9 days

B. The number of electrons ejected is given by the relation;

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

[tex]N = \dfrac{2.0 \times 10^{-3} \times 400 \times 10^{-9}}{6.626 \times 10^{-34} \times 3 \times 10^8} = 4.025 \times 10^{15}/s[/tex]

(ii) The power carried by the electron

The energy carried away by the electrons is given by the relation;

[tex]KE_e = hv - \Phi[/tex]

[tex]KE_e = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} - 2.31 \times \frac{1.6 \times 10 ^{-19} }{1}[/tex]

[tex]KE_e = 4.9695 \times 10^{-19} - 3.696 \times 10 ^{-19} = 1.2735 \times 10^{-19} J[/tex]

Power, P[tex]_e[/tex], carried away by the electron = 4.025 × 10¹⁵ × 1.2735 × 10⁻¹⁹ = 0.512 mW

C. The given parameters are;

d = 1.19 mm, ∴ r = 1.19/2 = 0.595 × 10⁻³ m

l = 50 mm = 5 × 10⁻³ m

V = 500 ml = 5 × 10⁻⁴ m³

η = 0.0027 Pa

p = 1,900 Pa.

[tex]\dfrac{V}{t} = \dfrac{\pi }{8} \times \dfrac{P/l}{\eta } \times r^4[/tex]

[tex]t = \dfrac{8\times \eta\times V\times l }{\pi \times P \times r^4}[/tex]

[tex]t = \dfrac{8\times 0.0027 \times 5 \times 10^{-4} \times 5 \times 10^{-2} }{\pi \times 1900 \times (0.595 \times 10^{-4} )^4}[/tex]

t = 7218092.2 seconds

D) i. Energy absorbed is given by the relation;

E = m×D

Where:

D = 35 Gray = 35 J/kg

m = 18 g = 18 × 10⁻³ kg

∴ E = 35 × 18 × 10⁻³ = 6.3 × 10⁻¹ J

ii. Total time for treatment = 15 × 5 = 75 minutes

Energy absorbed = 6.3 × 10⁻¹ × 100 = 63 J

Power = Energy(in Joules)/Time (in seconds)

∴ Power = 63/(75×60) = 1.4 × 10⁻² W

iii. Whereby the power is provided by 0.5% of the photons emitted by the source, we have;

[tex]P_{source}= \frac{P_{beam}}{0.005} =\frac{0.0014}{0.005} =0.28 \, W[/tex]

1 MeV = 1.60218 × 10⁻¹³ J

0.03 MeV = 0.03 × 1.60218 × 10⁻¹³ J = 4.80654 × 10⁻¹⁵ J/photon

Therefore, the number of disintegration per second = 0.28 J/s ÷  4.80654 × 10⁻¹⁵ J/photon = 5.83 × 10¹³ disintegrations per second

1 Curie = 3.7 × 10¹⁰  disintegrations per second

Hence, 5.83 × 10¹³ disintegrations per second = (5.83 × 10¹³)/(3.7 × 10¹⁰) Curie

= 1.57 × 10³ Curie

E. The parameters given are;

Density of water = 1000 kg/m³

Volume of water = 250 ml = 0.00025 m³

Initial temperature, T₁, = 25°C

Final temperature, T₂, = 100°C

Change in temperature, ΔT = 100 - 25 = 75°

Specific heat capacity of the water = 4200 J/kg/°C

Mass of water = Density × Volume = 1000 × 0.00025 = 0.25 kg

∴ Heat supplied = 4200 × 0.25 × 75 = 78,750 J

Time to heat the water = 45.0 sec

Therefore, power = Energy/time = 78750/45 = 1750 W

The formula for electrical power = I²R =VI = V²/R

Therefore, where V = 15.0 V, we have;

15²/R = 1750

R = 15²/1750 = 0.129 Ω.

The resistance of the heater = 0.129 Ω.

Answer:30

Explanation:

Current=60 milliamps

Current=(voltage)/(resistance)

60=(voltage)/(resistance)

Doubling the resistance means multiplying both sides by 1/2

60x1/2=(voltage)/(resistance) x 1/2

30=(voltage)/2(resistance)

Therefore the resistance would be 30 milliamp if we double the resistance

Answer:

Materials that allow electricity to pass through them are called conductors.

Copper wire is a good conductor. Materials that do not allow electricity to pass through them are called insulators.

Explanation:

Can I have brainliest?

Answer:I=V/R

Explanation:

V=IR

Divide both sides by R

V/R=IR/R

V/R=I

I=V/R

The voltage in a circuit is given by the equation V= IR, in this equation v is the voltage Iis correct and R is the resistance the solution for the current is given as follows,

V= IR

I = V/R

Resistance is the obstruction of electrons in an electrically conducting material. The mathematical relation for resistance can be understood with the help of the empirical relation provided by Ohm's law.

V=IR

As for the given problem  if we have to solve for the current from the equation V=IR

V=IR

I = V/R

Let us suppose a 60-volt battery connected in a closed circuit with a resistor of 15 ohms then we have o find out the amount of current flowing in the circuit,

Voltage = 60V

Resistance = 15 Ohm

Current =?

By using Ohm,s Law,

V=IR

I = V/R

By substituting the respective values,

I = 60/15

I = 4 Ampere

Hence, we solved for the current from the equation V=IR.

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