A student is skateboarding down a ramp that is 8.51 m long and inclined at 23.7° with respect to the horizontal. The initial speed of the skateboarder at the top of the ramp is 4.97 m/s. Neglect friction and find the speed at the bottom of the ramp.

Answer:

286

Explanation:

p1v1/T1=p2v2/T2

then subtitude your values

T2=760*0.65*273/210*0.040

T2=135/8.4=16+273=289

The pressure-volume work done by the system during the expansion is 5.25 x 10³J.

The work that is done when a fluid is compressed or expanded is known as pressure-volume work. Pressure-volume work occurs whenever there is a change in volume but the outside pressure stays the same.

Here,

The pressure of the system of gas, P = 210 kPa

Initial volume of the system of gas, V₁ = 0.04 m³

Final volume of the system of gas, V₂ = 0.065 m³

The expression for the pressure-volume work done by a system of gas is given by,

Work done,

W = PΔV

W = P(V₂ - V₁)

Applying the values of P, V₁ and V₂,

W = 210 x 10³(0.065 - 0.04)

W = 210 x 10³x 0.025

W = 5.25 x 10³J

Hence,

The pressure-volume work done by the system during the expansion is 5.25 x 10³J.

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Answer:

[tex]log_28=3[/tex]

Explanation:

To calculate the value of [tex]log_28[/tex]

Also,Some relation for log values:

[tex]loga^m= mloga[/tex]  -----------------------------------------------------1

[tex]log_aa= 1[/tex]---------------------------------------------------------------2

Also, [tex]2^3= 8[/tex]

Applying in the question as:

[tex]log_28=log_2(2^3)[/tex]

Using Relation-1 mentioned above:

[tex]log_28=3\times log_2(2)[/tex]

Using Relation-2 mentioned above:

[tex]log_28=3\times 1[/tex]

Thus,

[tex]log_28=3[/tex]

Answer:

The body will strike at angle 60.46°

Explanation:

Vertical motion of body:

 Initial speed, u = 40sin30 = 20m/s

 Acceleration, a = 9.81 m/s²

 Displacement, s = 170 m

 We have equation of motion, v² = u² + 2as

 Substituting

   v² = 20² + 2 x 9.81 x 170

    v = 61.12 m/s

 Final vertical speed = 61.12 m/s

 Final horizontal speed = initial horizontal speed = 40cos30= 34.64m/s

 Final velocity = 34.64 i - 61.12 j m/s

 Magnitude

     [tex]v=\sqrt{34.64^2+(-61.12)^2}=70.25m/s[/tex]

 Direction

      [tex]\theta =tan^{-1}\left ( \frac{-61.12}{34.64}\right )=-60.46^0[/tex]

 The body will strike at angle 60.46°

Answer:

True

Explanation:

The force on the electron when it enters in a magnetic field is given by

F = q ( v x B)

F = -e x V x B x Sin∅

here, F is the force vector, B be the magnetic field vector and v be the velocity vector.

If the angle between the velocity vector and the magnetic field vector is 0 degree, then force is zero.

When the electrons enters in the magnetic field at any arbitrary angle, it experiences a force and hence it accelerate up.

The change in momentum of the ball is 24 kg·m/s.

The change in momentum of the ball can be found by subtracting the initial momentum from the final momentum. The initial momentum is given by the formula p1 = m * v1, where m is the mass of the ball and v1 is its initial velocity. In this case, the initial velocity is 12 m/s downward, so the initial momentum is -12 kg·m/s. The final momentum is given by the formula p2 = m * v2, where v2 is the velocity of the ball after the bounce. Since the motion after the bounce is the mirror image of the motion before the bounce, the final velocity is 12 m/s upward. Thus, the final momentum is 12 kg·m/s. Subtracting the initial momentum from the final momentum, we get the change in momentum to be 24 kg·m/s.

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The change in momentum of a 1.0-kg ball that bounces with a velocity of 12 m/s upward after striking the ground with a velocity of 12 m/s downward is 24 kg·m/s. This is calculated using the formula for momentum, p = mv, and finding the difference between initial and final momentum.

Momentum (old{p}old{) is calculated by the formula:

p = mv

Step-by-Step Explanation:

The initial momentum just before striking the ground (p₁) is:

p₁ = mass × velocity

p₁ = 1.0 kg × (-12 m/s)

p₁ = -12 kg·m/s

The final momentum just after bouncing up (p₂) is:

p₂ = mass × velocity

p₂ = 1.0 kg × 12 m/s

p₂ = 12 kg·m/s

The change in momentum (Δp) is:

Δp = p₂ - p₁

Δp = 12 kg·m/s - (-12 kg·m/s)

Δp = 12 kg·m/s + 12 kg·m/s

Δp = 24 kg·m/s

So, the change in momentum of the ball is 24 kg·m/s.

Answer:

The ball must be tossed to a height of 4 times the initial height H

Explanation:

We have equation of motion S = ut + 0.5at²

A juggler tosses balls vertically to height H.

That is

      H = 0 x t + 0.5 x a x t²

      H = 0.5at²

To what height must they be tossed if they are to spend twice as much time in the air.

      H' = 0 x 2t + 0.5 x a x (2t)²

      H' = 2at² = 4 H

So the ball must be tossed to a height of 4 times the initial height H

Answer:

2.6 m/s²

Explanation:

m = mass of the object = 7.8 kg

Consider the right direction as positive and left direction as negative

F = horizontal applied force towards right = 50 N

f = frictional force acting towards left = - 30 N

a = acceleration

Force equation for the motion of the object is given as

F + f = ma

50 - 30 = (7.8) a

a = 2.6 m/s²

Answer:

The difference in the length of the two rods when compressed is [tex]5.4\times10^{-5}\ m[/tex].

Explanation:

Given that,

Length = 0.780 m

Diameter = 1.50 cm

Force = 4350 N

(a). For steel rod

We know ,

The young modulus for steel rod

[tex]Y=2\times10^{11}[/tex]

Using formula of young modulus

[tex]e_{s}=\dfrac{Fl}{AY}[/tex]

[tex]e_{s}=\dfrac{4350\times0.780}{3.14\times(0.75\times10^{-2})^2\times2\times10^{11}}[/tex]

[tex]e_{s}=9.6\times10^{-5}\ m[/tex]

(b). For copper rod

We know ,

The young modulus for steel rod

[tex]Y=1.1\times10^{11}[/tex]

Using formula of young modulus

[tex]e_{c}=\dfrac{Fl}{AY}[/tex]

[tex]e_{c}=\dfrac{4350\times0.780}{3.14\times(0.75\times10^{-2})^2\times1.1\times10^{11}}[/tex]

[tex]e_{c}=1.5\times10^{-4}\ m[/tex]

The difference in the length of the two rods when compressed is

[tex]difference\ in\ length=e_{c}-e_{s}[/tex]

[tex]difference\ in\ length=1.5\times10^{-4}-9.6\times10^{-5}[/tex]

[tex]difference\ in\ length =5.4\times10^{-5}\ m[/tex]

Hence, The difference in the length of the two rods when compressed is [tex]5.4\times10^{-5}\ m[/tex].

Answer:

2.5 min

Explanation:

Hello

by definition  the power   is the amount of work done per unit of time. in this case, we know  the total power and the work developed

[tex]1 Watt= \frac{Joule }{sec}[/tex]

Let

[tex]P=1200 W\\E=1.8 *10^{5}\\X=\frac{1.8 *10^{5} j }{1200 W}\\X= unknown\ time\ (sec) \\\\we\ need\ the\ answer\ in\ minutes,\\ 1\ min=60\ sec\\\\x=150 sec \\150\ sec*(\frac{1 min}{ 60 sec})=2.5 min\\[/tex]

hence , the data is not altered, we did it like this to eliminate the sec units.

Answer 2.5 min

I hope it helps

The food was in the 1200 W microwave for 150 seconds, which is equivalent to 2.5 minutes.

To calculate how long the food was in the microwave, we use the formula for power, which is the rate at which work is done or energy is transferred:

Power (P) = Energy (E) / Time (t)

First, rearrange this formula to solve for time (t):

t = E / P

Plugging in the given values:

So:

t = (1.8 times 10⁵J) / (1200 W)

Calculate this to find the time in seconds, and then convert it to minutes as follows:

t = 150 seconds

Divide by 60 to get minutes:

t = 2.5 minutes

Answer:

21813.46 J

Explanation:

The Potential energy stored at the time it is hanging freely is mgl.

the potential energy stored at the time when it makes a angle 23 degree from the vertical is mglCos 23

Change in gravitational potential energy

U = mgl = mgl cos 23

U = mgl (1 - cos23)

U = 1400 x 9.8 x 20 ( 1 - Cos 23)

U = 21813.46 J

Answer: [tex]22.8^0C[/tex]

Explanation:-

[tex]Q_{absorbed}=Q_{released}[/tex]

As we know that,  

[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]

[tex]m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)][/tex]      

where,

[tex]m_1[/tex] = mass of iron horseshoe = 0.35 kg = 350 g (1kg=1000g[/tex]

[tex]m_2[/tex] = mass of water = 21.9 kg = 21900 g

[tex]T_{final}[/tex] = final temperature = ?

[tex]T_1[/tex] = temperature of iron horseshoe = [tex]600^oC[/tex]

[tex]T_2[/tex] = temperature of water = [tex]21.8^oC[/tex]

[tex]c_1[/tex] = specific heat of iron horseshoe = [tex]0.450J/g^0C[/tex]

[tex]c_2[/tex] = specific heat of water =  [tex]4.184J/g^0C[/tex]

Now put all the given values in equation (1), we get

[tex]m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)][/tex]

[tex]350\times 0.450\times (T_{final}-600)^0C=-[21900g\times 4.184\times (T_{final}-21.8)][/tex]

[tex]T_{final}=22.8^0C[/tex]

Therefore, the final equilibrium temperature is [tex]22.8^0C[/tex].

Answer:

[tex]\mu = 1.39[/tex]

Explanation:

Since the reflection is nearly zero intensity

so here we will say that the reflected light must show destructive interference

so here we have

[tex]path \: difference = \frac{\lambda}{2}[/tex]

[tex]2 \mu t = \frac{\lambda}{2}[/tex]

[tex]t = \frac{\lambda}{4\mu}[/tex]

here we have

[tex]\lambda = 675 nm[/tex]

t = 121 nm

now from above equation we have

[tex]\mu = \frac{675 nm}{4(121 nm)}[/tex]

[tex]\mu = 1.39[/tex]

Answer:

3.91 N/C

Explanation:

u = 0, s = 5.50 m, t = 4 us = 4 x 10^-6 s

Let a be the acceleration.

Use second equation of motion

s = u t + 1/2 a t^2

5.5 = 0 + 1/2 a (4 x 10^-6)^2

a = 6.875 x 10^11 m/s^2

F = m a

The electrostatic force, Fe = q E

Where E be the strength of electric field.

So, q E = m a

E = m a / q

E = (9.1 x 10^-31 x 6.875 x 10^11) / ( 1.6 x 10^-19)

E = 3.91 N/C

Answer:

1.59 m/s^2, 65.2°

Explanation:

F1 = 390 N North

F2 = 180 N east

m = 270 kg

Net force is the vector sum of both the forces.

[tex]F = \sqrt{F_{1}^{2}+F_{2}^{2}}[/tex]

[tex]F = \sqrt{390^{2}+180^{2}}[/tex]

F = 429.53 N

Direction of force

tan∅ = F1 / F2 = 390 / 180 = 2.1667

∅ = 65.2°

The direction of acceleration is same as the direction of net force.

The magnitude of acceleration is

a = F / m = 429.53 / 270 = 1.59 m/s^2

Final answer:

The magnitude of the sailboat's acceleration is 1.59 m/s², and the direction is 25 degrees east of north. This is calculated using Newton's second law and vector addition of the orthogonal forces exerted by the wind and water.

Explanation:

To calculate the magnitude and direction of the sailboat's acceleration, we need to use Newton's second law, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Here, we combine the forces exerted by the wind and the water to obtain the net force. The net force can be calculated using vector addition, where the force due to the wind (390 N north) and the force due to the water (180 N east) are treated as orthogonal vectors.

The net force (Fnet) is the vector sum of the two individual forces. We calculate the net force using the Pythagorean theorem:

Fnet = √Fwind² + Fwater² = √390² + 180² = √152100 + 32400 = √184500 N

So, the magnitude of the net force is approximately 429.53 N. To find the acceleration (a), we use the formula:

a = Fnet / m

Substituting the mass of the sailboat (270 kg) and the net force, we get:

a = 429.53 N / 270 kg = 1.59 m/s²

To determine the direction of the acceleration, we take the arctangent of the ratio of the forces. Since arctan(180/390) equals approximately 25 degrees, this is the angle east of north.

The sailboat's acceleration has a magnitude of 1.59 m/s² and a direction of 25 degrees east of north.

Answer:

0.324×10⁻³ m/s²

Explanation:

G=Gravitational constant=6.67408×10⁻⁸ m³/kg s

m₁=m₂=Mass of the ships=42000×10³ kg

r=Distance between the ships=93 m

The ships are considered particles

From Newtons Law of Universal Gravitation

[tex]F=G\frac {m_1m_2}{r^2}\\Here\ m_1=m_2\ hence\ m_1\times m_2=m_1^2\\\Rightarrow F=G\frac {m_1^2}{r^2}\\\Rightarrow F=\frac{6.67408\times 10^{-8}\times (42000\times 10^3)^2}{93^2}\\\Rightarrow F=13612.0674\ N\\[/tex]

[tex]F=ma\\\Rightarrow 13612.0674=42000\times 10^3a\\\Rightarrow a=0.324\times 10^{-3}\ m/s^2[/tex]

∴Acceleration of one of the liners toward the other due to their mutual gravitational attraction is 0.324×10⁻³

Answer:

517.6 N

8.63 m/s²

Explanation:

M = mass of earth = 5.98 x 10²⁴ kg

m = mass of the person = 60 kg

W = weight of the person

R = radius of earth = 6.4 x 10⁶ m

d = distance of ISS above the surface of earth = 400 km = 4 x 10⁵ m

Weight of person on the ISS is given as

[tex]W = \frac{GMm}{(R+d)^{2}}[/tex]

[tex]W = \frac{(6.67\times 10^{-11})(5.98\times 10^{24})(60)}{((6.4\times 10^{6})+(4\times 10^{5}))^{2}}[/tex]

W = 517.6 N

Acceleration due to gravity is given as

[tex]g = \frac{GM}{(R+d)^{2}}[/tex]

[tex]g = \frac{(6.67\times 10^{-11})(5.98\times 10^{24})}{((6.4\times 10^{6})+(4\times 10^{5}))^{2}}[/tex]

g = 8.63 m/s²

The rotational kinetic energy of the pitcher's arm when throwing a softball at a speed of 139 km/h is calculated using the equation for rotational kinetic energy, and taking moment of inertia and angular velocity into account. The angular velocity is inferred from the linear speed of the ball and the distance it leaves the pitcher's hand from the pivot at the shoulder. The rotational kinetic energy is found to be approximately 1491 Joules.

This question pertains to the rotational kinetic energy of the pitcher's arm during the act of throwing a softball. By definition, the kinetic energy associated with rotational motion (rotational kinetic energy) can be given by the equation: K_rot = 0.5 * I * ω² where 'I' is the moment of inertia and 'ω' is the angular velocity.

To calculate the angular velocity 'ω', we can infer it from the linear speed of the ball when it leaves the pitcher's hand, as 'ω = v/r', v = 139 km/h = 38.6 m/s, and r = 0.600 m (distance ball leaves hand from pivot at shoulder). Therefore, 'ω' is approximately 64.4 rad/s.

Substituting 'I' and 'ω' into the equation above, we get K_rot = 0.5 * 0.720 kg*m² * (64.4 rad/s)² = 1491 Joules, which is the rotational kinetic energy of the pitcher's arm.

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Answer:

The speed of the ball is 0.8 m/s.

(D) is correct option.

Explanation:

Given that,

mass of ball = 3.0 kg

length = 50 cm

We need to calculate the speed of the ball

Using relation between the centripetal force and tension

[tex]\dfrac{mv^2}{r}=T[/tex]

[tex]v^2=\dfrac{T\times r}{m}[/tex]

Where, m = mass

T = tension

r = radius

Put the value into the formula

[tex]v=\sqrt{\dfrac{3.8\times50\times10^{-2}}{3.0}}[/tex]

[tex]v=0.8\ m/s[/tex]

Hence, The speed of the ball is 0.8 m/s.

Answer:

The inductance is 0.021 H.

Explanation:

Given that,

Number of turns = 16

Current = 3 A

Total flux [tex]\phi=4\times10^{-3}\ Tm^2[/tex]

We need to calculate the inductance

Using formula of total flux

[tex]N\phi = Li[/tex]

[tex]L=\dfrac{N\phi}{i}[/tex]

Where, i = current

N = number of turns

[tex]\phi[/tex] = flux

Put the value into the formula

[tex]L=\dfrac{16\times4\times10^{-3}}{3}[/tex]

[tex]L=0.021\ H[/tex]

Hence, The inductance is 0.021 H.

Mario's velocity vector relative to the ice is

[tex]\vec v_{M/I}=-4.79\,\vec\jmath[/tex]

(note that all velocities mentioned here are given in m/s)

The puck is moving in a direction of 17.8 degrees west of south, or 252.2 degrees counterclockwise relative to east. Its velocity vector relative to the ice is then

[tex]\vec v_{P/I}=9.13(\cos252.2^\circ\,\vec\imath+\sin252.2^\circ\,\vec\jmath)=-2.79\,\vec\imath-8.69\,\vec\jmath[/tex]

The velocity of the puch relative to Mario is

[tex]\vec v_{P/M}=\vec v_{P/I}+\vec v_{I/M}=\vec v_{P/I}-\vec v_{M/I}[/tex]

[tex]\vec v_{P/M}=-2.79\,\vec\imath-3.90\,\vec\jmath[/tex]

Then, relative to Mario,

a. the puck is traveling at a speed of [tex]\boxed{\|\vec v_{P/M}\|=4.80}[/tex], and

b. is moving in a direction [tex]\theta[/tex] such that

[tex]\tan\theta=\dfrac{-3.90}{4.80}\implies\theta=-126^\circ[/tex]

which is about [tex]\boxed{35.6^\circ}[/tex] west of south.

The magnitude of the puck's velocity as observed by Mario is 4.9 m/s and the direction is 35.6 degrees west of south.

When addressing this problem, we're dealing with relative velocity in two dimensions. Since we need to find the puck's velocity as seen by Mario, we should use the principle that the relative velocity of the puck with respect to Mario is the vector difference between the velocity of the puck and Mario.

Given that Mario is travelling south at 4.79 m/s and the puck is travelling at an angle of 17.8 degrees west of south at a speed of 9.13 m/s, we can break the puck's velocity into southward and westward components. The southward component is 9.13 m/s × cos(17.8°) = 8.74 m/s and the westward component is 9.13 m/s × sin(17.8°) = 2.84 m/s.

The relative southward velocity of the puck with respect to Mario is 8.74 m/s - 4.79 m/s = 3.95 m/s. The westward component is unchanged since Mario has no westward velocity. So, the puck appears to Mario to be moving southwest at an angle of arctan(2.84 m/s / 3.95 m/s) = 35.6 degrees west of south.

The magnitude is given by the Pythagorean theorem, sqrt((2.84 m/s)² + (3.95 m/s)²) = 4.9 m/s.

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Answer:

The net force acting on this object is 180.89 N.

Explanation:

Given that,

Mass = 3.00 kg

Coordinate of position of [tex]x= 5t^3+1[/tex]

Coordinate of position of [tex]y=3t^2+2[/tex]

Time = 2.00 s

We need to calculate the acceleration

[tex]a = \dfrac{d^2x}{dt^2}[/tex]

For x coordinates

[tex]x=5t^3+1[/tex]

On differentiate w.r.to t

[tex]\dfrac{dx}{dt}=15t^2+0[/tex]

On differentiate again w.r.to t

[tex]\dfrac{d^2x}{dt^2}=30t[/tex]

The acceleration in x axis at 2 sec

[tex]a = 60i[/tex]

For y coordinates

[tex]y=3t^2+2[/tex]

On differentiate w.r.to t

[tex]\dfrac{dy}{dt}=6t+0[/tex]

On differentiate again w.r.to t

[tex]\dfrac{d^2y}{dt^2}=6[/tex]

The acceleration in y axis at 2 sec

[tex]a = 6j[/tex]

The acceleration is

[tex]a=60i+6j[/tex]

We need to calculate the net force

[tex]F = ma[/tex]

[tex]F = 3.00\times(60i+6j)[/tex]

[tex]F=180i+18j[/tex]

The magnitude of the force

[tex]|F|=\sqrt{(180)^2+(18)^2}[/tex]

[tex]|F|=180.89\ N[/tex]

Hence, The net force acting on this object is 180.89 N.

To find the magnitude of the net force acting on the object at t = 2.00 s, we need to calculate the x and y components of the force separately. Once we have both components, we can use the Pythagorean theorem to find the magnitude of the net force.

To find the magnitude of the net force acting on the object at t = 2.00 s, we need to calculate the x and y components of the force separately. The x-component of the force can be found by taking the derivative of the x-coordinate equation with respect to time and multiplying it by the mass of the object. The same process can be applied to find the y-component of the force. Once we have both components, we can use the Pythagorean theorem to find the magnitude of the net force.

Starting with the x-coordinate equation, x = 5t³ - 1, taking its derivative gives dx/dt = 15t². Multiplying the derivative by the mass of the object (3.00 kg) gives the x-component of the force: Fx = 3.00 kg * 15t².

Similarly, for the y-coordinate equation y = 3t² + 2, taking its derivative gives dy/dt = 6t. Multiplying the derivative by the mass of the object (3.00 kg) gives the y-component of the force: Fy = 3.00 kg * 6t.

Using the Pythagorean theorem, the magnitude of the net force Fnet can be calculated as: Fnet = sqrt(Fx² + Fy²). Substituting the values into the equation will give you the magnitude of the net force at t = 2.00 s.

Answer:

The time period of the wave is 0.002 sec.

Explanation:

Given that,

Frequency = 0.5 kHz

Phase difference [tex]\phi=\dfrac{\pi}{3}[/tex]

Path difference = 1.5 cm

We need to calculate the time period

Using formula of time period

The frequency is the reciprocal of time period.

[tex]f =\dfrac{1}{T}[/tex]

[tex]T=\dfrac{1}{f}[/tex]

Where, f = frequency

Put the value into the formula

[tex]T=\dfrac{1}{0.5\times10^{3}}[/tex]

[tex]T=0.002\ sec[/tex]

Hence, The time period of the wave is 0.002 sec.

Answer:

2.17 x 10^8 m/s

Explanation:

Angle of incidence, i = 30 degree, refractive index of mineral oil, n = 1.38

Let r be the angle of refraction.

By use of Snell's law

n = Sin i / Sin r

Sin r = Sin i / n

Sin r = Sin 30 / 1.38

Sin r = 0.3623

r = 21.25 degree

Let the speed of light in oil be v.

By the definition of refractive index

n = c / v

Where c be the speed of light

v = c / n

v = ( 3 x 10^8) / 1.38

v = 2.17 x 10^8 m/s

Answer:

3.62 m  and - 1.4 m

Explanation:

Consider a location towards the positive side of x-axis beyond the location of charge Q₂

x = distance of the location from charge Q₂

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

[tex]\frac{kQ_{1}}{(2 + x)^{2}}= \frac{kQ_{2}}{x^{2}}[/tex]

[tex]\frac{2.5\times 10^{-5}}{(2 + x)^{2}}= \frac{5 \times 10^{-6}}{x^{2}}[/tex]

x = 1.62 m

So location is 2 + 1.62 = 3.62 m

Consider a location towards the negative side of x-axis beyond the location of charge Q₁

x = distance of the location from charge Q₁

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

[tex]\frac{kQ_{1}}{(x)^{2}}= \frac{kQ_{2}}{ (2 + x)^{2}}[/tex]

[tex]\frac{2.5\times 10^{-5}}{(x)^{2}}= \frac{5 \times 10^{-6}}{(2+x)^{2}}[/tex]

x = - 1.4 m

Electric field is zero due to positive point charge Q1 and negative Q2 along the x axis is at the location of 3.62 meters.

The electric field is the field, which is surrounded by the electric charged. The electric field is the electric force per unit charge.

Given information-

The charge of the point 1 is [tex]2.5\times10^{-5} \rm C[/tex].

The charge of the point 2 is [tex]-5.0\times10^{-6} \rm C[/tex].

The distance between the point 1 and point 2 is 2 meters away from the x axis.

Let the position of the point 1 is at x.

Thus the electric force on point Q1 is,

[tex]F_1=\dfrac{kQ1}{x^2}[/tex]

As the distance between the point Q1 and point Q2 is 2 meters away from the x axis. Thus the position of it should be at (x+2). The electric force on point 2

[tex]F_2=\dfrac{kQ1}{(x+2)^2}[/tex]

As the force of two is equal and opposite thus,

[tex]\dfrac{kQ1}{x^2}=\dfrac{kQ2}{(x+2)^2}\\\dfrac{2.5\times10^{-5}}{x^2}=\dfrac{-5\times10^{-6}}{(x+2)^2}\\x=1.62[/tex]

Thus the position of point 2 is,

[tex]p_2=2+1.62\\p_2=3.62\rm m[/tex]

Thus, the location of the place along the x axis where the electric field due to these two charges is zero is 3.62 meters.

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The center of mass is given by:

∑mx/∑m

m is the mass of each object

x is the position of each object

We will assign x = 0m to the 18kg mass, therefore x = 6m for the 6kg mass.

∑mx/∑m = (18×0+6×6)/(18+6) = 1.5m

The center of mass is located 1.5m away from the 18kg mass.

The total moment of inertia of the system about the center of mass is given by:

I = ∑mr²

I is the moment of inertia

m is the mass of each object

r is the distance of each object from the center of mass

We know r = 1.5m for the 18kg mass and the rod is 6m long, therefore the 6kg mass must be r = 4.5m from the center of mass.

I = 18(1.5)² + 6(4.5)²

I = 162kg×m²

Answer:

3.014 x 10⁻⁸ N

Explanation:

q = magnitude of charge on the supersonic jet = 0.55 μC = 0.55 x 10⁻⁶ C

v = speed of the jet = 685 m/s

B = magnitude of magnetic field in the region = 8 x 10⁻⁵ T

θ = angle between the magnetic field and direction of motion = 90

magnitude of the magnetic force is given as

F = q v B Sinθ

F = (0.55 x 10⁻⁶) (685) (8 x 10⁻⁵) Sin90

F = 3.014 x 10⁻⁸ N

Answer:

1.7 m/s²

Explanation:

d = length of the ramp = 13.5 m

v₀ = initial speed of the skateboarder = 0 m/s

v = final speed of the skateboarder = 7.37 m/s

a = acceleration

Using the equation

v² = v₀² + 2 a d

7.37² = 0² + 2 a (13.5)

a = 2.01 m/s²

θ = angle of the incline relative to ground = 29.9

a' = Component of acceleration parallel to the ground

Component of acceleration parallel to the ground is given as

a' = a Cosθ

a' = 2.01 Cos29.9

a' = 1.7 m/s²

Answer:

kinematic viscosity in SUS is = 671.64 SUS

Explanation:

given data

kinetic viscosity = 145 mm^2/s

we know

1 mm = 0.1 cm

so kinetic viscosity in cm is [tex]\nu =145 (0.1)^{2} =1.45 cm^{2}/s[/tex]

other unit of kinetic viscosity is centistokes

[tex]1 cm^{2}/s = 100 cst[/tex]

so 1.45 cm^2/s will be 145 cst

if the temperature is 260°f , then cst value should be multiplied by 4.632. therefore kinematic viscosity in SUS is = 4.362 *145 = 671.64 SUS

Answer:

(a):The time what it takes his voice reach the earth via radio waves is t1= 1.28 seconds.

(b): The time what it takes his voice reach from Mars to the earth via radio waves is t2= 186.66 seconds.

Explanation:

d1= 3.85 *10⁸ m

d2= 5.6 *10¹⁰ m

C= 300,000,000 m/s

t1= d1/C

t1= 1.28 s

t2= d2/C

t2= 186.66 s

To find the electric field in the wire, we first find the cross-sectional area of the wire and then substitute the given values and the calculated area into the formula for electric field. The electric field in the gold wire is approximately 8.24 kN/C.

The question asks for the electric field in a gold wire of given dimensions and current. We'll solve this using the formula for electric field (E) in terms of resistivity (ρ),current (I) and area (A), which is given by E = ρI/A.

First, we need to find the cross-sectional area (A) of the wire using the formula for the area of a circle, since the wire is cylindrical. The area of a circle is given by A = π*(d/2)^2, where d is the diameter. Substituting d = 1.8 mm or 1.8 * 10^-3 m, we find A ≈ 2.54 * 10^-6 m^2.

Next, we need to substitute the known values into the formula E = ρI/A. Using the given values ρ = 2.44 *10^-8 Ω•m, I = 860 mA or 860 * 10^-3 A, and the calculated A ≈ 2.54 * 10^-6 m^2, we find E ≈ 8.24 * 10^3 N/C or 8.24 kN/C.

So, the electric field in the wire is approximately 8.24 kN/C.

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The electric field within the gold wire can be calculated utilizing resistivity formula, determining the cross-sectional area of the wire, and then computing the current density. After these intermediate calculations, the ultimate electric field in the wire is found to be approximately 72 mV/m.

The electric field in the gold wire can be calculated by using the formula for

resistivity: ρ=E/J where

ρ (rho) is the resistivity of the material,

E is the electric field, and

J is the current density.

Before we apply this formula, let's calculate the cross-sectional area of the wire (A) using the formula A=πr², where r is the radius of the wire. The diameter is given as 1.8mm, therefore

r=0.9mm = 0.9x10^-3 m.

Substituting this into the formula gives us

A=π(0.9x10^-3)²≈2.54x10^-6 m².

Now we calculate current density: J=I/A. where I is current, given as

860mA=860x10^-3 A.

Substituting values of I and A into the formula gives

J=860x10^-3/2.54x10^-6≈3.39x10^8 A/m².

Finally, the electric field E=ρ/J=2.44x10^-8/3.39x10^8 ≈ 7.2x10^-2 V/m or 72mV/m.

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