Can someone please help!!!

Answer:

Explanation:

Area of the loop = π x ( 55 x 10⁻³ )²

= 9.5 x 10⁻³ m²

Change in Magnetic flux dφ = 450 x 10⁻³ - 350 x 10⁻³ = 150x 10⁻³ Weber.

time dt =.10 s

emf induced = dφ / dt = 150x 10⁻³ Weber / .10 s

= 1.5 V .

b )

Magnetic field is directed outwards and it is increasing so according to Lenz's law , direction of induced current will be clockwise in the loop.

Answer:

(a) 9.5 mV

(b) clockwise

Explanation:

Radius, r = 55 mm

Time, t = 0.1 s

Change in magnetic field, B = 450 - 350 = 100 mT =0.1 T

(a) induced emf is given by

[tex]e = A \frac{dB}{dt}[/tex]

[tex]e = A \frac{dB}{dt}\\\\e=3.14\times 0.055\times0.055\times \frac{0.1}{0.1}\\\\e= 9.5 \times 10^{-3} V = 9.5 mV[/tex]

(b) According to the Lenz law, the direction of current is clockwise.

Answer:

19,224 N

Explanation:

The given parameters are;

The mass limit of the elevator = 2,400 kg

The maximum ascent speed = 18.0 m/s

The maximum descent speed = 10.0 m/s

The maximum acceleration = 1.80 m/s²

Given that the acceleration due to gravity, g ≈ 9.81 m/s²

The minimum upward force that the elevator cable exert on the elevator car, [tex]F_{min}[/tex] , is given in the downward motion as follows;

[tex]F_{min}[/tex] = m·g - m·a

∴ [tex]F_{min}[/tex] = 2,400 kg × 9.81 m/s² - 2,400 kg × 1.80 m/s² = 19,224 N

The minimum upward force that the elevator cable exert on the elevator car, [tex]F_{min}[/tex] = 19,224 N

Answer:

[tex]T=8.1N[/tex]

Explanation:

From the question we are told that:

Mass m=0.40

Radius r=1.8m

Angle Beneath the Horizontal \theta =40 \textdegree

Speed v=5.0m/s

The Tension Angle

 [tex]\alpha=90-\theta\\\\\alpha=90-40[/tex]

 [tex]\alpha=50 \textdegree[/tex]

Generally the equation for Tension is is mathematically given by

 [tex]T=\frac{mv^2}{r}+mgcos \alpha[/tex]

 [tex]T=\frac{0.40*5^2}{1.8}+0.40*5cos50[/tex]

 [tex]T=8.1N[/tex]

Answer:

8 cm

Explanation:

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

Initial velocity (u) = 20 cm/s

Height (h) = 80 cm

Horizontal distance (s) =?

Next, we shall determine the time taken for marble to get to the ground. This can be obtained as follow:

Height (h) = 80 cm

Acceleration due to gravity (g) = 1000 cm/s²

Time (t) =?

t = √(2h/g)

t = √[(2 × 80)/1000]

t = √(160/1000)

T = √0.16

t = 0.4 s

Finally, we shall determine the horizontal distance travelled by the marble. This can be obtained as illustrated below:

Initial velocity (u) = 20 cm/s

Time (t) = 0.4 s

Horizontal distance (s) =?

s = ut

s = 20 × 0.4

s = 8 cm

Thus, the horizontal distance travelled by the marble is 8 cm.

The horizontal distance traveled by the marble is 8 cm.

The given parameters;

The time of motion of the marble is calculated as follows;

[tex]h = ut + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 0.8}{9.8} }\\\\t = 0.4 \ s[/tex]

The horizontal distance traveled by the marble is calculated as follows;

[tex]X = v_0_x t\\\\X = (20 \times 0.4)\\\\X = 8 \ cm[/tex]

Thus, the horizontal distance traveled by the marble is 8 cm.

Learn more here:https://brainly.com/question/2411455

Answer:

the correct one is D

Explanation:

The velocity of the center of mass of a body is

          [tex]v_{cm} = \frac{1}{M } \ \sum m_i v_i[/tex]vcm = 1 / M sum mi vi

          M = [tex]\sum m_i[/tex] sum my

in this case we have two bodies

          v_{cm} = [tex]\frac{1}{m1+m2} \ ( m_1v_1 +m_2v_2)[/tex]1 / m1 + m2 (m1 v1 + m² v2)

          v_{cm} = [tex]\frac{m_1}{m1+m2} \ v_1 + \frac{m2}{m_1+m_2} \ v_2[/tex]

we can see that the velocity is an average between v₁ and v₂, but the exact value depends on the ratio of the masses of the bodies

when checking the answers the correct one is D

Answer:

b) 450 V

Explanation:

We are given that

Total charge, q=10nC=[tex]10\times 10^{-9} C[/tex]

[tex]1nC=10^{-9}C[/tex]

Radius, r=20 cm=[tex]\frac{20}{100}=0.2m[/tex]

1 m=100 cm

x=15 cm=0.15 cm

We have to find the electrical potential 15 cm above  the center of a uniform charge density disk .

We know that

[tex]\sigma=\frac{q}{A}=\frac{q}{\pi r^2}[/tex]

[tex]\sigma=\frac{10\times10^{-9}}{3.14\times (0.2)^2}[/tex]

Where [tex]\pi=3.14[/tex]

[tex]\sigma=7.96\times 10^{-8}C/m^2[/tex]

Electric potential,[tex]V=\frac{\sigma}{2\epsilon_0}(\sqrt{x^2+r^2}-x)[/tex]

Where [tex]\epsilon_0=8.85\times 10^{-12}[/tex]

Using the formula

[tex]V=\frac{7.96\times 10^{-8}}{2\times 8.85\times 10^{-12}}(\sqrt{(0.15)^2+(0.2)^2}-0.15)[/tex]

[tex]V=449.7 V\approx 450V[/tex]

Hence, option b is correct.

Answer:

The potential is given by 449.7 V.

Explanation:

radius of disc, R = 20 cm = 0.2 m

distance, x = 15 cm = 0.15 m

charge, q = 10 nC

surface charge density

[tex]\sigma = \frac{q}{\pi R^2}\\\\\sigma = \frac{10\times 10^{-9}}{3.14\times 0.2\times 0.2 }\\\\\sigma = 7.96\times 10^{-8} C/m^2[/tex]

The electric potential is given by

[tex]V=\frac{\sigma}{2\varepsilon 0}\left ( \sqrt{R^2 + x^2} - x \right )\\\\V = \frac{7.96\times 10^{-8}}{2\times 8.85\times 10^{-12}}\left ( \sqrt{0.2^2 + 0.15^2} - 0.15 \right )\\\\V = 449.7 V[/tex]

Answer:

the instantaneous velocity is 51 m/s

Explanation:

Given;

acceleration, a = 2 + 5t²

Acceleration is the change in velocity with time.

[tex]a = \frac{dv}{dt} \\\\a = 2 + 5t^2\\\\The \ acceleration \ (a) \ is \ given \ so \ we \ have \ to \ find \ the \ velocity \ (v)\\\\To \ find \ the \ velocity, \ integrate\ both \ sides \ of \ the \ equation\\\\2 + 5t^2 = \frac{dv}{dt} \\\\\int\limits^3_0 {(2 + 5t^2)} \, dt = dv\\\\v = [2t + \frac{5t^3}{3} ]^3_0\\\\v = 2(3) + \frac{5(3)^3}{3} \\\\v = 6 + 5(3)^2\\\\v = 6 + 45\\\\v = 51 \ m/s[/tex]

Therefore, the instantaneous velocity is 51 m/s

Answer:

The displacement is constant when the rate of change in velocity is constant, for example, if a car’s speed increases by 5 km every 5 minutes it is called constant displacement

1000 watts

Joules are a measure of energy; Watts are a measure of the rate of energy generated or consumed. A Watt is defined as one joule per second. So if your 1000 Joules flowed in one second, the power would be 1000 Watts or 1 kW. But if it took 1000 seconds for the 1000Joules to flow, the power would be 1 Watt.

Answer:

option (b) is the correct answer

Answer:

B = 4.5 x 10⁻⁷ T = 0.45 μT

Explanation:

From Ampere's law the magnetic field at a certain radius from straight current-carrying wire is given by the following formula:

[tex]B = \frac{\mu_oI}{2\pi r}[/tex]

where,

B = magnetic field = ?

[tex]\mu_o[/tex] = permeability of free space = 4π x 10⁻⁷ H/m

I = current = 180 A

r = radius = 80 m

Therefore,

[tex]B=\frac{(4\pi\ x\ 10^{-7}\ H/m)(180\ A)}{2\pi (80\ m)}\\\\[/tex]

B = 4.5 x 10⁻⁷ T = 0.45 μT

Answer:

The final speed of the bullet is 480.4 m/s.

Explanation:

mass of bullet, m = 1.9 g

initial speed, u = 500 m/s

thickness, d = 0.75 inch = 0.01905 m

Force, F = 960 N

Let the final speed is v.

According to the work energy theorem,

Work = change in kinetic energy

[tex]W = F d = 0.5 m{\left (v^2 - u^2 \right )}[/tex]

-960 x 0.01905 = 0.5 x 0.0019 x (v^2 - 500 x 500)

-18.288 = 0.00095 (v^2 - 250000)

v = 480.4 m/s

Answer:

energy

Explanation:

I am not sure but I think it's electric energy

Answer:

i can help you i know this answer

Answer: the side two are 50 then the other two are 140  i thank

Explanation:

Answer:

0.83 J of work

Explanation:

2 J of work is required to stretch a spring from 34cm to 46cm

So that is 12cm stretched with 2 J of work

We can make that 6cm for 1 J of work

So, we need the find the work for stretching 36cm to 41cm

Which is 5cm

So, What is the work required to stretch 5cm?

1 J of work for 6cm

x work for 5cm

So, by proportion method

1 : 6 :: x : 5

6 * x = 1 * 5

6x = 5

x = 5/6

= 0.83

So to stretch 36cm to 41cm we need 0.83 J of work

Answer:

The speed of car A before collision is 3.5 km/h.

Explanation:

Mass of car A = 690 kg eastwards

Mass of car B = 520 kg at 74 km/h west wards

Distance, s = 6 m

coefficient of friction = 0.75

Let the speed after collision is v.

Use third equation of motion

[tex]v^2 = u^2 + 2 as \\\\0 =v^2- 2 \times 0.75\times9.8\times 6\\\\v = 9.4 m/s = 33.84 km/h[/tex]

Let the initial speed of car A is v'.

Use conservation of momentum

690 x v' - 520 x 74 = (690 + 520) x 33.8

690 v' + 38480 = 40898

v' = 3.5 km/h

Answer:

  f = 276.6 Hz

Explanation:

This musical instrument can be approximated to a tube system where each tube has one end open and the other closed.

In the closed part there is a node and in the open part a belly or antinode. Therefore the wavelength is

          L = λ/ 4

speed is related to wavelength and frequency

          v = λ f

          λ = v / f

we substitute

          L = v / 4f

          f = v / 4L

the speed of sound at 20ºC is

          v = 343 m / s

let's calculate

          f = [tex]\frac{343 }{4 \ 0.31}[/tex]

          f = 276.6 Hz

Answer:

the magnetic field must double

Explanation:

For this exercise we use Ampere's law

         ∫ B . ds = μ₀ I

Where the bold indicate vectors

With this expression we can see that if we double the current, keeping the same trajectory, the magnetic field must double

Answer:

The gauge pressure is equal to 147 kPa.

Explanation:

The pressure exerted by fluid is given by :

[tex]P=\rho gh[/tex]

Where

[tex]\rho[/tex] is density of water

h is height

So, put all the values,

[tex]P=1000\times 9.8\times 15\\\\P=147000\ Pa[/tex]

or

P = 147 kPa

So, the gauge pressure is equal to 147 kPa.

Answer:

The gauge pressure is 147000 Pa.

Explanation:

Height, h = 15 m

density of water, d= 1000 kg/m^3

gravity, g = 9.8 m/s^2

The gauge pressure is the pressure exerted by the fluid.

The pressure exerted by the fluid is given by

P  = h d g

P = 15 x 1000 x 9.8 =  147000 Pa

Answer:

220V a.c is more dangerous than 220V d.c because of the peak voltage of 220V a.c. which is much larger.

Answer:

[tex]B_1=1.29967T[/tex]

Explanation:

From the question we are told that:

Relative permeability Platinum [tex]\mu= 1.00026[/tex]

Magnetic field [tex]B_e= 1.3000 T[/tex]

Generally the equation for Magnetic field  is mathematically given by

[tex]B_i=\frac{B_e}{\mu}[/tex]

[tex]B_i=\frac{1.3}{1.00026}[/tex]

[tex]B_1=1.29967T[/tex]

Answer:

 A_resulting = 0.2 m

Explanation:

Let's analyze the impact of the pulse with the pole, this is a fixed obstacle that does not move therefore by the law of action and reluctant, the force that the pole applies on the rope is of equal magnitude to the force of the rope on the pole (pulse), but opposite directional, so the reflected pulse reverses its direction and sense.

With this information we analyze a point on the string where the incident pulse is and each reflected with an amplitude A = 0.1 m, the resulting is

           A_res = 2A

           A_resultant = 2 .01

           A_resulting = 0.2 m

Explanation:

pls write full question

Answer:

Where is the figure ?????

In vacuum, going at 2.99×10^8 m/s.

Answer:

The velocity becomes [tex]v\sqrt 2[/tex].

Explanation:

The force acting on the bobber is centripetal  force.

The centripetal force is given by

[tex]F =\frac{mv^2}{r}[/tex]

when mass remains same, radius is doubled and the force is same, so the velocity is v'.

[tex]F =\frac{mv^2}{r}=\frac{mv'^2}{2r}\\\\v'=v\sqrt 2[/tex]

Answer:

b. parallel

Explanation:

Answer:

f = 106.3 N

Explanation:

The force applied on the sled must be equal to the static frictional force to move the sled:

Tension Force Horizontal Component = Static Frictional Force

[tex]TCos\theta = \mu W\\TCos\theta = \mu mg[/tex]

where,

T = Tension = 120 N

θ = angle of rope = 37°

μ = coefficient of static friction = ?

m = mass of children plus sled = 55 kg

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex](120\ N)Cos\ 37^o = \mu (55\ kg)(9.81\ m/s^2)\\\\\mu = \frac{95.84\ N}{(55\ kg)(9.81\ m/s^2)}\\\\\mu = 0.18[/tex]

Now, the static friction acting on the mother will be:

[tex]f = \mu mg = (0.18)(61\ kg)(9.81\ m/s^2)\\[/tex]

f = 106.3 N

Answer:

the final steady temperature of the mixture is 46.67 ⁰C

Explanation:

Given;

mass of first water,m₁ = 6 g = 0.006 kg

initial temperature, t₁ = 50 ⁰C

mass of the second water, m₂ = 3.0 g = 0.003 kg

initial temperature of the second water, t₂ = 40 ⁰C

specific heat capacity of water, C = 4,200 J/kg.K

Let the final temperature of the mixture = T

Based on Law of conservation of energy, the temperature of the final mixture is calculated as follows;

m₁c(t₁ - T) = m₂c(T - t₂)

0.006 x 4200 x (50 - T) = 0.003 x 4200 x (T - 40)

1,260 - 25.2T = 12.6T - 504

1,260 + 504 = 12.6T + 25.2T

1764 = 37.8T

T = 1764/37.8

T = 46.67 ⁰C

Therefore, the final steady temperature of the mixture is 46.67 ⁰C