The force exerted by the wind on the sails of a sailboat is 390 N north. The water exerts a force of 180 N east. If the boat (including its crew) has a mass of 270 kg, what are the magnitude and direction of its acceleration? Serway, Raymond A.; Jewett, John W.. Physics for Scientists and Engineers (MindTap Course List) (Page 121). Cengage Learning. Kindle Edition.

Answer:

F = 3482.9 N

Explanation:

Change in velocity of the Parachutist is given as

[tex]v_f = 15 mph = 6.675 m/s[/tex]

[tex]v_i = 120 mph = 53.4 m/s[/tex]

now it is given as

[tex]\Delta v = v_f - v_i [/tex]

[tex]\Delta v = 120 - 15 = 105 mph[/tex]

[tex]\Delta v = 46.7 m/s[/tex]

now the acceleration of the parachutist is given as

[tex]a  = \frac{v_f^2 - v_i^2}{2d}[/tex]

distance moved by the parachutist is given as

[tex]d = 120 ft = 36.576 m[/tex]

now we have

[tex]a = \frac{6.675^2 - 53.4^2}{2(36.576)}[/tex]

[tex]a = - 38.4m/s^2[/tex]

Now the mass of parachutist is given as

[tex]m = 200 lb = 90.7 kg[/tex]

now we have

[tex]F = ma[/tex]

[tex]F = (90.7 kg)(38.4) = 3482.9 N[/tex]

Answer:

Ek= 35.392 kJ or 35.392 10³ J

Explanation:

Formula for Kinetic energy:

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

since m= 65kg and v= 33m/s

Ek= [tex]\frac{1}{2}[/tex]× 65×[tex]33^{2}[/tex]

Ek= 35.392 kJ or 35.392 10³ J

Final answer:

The kinetic energy of a 65 kg cheetah running at its top speed of 33 m/s is calculated using the formula KE = 1/2 mv², resulting in a kinetic energy of 35442.5 Joules.

Explanation:

The question asks for the kinetic energy of a 65 kg cheetah running at its top speed, which is given to be 33 m/s. To calculate kinetic energy, we use the formula KE = 1/2 mv², where m is the mass of the object (in this case, the cheetah) and v is the velocity of the object. Substituting the given values, we get:

KE = 1/2 × 65 kg × (33 m/s)²

KE = 1/2 × 65 × 1089

KE = 1/2 × 70885

KE = 35442.5 Joules

Therefore, the kinetic energy of a 65 kg cheetah running at its top speed of 33 m/s is 35442.5 Joules.

Answer:

544.65 ohm and 179.84 ohm

Explanation:

Hello

Let

Resistor 1 (R1) and Resistor 2 (R2)

in series [tex]R_{eq} =R1+R2[/tex]      

in parallel

[tex]\frac{1}{R_{eqp} } =\frac{1}{R1} +\frac{1}{R2}\\\\R1=R_{eqs} -R2\\\\R1 =724.5-R2 (equation 1)\\ R_{eqp}}=\frac{R1*R2}{R1+R2}\\\\\ 135.2 = \frac{R1*R2}{R1+R2}(equation 2)\\\\\\\\replacing 1 in \ 2\\\\probema data[/tex]

[tex]135.2=\frac{(724.5-R2)(R2)}{724.5-R2+R2} }\\135.2=\frac{(-R_{2} ^{2}+724.5R2) }{724.5}\\ 135.2*724.5=-R_{2} ^{2}+724.5R_{2}\\\\R_{2} ^{2} -724.5R_{2}  +97952.4 =0\\R_{2} =\frac{724.5 \±\sqrt{(-724.5^{2})-4(1)(97952.4) } }{2(1)}\\R_{2} =\frac{724.5\±\sqrt{(133092.25) } }{2(1)}\\\\R_{2} =\frac{724.5+364.81810}{2} \\R_{2} =544.65 ohm\\\\R_{2} =\frac{724.5-364.81810}{2} =179.84 ohm[/tex]\\\\

let R2=544.65 and replace in equation 1

R1=724.5-544.65

R1=179.85

so, the resistors are 544.65 ohm and 179.84 ohm

Have a  great day

Answer:

The value of 2.2 radians in degree is 126.11°.

Explanation:

Given that,

Angle = 2.2 radians

We need to calculate the value in degrees

We know that,

[tex]1\ radians = \dfrac{180}{\pi}[/tex]

[tex]2.2\ radians = \dfrac{180}{3.14}\times2.2[/tex]

[tex]2.2\ radians = 126.11^{\circ}[/tex]

Hence, The value of 2.2 radians in degree is 126.11°.

The person must exert a force of 8.32 N to accelerate the lawn mower from rest to 1.3 m/s in 2.5 seconds.

When an object is moving at a constant speed, the net force acting on it is zero.

In this case, the force applied by the person along the handle is balanced by the frictional force opposing the motion.

So, the frictional force [tex](\(F_{\text{friction}}\))[/tex] is equal in magnitude but opposite in direction to the force applied by the person.

Given:

- Force applied by the person  = 87.5 N

- The angle of the handle = 45.0 degrees to the horizontal

To find the horizontal component of the force applied by the person, use trigonometry:

[tex]\[F_{\text{person-horizontal}} = F_{\text{person}} \cdot \cos(\theta)\][/tex]

[tex]\[F_{\text{person-horizontal}} = 87.5 \, \text{N} \cdot \cos(45.0^\circ)\][/tex]

                          [tex]= 87.5 \, \text{N} \cdot 0.707 = 61.29 \, \text{N}[/tex]

So, the person exerts a horizontal force of 61.29 N.

Since the lawn mower moves at a constant speed, the frictional force must be equal in magnitude and opposite in direction to this horizontal force:

[tex]\[F_{\text{friction}} = -61.29 \, \text{N}\][/tex]

To calculate this force [tex](\(F_{\text{acceleration}}\))[/tex], we can use Newton's second law:

[tex]\[F = m \cdot a\][/tex]

Given:

Mass of the lawn mower = 16.0 kg

Acceleration = [tex]\(\frac{\Delta v}{\Delta t}\)[/tex],

Now, calculate [tex]\(\Delta v\):[/tex]

[tex]\[\Delta v = 1.3 \, \text{m/s} - 0 \, \text{m/s} = 1.3 \, \text{m/s}\][/tex]

Now, using Newton's second law:

[tex]\[F_{\text{acceleration}} = m \cdot a = (16.0 \, \text{kg}) \cdot \left(\frac{1.3 \, \text{m/s}}{2.5 \, \text{s}}\right)\][/tex]

[tex]\[F_{\text{acceleration}} = 8.32 \, \text{N}\][/tex]

Thus, the person must exert a force of 8.32 N.

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To accelerate a 16.0 kg lawn mower from rest to 1.3 m/s in 2.5 seconds, a person needs to exert a force of 136 N along the handle, assuming that the same friction force is present as when the mower is pushed at constant speed.

In order to find the force that the person must exert on the lawn mower to accelerate it from rest to 1.3 m/s in 2.5 seconds, we can use the second law of motion that states that the net force acting on an object is equal to its mass times its acceleration: F = m*a.

In this case, the acceleration would be the change in speed over time, which is (1.3 m/s - 0 m/s) / 2.5 s = 0.52 m/s². Consequently, F = (16.0 kg) * (0.52 m/s²) = 8.3 N.

It's important to mention this force is in addition to the friction force and the force to move the mower horizontally. Asume that the 87.5 N force previously exerted was just overcoming friction, then the total horizontal force required would be 87.5 N (for friction) + 8.3 N (for acceleration) = 95.8 N. However, since this force makes an angle of 45.0 with the horizontal, the person must exert a larger force along the handle, about 135.5 N. Hence, the correct answer, expressed to three significant figures, is 136 N.

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Answer:

Yes it is related

Explanation:

Consider 2 points in a rectangular co-ordinate system

[tex](x_{1},y_{1}),(x_{2},y_{2})[/tex]

The distance between them can be found by

d = [tex]\sqrt{(x_{1}-x_{2})^{2}-(y_{2}-y_{1})^{^{2}}}[/tex]

This is clear from the figure attached below

Answer:

Explanation:

Let E be the strength of electric field and m be the mass of plastic sphere.

charge on plastic sphere, q = - 3 n C = - 3 x 10^-9 C

The force due to electric field on the charged plastic sphere = q E

The weight of the plastic sphere = m g

where, g is the acceleration due to gravity.

Now the force due to electrostatic field is balanced by teh weight of the plastic sphere.

q E = m g

E = m g / q

E = m x 9.8 / (3 x 10^-9) = 3.26 x 10^9 x m

Substitute the value of m and get the value E.

The magnitude of the vertical electric field required to balance the weight of the charged plastic sphere with a mass of 2.1 g and a charge of -3.0 nC is [tex]6.86 * 10^6 N/C[/tex].

To find the magnitude of the vertical electric field that will balance the weight of a charged plastic sphere, we need to equate the electrostatic force on the sphere due to the electric field with the weight of the sphere.

Given information:

- Mass of the plastic sphere, m = 2.1 g = [tex]2.1 * 10^-3 kg[/tex]

- Charge on the plastic sphere, q = -3.0 nC = [tex]-3.0 * 10^-9 C[/tex]

- Acceleration due to gravity, g = [tex]9.8 m/s^2[/tex]

- Coulomb's constant, k = 1/(4πε₀) =[tex]9.0 * 10^9 N.m^2/C^2[/tex]

Step 1: Calculate the weight of the plastic sphere.

Weight of the sphere = mg

Weight = [tex](2.1 * 10^-3 kg)[/tex] × [tex](9.8 m/s^2)[/tex]

Weight = 2.058 × [tex]10^{-2}[/tex] N

Step 2: Equate the electrostatic force on the sphere to its weight.

Electrostatic force = qE

Weight = qE

[tex]2.058 * 10^{-2} N = (-3.0 * 10^{-9} C) * E[/tex]

Step 3: Rearrange the equation to find the electric field magnitude E.

Equation to find the electric field magnitude E.

E = (2.058 × [tex]10^{-2}[/tex] N) / (-3.0 × [tex]10^{-9}[/tex]C)

[tex]E = 6.86 * 10^6 N/C[/tex]

Complete question:

what's the magnitude of a vertical electric field that will balance the weight of a plastic sphere of mass 2.1g that has been charged to -3.0nC? (k=[tex]1/4\pi[/tex], [tex]E_0[/tex]=[tex]9.0*10^9 N*m^2/C^2[/tex])

Answer:

energy decreases

Explanation:

The energy of a wave is directly proportional to the square of amplitude of a wave.

Amplitude is defined as the maximum displacement of the wave particle from its mean position.

So, as the amplitude goes on decreasing, the energy of the wave is also decreasing.

If the amplitude of a wave decreases gradually as the wave travels down, the energy of the wave increases.

A wave can be defined as a disturbance in a medium that progressively transport energy from a source to another location, especially without the transportation of matter.

In Science, some of the characteristics of a wave include the following:

Generally, the energy of a wave is directly proportional to the square of wave amplitude and the square of its frequency. Thus, the energy of the wave increases when the amplitude of a wave decreases gradually as the wave travels down.

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Answer:

7174.09 N

Explanation:

Area = 7 ft x 4 x ft = 28 ft^2 = 2.60129 m^2

Pressure = 0.4 psi = 2757.9 Pa

Force = Pressure x Area

Force = 2757.9 x 2.60129 = 7174.09 N

Answer:

3.43 x 10⁵ N/m

Explanation:

M = mass of the car alone = 2000 kg

m = mass of the four people combined = 315 kg

x = compression of the spring when there is car alone

x' = compression of the spring when there is car and four people = x + 0.009

k = spring constant

When there is car, the weight of the car is balanced by the spring force and the force equation is given as

k x = Mg

k x = (2000)(9.8)

k x = 19600                                                eq-1

When there is car and four people, the weight of the car and four people is balanced by the spring force and the force equation is given as

k x' = (M + m) g

k (x + 0.009) = (2000 + 315) (9.8)

k x + 0.009 k = (2000 + 315) (9.8)

using eq-1

19600 + 0.009 k = 22687

k = 3.43 x 10⁵ N/m

Final answer:

The effective force constant of the springs k = 343000 N/m

Explanation:

To solve for the effective force constant of the springs in a car's suspension system, we can apply Hooke's Law, which is stated as F = -kx. Here, F represents the restoring force supplied by the springs, k is the spring constant or force constant, and x is the displacement of the springs from their equilibrium position.

When the four people with a combined mass of 315 kg sit in the car, their weight causes an additional displacement of 0.90 cm, or 0.009 m.

Since weight acts as the restoring force F, we can calculate it as F = mg, where m is the total mass and g is the acceleration due to gravity (9.80 m/s²). So, F = (315 kg)(9.80 m/s²) = 3087 N.

Using Hooke's Law, we can rearrange the formula to solve for k: k = F/x. Plugging in the values gives us k = 3087 N / 0.009 m, which provides the spring constant for the suspension system.

The effective force constant of the springs k = 343000 N/m

Answer:

6.4 J

Explanation:

For cart A

mA = 0.4 kg, uA = 6 m/s, vA = - 2 m/s

For cart B

mB = 0.8 kg, uB = 0, vB = ?

use the law of conservation of momentum

momentum of system before coliision = momentum of system after collision

mA uA + mB uB = mA vA + mB vB

0.4 x 6 + 0 = 0.4 x (- 2) + 0.8 x vB

vB = 4 m/s

Kinetic energy of Cart B after collision = 1/2 mB vB^2

                                                                = 1/2 x 0.8 x 4 x 4 = 6.4 J

Try this option; answers are marked with red colour: a) 19.215 GJ; b) 12.81 kN.

All the details are in the attached picture.

Answer:

14.4 Nm

Explanation:

Moment of Inertia, I = 0.9 kg m^2, w0 = 0, w = 8 rad/s, t = 0.5 second

Use first equation of motion for rotational motion

w = w0 + α t

where, α be the angular acceleration

8 = 0 + α x 0.5

α = 16 rad/s^2

Now Torque = Moment of inertia x angular acceleration

τ = I x α

τ = 0.9 x 16 = 14.4 Nm

Answer:

[tex]n_{glass}[/tex] = 1.6

Explanation:

[tex]\theta _{i}[/tex] = Angle of incidence = 65.9°

[tex]\theta _{r}[/tex] = Angle of refraction = 34.8°

[tex]n_{air}[/tex] = Index of refraction of air = 1

[tex]n_{glass}[/tex] = Index of refraction of glass = ?

Using Snell's law

[tex]n_{air}[/tex] Sin[tex]\theta _{i}[/tex] = [tex]n_{glass}[/tex] [tex]\theta _{r}[/tex]

(1) Sin65.9 =  [tex]n_{glass}[/tex] Sin34.8

[tex]n_{glass}[/tex] = 1.6

Since the index of refraction of air is 1, the refractive index of the glass is 1.6 approximately

Refractive Index is the measure of refraction or bending when light passes from one medium to another.

Given that a ray of light traveling in air is incident on the flat surface of a piece of glass at an angle of 65.9° with respect to the normal to the surface of the glass. If the ray refracted into the glass makes an angle of 34.8° with respect to the normal, the refractive index of the glass can be calculated with the formula below

n = sin i / sin r

Where

Substitute all the parameters

n = sin 65.9 / sin 34.8

n = 0.913 / 0.5707

n = 1.599

n = 1.6 approximately

Therefore, the refractive index of the glass is 1.6 approximately.

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Answer:

The electric field is [tex]-2.14\times10^{5}\ N/C[/tex]

Explanation:

Given that,

Distance = 2.3 m

Charge [tex]q= -20.0\ \muC[/tex]

We need to calculate the electric field

Using formula of electric field

[tex]E=\dfrac{\sigma}{2\epsilon_{0}}[/tex]

Where, [tex]\sigma=\dfrac{Q}{A}[/tex]

Q = charge

A = area

Put the value into the formula

[tex]E=\dfrac{Q}{2A\epsilon_{0}}[/tex]

[tex]E=\dfrac{-20.0\times10^{-6}}{2\times2.3\times2.3\times8.85\times10^{-12}}[/tex]

[tex]E=-213599.91\ N/C[/tex]

[tex]E=-2.14\times10^{5}\ N/C[/tex]

Negative sign shows the direction of the electric field.

The direction of electric field is toward the plates.

Hence, The electric field is [tex]-2.14\times10^{5}\ N/C[/tex]

Final answer:

The electric field 1.0 cm above a thin conducting plate with a side length of 2.3 m and a total charge of -20.0 µC is -213,559.32 N/C, indicating a downward direction.

Explanation:

A thin conducting plate 2.3 m on a side is given a total charge of -20.0 µC. To find the electric field 1.0 cm above the plate, we use the concept of surface charge density (σ) and the formula for the electric field near an infinite plane sheet of charge, E = σ / (2ε0), where ε0 is the vacuum permittivity (ε0 = 8.85 x [tex]10^-^1^0[/tex] C2/N·m²). The surface charge density is σ = Q / A, with Q being the charge and A the area of the plate.

The area of the plate, A, is 2.3 m × 2.3 m = 5.29 m2. Therefore, σ = (-20.0 x 10-6 C) / (5.29 m2) = -3.78 x [tex]10^-^6[/tex] C/m². Substituting σ into the electric field equation yields E = -3.78 x [tex]10^-^6[/tex] C/m² / (2 × 8.85 x [tex]10^-^1^2[/tex] C2/N·m2) = -213,559.32 N/C, indicating the direction is downward due to the negative sign.

Answer:

The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

Explanation:

We know that,

Mass of electron [tex]m_{e}=9.11\times10^{-31}\ kg[/tex]

Rest mass energy for electron = 0.511 Mev

(a). The energy required to accelerate an electron from 0.500c to 0.900c Mev

Using formula of rest,

[tex]E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}[/tex]

[tex]E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.500c)^2}{c^2}}}[/tex]

[tex]E=0.582\ Mev[/tex]

(b). The energy required to accelerate an electron from 0.900c to 0.942c Mev

Using formula of rest,

[tex]E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}[/tex]

[tex]E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.942c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}[/tex]

[tex]E=0.350\ Mev[/tex]

Hence, The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

Answer:

[tex] \frac{F}{64}[/tex]

Explanation:

m = product of masses of two objects

r = distance between the two objects = 100 m

F = initial force between the two object

r' = new distance between the two objects = 800 m

F' = new force between the two objects = ?

k = constant of proportionality

initial force between the two object is given as

[tex]F=\frac{km}{r^{2}}[/tex]                                     eq-1

new force between the two objects is given as

[tex]F'=\frac{km}{r'^{2}}[/tex]                                  eq-2

Dividing eq-2 by eq-1

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

Inserting the values

[tex]\frac{F'}{F}=\frac{100^{2}}{800^{2}}[/tex]

[tex]F' = \frac{F}{64}[/tex]

Answer:

The car moves from start to stop 328.88 m in total.

Explanation:

Vo= 0 m/s

V1= 21.1 m/s

V2= 0 m/s

a1= 2,03 m/s²

a2= -1.015 m/s²

Speed Up:

Speed up time:

V1= Vo + a1 * t1

t1= V1/a1

t1= 10.39 sec

total distance of speed up:

d1= Vo * t1 + (a1 * t1²)/2

d1= 109.57m

Slow Down:

V2= V1 - a2 * t2

t2= V1/a2

t2= 20.78 sec

total distance of slow down:

d2= V1 * t2 - (a2 * t2²)/2

d1= 219.31m

Total Distance:

TD= d1+d2= 109.57m + 219.31m

TD= 328.88 m

Answer:

Work one by the force, W = 1378.87 lb-ft

Explanation:

It is given that,

Force acting on the sled, F = 20 lb

Angle between the force and the horizontal, θ = 40°

Distance moved, d = 90 ft

We need to find the work done by the force. We know that the work done can be calculated as :

[tex]W=Fd\ cos\theta[/tex]

[tex]W=20\ lb\times 90\ ft\ cos(40)[/tex]

W = 1378.87 lb-ft

So, the work done by the force is 1378.87 lb-ft. Hence, this is the required solution.

Answer: 25 m

Explanation:

The resolving power of a telescope is given by:

[tex] \theta = 1.22\frac{\lambda}{diameter}[/tex]

Diameter of the telescope = 10.0 m

Wavelength of the telescope = 550 nm = 550 ×10⁻⁹m

Thus, the resolving power of one of telescopes at the Keck Observatory is:

[tex] \theta = 1.22 \frac{550\times 10^{-9}}{10.0} =671 \times 10^{-10} rad = 3.85 \times 10^{-6} deg[/tex]

Also, [tex] tan\theta = \frac {\text {separation between two objects}}{\text{distance to the objects}} [/tex]

separation between objects = tan (3.85×10⁻⁶)° × 3.8×10⁵km = 2.5 × 10⁻²km =25 m  

Answer:

0.2 m

Explanation:

Diameter = 0.3 m

radius, r = 0.15 m

Length, H = 0.4 m

density of wood, d = 0.5 g/cm^3 = 500 kg/m^3

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

Let h be the depth of cylinder immersed in water.

By the principle of floatation.

Buoyant force = Weight of cylinder

Volume immeresed x density of water x g = Volume of cylinder x density of wood x g

A x h x 1000 x g = A x H x 500 x g

1000 h = 500 x 0.4

h = 0.2 m

Answer:

The magnitude of the force per unit of lenght between the wires are of  F/L= 3.17 * 10⁻⁵ N/m.

Explanation:

d=0.0665m

I1=I2= 3.25A

μo= 4π * 10⁻⁷ N/A²

F/L= (μo * I1 * I2) / (2π * d)

F/L= 3.17 * 10⁻⁵ N/m

Answer:

Ball hit the tall building 50 m away below 10.20 m its original level

Explanation:

Horizontal speed = 20 cos40 = 15.32 m/s

Horizontal displacement = 50 m

Horizontal acceleration = 0 m/s²

Substituting in s = ut + 0.5at²

    50 = 15.32 t + 0.5 x 0 x t²

     t = 3.26 s

Now we need to find how much vertical distance ball travels in 3.26 s.

Initial vertical speed  = 20 sin40 = 12.86 m/s

Time = 3.26 s

Vertical acceleration = -9.81 m/s²

Substituting in s = ut + 0.5at²

    s = 12.86 x 3.26 + 0.5 x -9.81 x 3.26²

    s = -10.20 m

So ball hit the tall building 50 m away below 10.20 m its original level

The distance the ball will strike the opposite wall is 32.79 m.

The time of motion of the ball from the given height is calculated as follows;

h = vsinθ(t) + ¹/₂gt²

50 = 20 x sin(40)t + 0.5(9.8)t²

50 = 12.86t + 4.9t²

4.9t² + 12.86t - 50 = 0

solve the quadratic equation using formula method,

t = 2.14 s

The horizontal distance of the ball from the initial position is calculated as follows;

X = vcosθ(t)

X = 20 x cos(40) x 2.14

X = 32.79 m

Thus, the distance the ball will strike the opposite wall is 32.79 m.

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Answer:

Work done is zero

Explanation:

given data

Angle of kite with horizontal =  30 degree

tension in the string =  4.5 N

WE KNOW THAT

Work =  force * distance

horizontal force =  [tex]Tcos\theta = 4.5*cos30 = 3.89 N[/tex]

DISTANCE = 0 as boy stands still. therefore

work done = 3.89 *0 = 0

Answer:

The electric field always decreases.

Explanation:

The electric field due to a point charge is given by :

[tex]E=\dfrac{kq}{r^2}[/tex]

Where

k = electric constant

q = charge

r = distance from the charge

It is clear from the above equation that as the distance from the charge particle increases the electric field decreases. As you move away from a positive charge distribution, the electric field always decreases. Hence, the correct option is (c) "Always decreases".

Answer:

Sound intensity level of the sound = 140 dB

Explanation:

We have expression for sound intensity level

             [tex]L=10log_{10}\left ( \frac{I}{I_0}\right )[/tex]

We have sound intensity, I = 100 W/m²

                                          I₀ = 10⁻¹² W/m²

We need to find sound intensity level

Substituting

            [tex]L=10log_{10}\left ( \frac{I}{I_0}\right )=10log_{10}\left ( \frac{100}{10^{-12}}\right )=10x14=140dB[/tex]

Sound intensity level of the sound = 140 dB

Answer:

65.75 deg

Explanation:

v = initial speed of launch of projectile

θ = initial angle of launch

H = maximum height of the projectile

maximum height of the projectile is given as

[tex]H=\frac{v^{2}Sin^{2}\theta }{2g}[/tex]         eq-1

D = horizontal range of the projectile

horizontal range of the projectile is given as

[tex]D=\frac{v^{2}Sin{2}\theta }{g}[/tex]                     eq-2

It is also given that

D = 1.8 H

using eq-1 and eq-2

[tex]\frac{v^{2}Sin{2}\theta }{g} = (1.8) \frac{v^{2}Sin^{2}\theta }{2g}[/tex]

[tex]Sin{2}\theta = (1.8) \frac{Sin^{2}\theta }{2}[/tex]

[tex]2 Sin\theta Cos\theta= (0.9) Sin^{2}\theta[/tex]

[tex]2 Cos\theta = (0.9) Sin\theta[/tex]

tanθ = 2.22

θ = 65.75 deg

Explanation:

Electric field intensity E is defined as the electric (Coulomb) force on a unit test (1 C) charge. Mathematically, it is given by :

[tex]E=\dfrac{F}{q}[/tex]

The electric force is given by :

[tex]F=k\dfrac{qQ}{d^2}[/tex]

Where

Q and q are electric charges

d is the distance between charges

The electric field intensity at a distance d from the center is given by :

[tex]E=\dfrac{k\dfrac{qQ}{d^2}}{q}[/tex]

So,

[tex]E=\dfrac{kQ}{d^2}[/tex]

Hence, this is the required solution.

Answer:

8377 turns

Explanation:

Given:

Mean radius of toroid = 26 cm

Circular cross-section radius, r = 1.9 cm = 0.019

Length of the wire, l = 1000 m

The number of turns (n) on the coil is given by  

[tex]n = \frac{l}{s}[/tex]

where,

l is the length of the wire

s is the circumference of the circular wire

⇒[tex]n = \frac{1000}{2\pi r}[/tex]

or

[tex]n = \frac{1000}{2\pi \times 0.019}[/tex]

or

n = 8376.57 or 8377 turns

hence, the number of turns on the coils is 8377

The number of turns of the coil is determined as 8377 turns.

The number of turns of the superconducting wire is calculated as follows;

n = L/2πr

where;

Substitute the given parameters and solve for the number of turns of the coil,

n = 1000/(2π x 0.019)

n = 8377 turns

Thus, the number of turns of the coil is determined as 8377 turns.

Learn more about number of turns here: https://brainly.com/question/25886292

Answer:

For proton: 2592 m/s In the same direction of electric field.

For electron: 4752000 m/s In the opposite direction of electric field.

Explanation:

E = 500 N/C, t = 54 ns = 54 x 10^-9 s,

Acceleration = Force /mass

Acceleration of proton, ap = q E / mp

ap = (1.6 x 10^-19 x 500) / (1.67 x 10^-27) = 4.8 x 10^10 m/s^2

Acceleration of electron, ae = q E / me

ae = (1.6 x 10^-19 x 500) / (9.1 x 10^-31) = 8.8 x 10^13 m/s^2

For proton:

u = 0, ap = 4.8 x 10^10 m/s^2, t = 54 x 10^-9 s

use first equation of motion

v = u + at

vp = 0 + 4.8 x 10^10 x 54 x 10^-9 = 2592 m/s In the same direction of electric field.

For electron:

u = 0, ae = 8.8 x 10^13 m/s^2, t = 54 x 10^-9 s

use first equation of motion

v = u + at

vp = 0 + 8.8 x 10^13 x 54 x 10^-9 = 4752000 m/s In the opposite direction of electric field.