A 4.0 µC point charge and a 3.0 µC point charge are a distance L apart. Where should a third point charge be placed so that the electric force on that third charge is zero? (Use the following as necessary: L.)

Explanation:

Instantaneous velocity is specific rate of the change of the position or the displacement with respect to the time at a particular single point (x,t)

[tex]v(t)=\frac{d}{dt}x(t)[/tex]

Average velocity is average rate of the change of the position or the displacement with respect to the time over an particular interval.

[tex]{v}=\frac {\Delta x}{\Delta t}=\frac{{x}_{\text{f}}-{x}_{\text{i}}}{{t}_{\text{f}}-{t}_{\text{i}}}[/tex]

Answer:

C. The total distance is 105 kilometers and the total displacement is 15 kilometers east.

Explanation:

The main difference between distance and displacement is that: distance is a measure of the total length traveled along the road, the displacement only takes into account the length between the initial (departure) and final (arrival) position.

So the boat traveled 105 km, but the displacement between the start and finish point is only 15 km.

:)

Answer:

zero, acceleration due to gravity = 9.8 m/s^2

Explanation:

When an object throws vertically upwards, the acceleration acting on the object is acceleration due to gravity which is acting in vertically downwards direction.

As the object moves upwards, its velocity goes on decreasing because the direction of velocity and the acceleration both are opposite to each other. At maximum height, the velocity of object becomes zero and then object starts moving in downwards direction.

Thus, the value of instantaneous velocity at maximum height is zero but the vale of acceleration is acceleration due to gravity which is acting vertically downward direction.

Answer:

The magnitude of electric field is [tex]1.25\times10^{14}\ N/C[/tex] in downward.

Explanation:

Given that,

Force [tex]F= 4.00\times10^{-5}\ N[/tex]

Charge q= -2.00 μC

We know that,

Charge is negative, then the electric field in the opposite direction of the exerted force.

We need to calculate the magnitude of electric field

Using formula of electric force

[tex]F = qE[/tex]

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

[tex]E=\dfrac{4.00\times10^{-5}}{-2.00\times1.6\times10^{-19}}[/tex]

[tex]E=-1.25\times10^{14}\ N/C[/tex]

Negative sign shows the opposite direction of electric force.

Hence, The magnitude of electric field is [tex]1.25\times10^{14}\ N/C[/tex] in downward.

Answer:

h=5.04m

Explanation:

At the bigining he has only kinetic energy because he is at the lowest point. When he is at the highest point, he is no longer moving because he will start moving downwards: all his kinetic energy transformed into potential gravitational energy:

[tex]E_k=E_p[/tex]

[tex]\frac{mv^2}{2}=mgh\\ h= \frac{v^2}{g}[/tex]

h=5.04m

Answer:

The height of the highest point reached by the skateboarder on the right side of the ramp is [tex]h=2.52m[/tex]

Explanation:

In this problem, we use the Energy Conservation Principle, as there are not nonconservative forces, the energy is constant and we will call it E.  Then, we choose two different moments that will be advantageous for our analysis.

The first moment is the moment where the skateboarder is situated at the height 0 meters (meaning that the potential energy is zero), i.e, the energy is totally kinetic. We will call it K.

The second moment is the moment where the skateboarder is situated at the highest point reached, this means that all energy will be potential, because for an instant, the velocity of the skateboarder is zero (and in the following instant he will go back the way he came due to the pull of gravity). We will call it V. This is the point with the height that we want to calculate.

Therefore, we can equalize both energies, as they are constant and equal to E. We write

[tex]E=K=V=\frac{1}{2}mv^2 =mgh\Leftrightarrow \frac{1}{2}v^2=gh\Leftrightarrow h=\frac{v^2}{2g}[/tex]

Now, we can replace the given data, to obtain

[tex]h=\frac{7.03^2}{2*9.80}m=2.52m[/tex]

which is the outcome.

Answer:

106.52 minutes

Explanation:

Given:

Initial distance between Carl and Isaac = 175 miles

speed of Isaac = 65 mph

Speed of Carl = 50 mph

Now, the Carl starts after 35 minutes, but the Isaac has already started so the distance covered by Isaac in 35 minutes will be

= [tex]\frac{35}{60}\times65[/tex]

= 37.91 miles

Therefore,

the distance left between Isaac and Carl = 175 - 37.91 = 137.09 miles

Since Isaac and Carl are moving towards each other,

therefore the relative speed between the both = 65 + 50 = 115 mph

Hence, the time taken to meet = [tex]\frac{137.09}{115}[/tex]

or

The time taken to meet = 1.192 hours

or

The time taken = 1.192 × 60 = 71.52 minutes

Therefore the total time Isaac have been travelling = 71.52 + 35

= 106.52 minutes

Answer:

yield strength is 64000 lb/ in²

modulus of elasticity is 29.76 ×[tex]10^{6}[/tex] lb/in²

Explanation:

given data

length L = 2 in

area A = 0.5 in²

load = 32000 lb

gage length L1 = 2.0083 in

yield point = 0.2%

maximum load = 60000 lb

to find out

yield strength and modulus of elasticity

solution

we apply here yield strength that is express as

yield strength = [tex]\frac{load}{area}[/tex]  ...........1

yield strength =  [tex]\frac{32000}{0.5}[/tex]

yield strength = 64000 lb/ in²

and

modulus of elasticity  is calculated as

modulus of elasticity =  [tex]\frac{yield strength}{strain}[/tex]    ..........2

here strain = [tex]\frac{L1 - L}{L}[/tex]  

strain = [tex]\frac{2.0083 - 2}{2}[/tex]  

strain = 0.00415

so new strain after offset is here 0.00415 - 0.002

new strain = 0.00215

so from equation 2

modulus of elasticity =  [tex]\frac{yield strength}{strain}[/tex]  

modulus of elasticity =  [tex]\frac{64000}{0.00215}[/tex]  

modulus of elasticity is 29.76 ×[tex]10^{6}[/tex] lb/in²

Final answer:

The yield strength is 64,000 psi, modulus of elasticity is approximately 15.422 x 106 psi, and the tensile strength is 120,000 psi based on the data given from the tensile test of a specimen.

Explanation:

Yield Strength, Modulus of Elasticity, and Tensile Strength

To address the given tensile test problem, we first need to determine the yield strength, which is found by dividing the load at yield by the original area. Hence, the yield strength (\(\sigma_y\)) is 32,000 lb divided by 0.5 in2, which equals 64,000 psi.

The modulus of elasticity (E) can be calculated using Hooke's Law, which is the stress over the strain. In this scenario, the initial stress (\(\sigma_i\)) is the yield load (32,000 lb) over the area (0.5 in2) and the initial strain (\(\epsilon_i\)) is the change in length (0.0083 in) over the original gage length (2.0 in). Therefore, E is 64,000 psi divided by 0.00415, which equals approximately 15,422,000 psi or 15.422 x 106 psi.

As for the tensile strength, it is the maximum stress that the material can withstand while being stretched or pulled before necking, which is the maximum load (60,000 lb) divided by the original cross-sectional area (0.5 in2), which gives us 120,000 psi.

Answer:

Explanation:

GIVEN DATA:

Distance between keisha and her friend 8.3 m

angle made by keisha toside building 30 degree

height of her friend monique is 1.5 m

from the figure

[tex]\Delta ACB[/tex]

[tex]tan 30 = \frac{8.3}{h}[/tex]

[tex]h= \frac{8.3}{tan 30} = 14.376 m[/tex]

therefore

height of keisha is [tex]= h  + 1.5 m[/tex]

                               = 14.376 + 1.5

[tex]= 15.876 \simeq 16 m[/tex]

therefore option c is correct

Answer:

Speed of the cosmic rays, [tex]v=4.64\times 10^8\ km/hr[/tex]

Explanation:

It is given that, Cosmic rays are highly energetic particles which can travel at great speeds through outer space.

The speed of the cosmic rays, [tex]v=1.29\times 10^8\ m/s[/tex]

We need to convert the speed of cosmic rays to kilometers per hour. We know that :

1 kilometers = 1000 meters

1 hour = 3600 seconds

[tex]v=1.29\times 10^8\ m/s=\dfrac{1.29\times 10^8\times (1/1000\ km)}{(1/3600\ h)}[/tex]

On solving the above expression,

[tex]v=464400000\ km/h[/tex]

or

[tex]v=4.64\times 10^8\ km/hr[/tex]

So, the speed of the cosmic rays is [tex]4.64\times 10^8\ km/hr[/tex]. Hence, this is the required solution.

Answer:

No change

Explanation:

The time period of an oscillating body depends on the mass of the body attached and the spring constant of the spring.

The time taken by the oscillating body to complete one vibration is called the time period of teh body.

As the time period does not depend on the amplitude of the oscillations so the time period does not change as the amplitude changes.

Thus, the time taken by the mass to reach to the equilibrium position remains  same.

Answer:3.75 s

Explanation:

Given Body travels half of its motion in last 1.1 sec

Let h be the height and t be the total time taken

here initial velocity is zero

[tex]h=ut+\frac{gt^2}{2}[/tex]

[tex]h=0+\frac{gt^2}{2}[/tex]

[tex]h=\frac{gt^2}{2}------1[/tex]

Now half of the distance traveled will be in t-1.1 s and half distance traveled is in last 1.1 s

[tex]\frac{h}{2}=\frac{g\left ( t-1.1\right )^2}{2}-----2[/tex]

from 1 & 2 we get

[tex]gt^2=2g\left ( t-1.1\right )^2[/tex]

[tex]t^2-4.4t+2.42=0[/tex]

[tex]t=\frac{4.4\pm \sqrt{4.4^2-4\left ( 1\right )\left ( 2.42\right )}}{2}[/tex]

[tex]t=\frac{4.4\pm 3.11}{2}[/tex]

Therefore two value of t is satisfying the equation  but only one value is possible

therefore t=3.75 s

The resultant electric field (Er) is approximately 231.76 V/m, directed at an angle of about 62.76° to the left of the vertical E1 direction.

Find the resultant electric field when there are two fields, E1 and E2:

Step 1: Resolve the electric fields into their x and y components.

E1:

Ex1 = 0 V/m (because E1 is directed vertically upward)

Ey1 = 100 V/m

E2:

Since E2 is directed 45 degrees to the left of E1, we can use trigonometry to find its x and y components.

Ex2 = 150 V/m * sin(45°) = 106.07 V/m (directed to the left, so negative)

Ey2 = 150 V/m * cos(45°) = 106.07 V/m

Step 2: Add the x and y components of each electric field separately.

Σ Ex = Ex1 + Ex2 = 0 V/m - 106.07 V/m = -106.07 V/m

Σ Ey = Ey1 + Ey2 = 100 V/m + 106.07 V/m = 206.07 V/m

Step 3: Find the magnitude of the resultant electric field (Er) using the Pythagorean theorem.

Er = √(Σ Ex)² + (Σ Ey)²

Er = √(-106.07 V/m)² + (206.07 V/m)²

Er ≈ 231.76 V/m

Step 4: Find the direction of the resultant electric field using arctangent.

tan(θ) = Σ Ey / Σ Ex

tan(θ) = 206.07 V/m / -106.07 V/m

θ ≈ arctan(-1.94) ≈ -117.24°

Since the arctangent function only outputs values between -90° and 90°, we need to add 180° to get the angle within the range of 0° to 180°.

θ = -117.24° + 180° ≈ 62.76°

Therefore, the resultant electric field has a magnitude of approximately 231.76 V/m and is directed approximately 62.76° to the left of E1.

Answer:

[tex]h=53.09m[/tex]         (2)

[tex]v_{min}>5.05m/s[/tex]

[tex]v_{max}<10.4m/s[/tex]

Explanation:

a)Kinematics equation for the first ball:

[tex]v(t)=v_{o}-g*t[/tex]

[tex]y(t)=y_{o}+v_{o}t-1/2*g*t^{2}[/tex]

[tex]y_{o}=h[/tex]       initial position is the building height

[tex]v_{o}=8.9m/s[/tex]      

The ball reaches the ground, y=0, at t=t1:

[tex]0=h+v_{o}t_{1}-1/2*g*t_{1}^{2}[/tex]

[tex]h=1/2*g*t_{1}^{2}-v_{o}t_{1}[/tex]           (1)

Kinematics equation for the second ball:

[tex]v(t)=v_{o}-g*t[/tex]

[tex]y(t)=y_{o}+v_{o}t-1/2*g*t^{2}[/tex]

[tex]y_{o}=h[/tex]       initial position is the building height

[tex]v_{o}=0[/tex]       the ball is dropped

The ball reaches the ground, y=0, at t=t2:

[tex]0=h-1/2*g*t_{2}^{2}[/tex]

[tex]h=1/2*g*t_{2}^{2}[/tex]         (2)

the second ball is dropped a time of 1.03s later than the first ball:

t2=t1-1.03              (3)

We solve the equations (1) (2) (3):

[tex]1/2*g*t_{1}^{2}-v_{o}t_{1}=1/2*g*t_{2}^{2}=1/2*g*(t_{1}-1.03)^{2}[/tex]

[tex]g*t_{1}^{2}-2v_{o}t_{1}=g*(t_{1}^{2}-2.06*t_{1}+1.06)[/tex]

[tex]g*t_{1}^{2}-2v_{o}t_{1}=g*(t_{1}^{2}-2.06*t_{1}+1.06)[/tex]

[tex]-2v_{o}t_{1}=g*(-2.06*t_{1}+1.06)[/tex]

[tex]2.06*gt_{1}-2v_{o}t_{1}=g*1.06[/tex]

[tex]t_{1}=g*1.06/(2.06*g-2v_{o})[/tex]

vo=8.9m/s

[tex]t_{1}=9.81*1.06/(2.06*9.81-2*8.9)=4.32s[/tex]

t2=t1-1.03              (3)

t2=3.29sg

[tex]h=1/2*g*t_{2}^{2}=1/2*9.81*3.29^{2}=53.09m[/tex]         (2)

b)[tex]t_{1}=g*1.06/(2.06*g-2v_{o})[/tex]

t1 must :   t1>1.03  and t1>0

limit case: t1>1.03:

[tex]1.03>9.81*1.06/(2.06*g-2v_{o})[/tex]

[tex]1.03*(2.06*9.81-2v_{o})<9.81*1.06[/tex]

[tex]20.8-2.06v_{o}<10.4[/tex]

[tex](20.8-10.4)/2.06<v_{o}[/tex]

[tex]v_{min}>5.05m/s[/tex]

limit case: t1>0:

[tex]g*1.06/(2.06*g-2v_{o})>0[/tex]

[tex]2.06*g-2v_{o}>0[/tex]

[tex]v_{o}<1.06*9.81[/tex]

[tex]v_{max}<10.4m/s[/tex]

Answer:

129 m because the average velocity is 13.9 m/s, the change in velocity

divided by the acceleration is the time, and the time multiplied by the

average velocity is the distance. ⇒ answer C

Explanation:

Lets explain how to solve the problem

The given is:

The care is able to stop with an acceleration of -3 m/s²

→ The final velocity = 0 and acceleration = -3 m/s²

Calculate the distance required to stop from a velocity of 100 km/h

→ Initial velocity = 100 km/h

At first we must to change the unite of the initial velocity from km/h

to m/s because the units of the acceleration is m/s²

→ 1 km = 1000 meters and 1 hr = 3600 seconds

→ 100 km/h = (100 × 1000) ÷ 3600 = 27.78 m/s

The initial velocity is 27.78 m/s

Acceleration is the rate of change of velocity during the time,

then the time is the change of velocity divided by the acceleration

→ [tex]t=\frac{v-u}{a}[/tex]

where v is the final velocity, u is the initial velocity, t is the time and

a is the acceleration

→ v = 0 , u = 27.78 m/s , a = -3 m/s²

Substitute these values in the rule

→ [tex]t=\frac{0-27.78}{-3}=9.26[/tex] seconds

The time to required stop is 9.26 seconds

We can calculate the distance by using the rule:

→ s = ut + [tex]\frac{1}{2}[/tex] at²

→ u = 27.78 m/s , t = 9.26 s , a = -3 m/s²

Substitute these values in the rule

→ s = 27.78(9.26) + [tex]\frac{1}{2}[/tex] (-3)(9.26) = 128.6 ≅ 129 m

The distance required to stop is 129 m

Average velocity is total distance divided by total time

→ Total distance = 129 m and total time = 9.26 s

→ average velocity = 129 ÷ 9.26 = 13.9 m/s

The average velocity is 13.9 m/s

So the time multiplied by the average velocity is the distance

The answer is C

129 m because the average velocity is 13.9 m/s, the change in

velocity divided by the acceleration is the time, and the time

by the average velocity is the distance.

Answer:

Average speed 21.67 m/s

Average velocity=8.33 m/s

Explanation:

In order to solve this, We have to know the difference between speed and velocity.

Velocity is a vector value, so it means we have to specify the magnitud and direction, we take the displacement instead of the distance.

speed is an scalar value, we only care about the magnitud, we take in count the total distance.

first we have to get the total amount of time

Total time= 30min+10min+20min=60min

Total time is= 3600 seconds

the distance on the first trame is:

[tex]X1=v*t\\\\X1=30m/s*(30min)(60sec/1min)\\X1=54km[/tex]

the distance on the second trame is:

[tex]X2=v*t\\\\X2=20m/s*(20min)(60sec/1min)\\X2=24km[/tex]

The total distance is the sum of both values

Td=78km

the Displacement is the vectorial sum of them, because the second trame is opposite the first, we have to substract the second distance

D=54-24=30km

[tex]Speed(avg)=\frac{distance}{time} \\\\Speed(avg)=\frac{78000m}{3600} \\\\Speed(avg)=21.67 m/s[/tex]

[tex]Velocity(avg)=\frac{displacement}{time} \\\\Velocity(avg)=\frac{30000m}{3600} \\\\Velocity(avg)=8.33 m/s[/tex]

Answer:

a) -180.7 kN/C

b) -474.3 kN/C

c) 180.7 kN/C

Explanation:

For infinite planes the electric field is constant on each side, and has a value of:

E  = σ / (2 * e0) (on each side of the plate the field points in a different direction, the fields point towards positive charges and away from negative charges)

The plate at -5 m produces a field of:

E1 = 2.6*10^-6 / (2 * 8.85*10^-12) = 146.8 kN/C into the plate

The plate at 3 m:

E2 = 5.8*10^-6 / (2 * 8.85*10^-12) = 327.5 kN/C away from the plate

At x < -5 m the point is at the left of both fields

The field would be E = 146.8 - 327.5 = -180.7 kN/C

At -5 m < x < 3 m, the point is between the plates

E = -146.8 - 327.5 = -474.3 kN/C

At x > 3 m, the point is at the right of both plates

E = -146.8 + 327.5 = 180.7 kN/C

The electric fields at the given locations are calculated by using the formula E=σ/2ε₀, and considering that the fields always point toward negative charges. Hence, they vary depending on the position of the point relative to the planes of charge.

The field contribution due to each individual infinite plane is E=σ/2ε₀, where ε₀ is the permittivity of the free space (8.85 x10⁻ⁱ² C²/Nm²). Now, we'll calculate the electric field for each given location:

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Answer:

588 J

Explanation:

mass of ball, m = 6 kg

Height from it dropped, h = 10 m

initial velocity, u = 0

acceleration due to gravity, g = 9.8 m/s^2

Let it hits the ground with velocity v.

Use third equation of motion

[tex]v^{2} = u^{2}+2as[/tex]

[tex]v^{2}=0^{2}+2\times 9.8\times 10[/tex]

v = 14 m/s

The formula for the kinetic energy is given by

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

where, m is the mass of the ball and v be the velocity of the ball as it hits the ground

K = 0.5 x 6 x 14 x 14

K = 588 J

Answer:

intensity of sound at level of microphone is 0.00139 W / m 2

sound intensity level at position of micro phone is 91.456 dB

EXPLANATION:

Given data:

power of sound   P = 31 W

distance betwen microphone & speaker is   42 m

a) intensity of sound at microphone is calculated as

                   [tex]I = \frac{P}{A}[/tex]

                   [tex]= \frac{34}{4 \pi ( 44m )^ 2}[/tex]

                   =  0.00139 W / m 2        

b) sound intensity level at position of micro phone is

                [tex]\beta = 10 log \frac{I}{I_o}[/tex]

    where I_o id reference sound intensity and taken as

                [tex]= 1 * 10^{-12} W / m 2[/tex]  

               [tex]\beta = 10 log\frac{0.00139}{10^[-12}}[/tex]

                     = 91.456 dB

Two points are given in polar coordinates,  the distance between the two points is approximately 3.12m.

You can use the polar-to-cartesian conversion formula to convert each point to Cartesian coordinates (x, y), then apply the distance formula in

Cartesian coordinates to determine the separation between two points supplied in polar coordinates.

The polar-to-cartesian conversion formulas are:

[tex]\[ x = r \cdot \cos(\theta) \][/tex]

[tex]\[ y = r \cdot \sin(\theta) \][/tex]

Given the points:

[tex]\( (r_1, \theta_1) = (2.60 \, \text{m}, 50.0^\circ) \)[/tex]

[tex]\( (r_2, \theta_2) = (3.60 \, \text{m}, -46.0^\circ) \)[/tex]

Converting these points to Cartesian coordinates:

For the first point:

[tex]\[ x_1 = 2.60 \, \text{m} \cdot \cos(50.0^\circ) \approx 1.66 \, \text{m} \][/tex]

[tex]\[ y_1 = 2.60 \, \text{m} \cdot \sin(50.0^\circ)\\\\ \approx 1.98 \, \text{m} \][/tex]

For the second point:

[tex]\[ x_2 = 3.60 \, \text{m} \cdot \cos(-46.0^\circ)\\\\ \approx 2.53 \, \text{m} \][/tex]

[tex]\[ y_2 = 3.60 \, \text{m} \cdot \sin(-46.0^\circ)\\\\ \approx -2.57 \, \text{m} \][/tex]

Now, you can use the distance formula in Cartesian coordinates to find the distance between the two points:

[tex]\[ \text{distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]

Plug in the values and calculate:

[tex]\[ \text{distance} = \sqrt{(2.53 \, \text{m} - 1.66 \, \text{m})^2 + (-2.57 \, \text{m} - 1.98 \, \text{m})^2}\\\\ \approx 3.12 \, \text{m} \][/tex]

Thus, the distance between the two points is approximately 3.12 m.

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Answer:

57300 N

Explanation:

The container has a mass of 5300 kg, the weight of the container is:

f = m * a

w = m * g

w = 5300 * 9.81 = 52000 N

However this container was moving with more acceleration, so dynamic loads appear.

w' = m * (g + a)

w' = 5300 * (9.81 + 1) = 57300 N

The rating for the cable was 50000 N

The maximum load was exceeded by:

57300 / 50000 - 1 = 14.6%

Answer:

[tex]a.b=|a||b|\ cos\theta[/tex]                                                                    

Explanation:

Let a and b are two vectors such that [tex]\theta[/tex] is the angle between them. Dot product is also known as scalar product.  It is used to find the angle between two vectors such that,

[tex]a.b=|a||b|\ cos\theta[/tex]

[tex]\theta[/tex] is the angle between a and b. It can be calculated as :

[tex]\theta=cos^{-1}(\dfrac{a.b}{|a||b|})[/tex]

[tex]|a|\ and\ |b|[/tex] are the magnitude of vectors a and b such that :

[tex]|a|=\sqrt{x^2+y^2+z^2}[/tex] if a = xi +yj +zk

and

[tex]|b|=\sqrt{p^2+q^2+r^2}[/tex] if a = pi +qj +rk            

So, the correct option is (a). Hence, this is the required solution.                                        

Answer:

58.703 ft/centuri

Explanation:

Length of the hair grow, l = 0.49 mm

Time to grow, t = 1 day

Convert mm into feet.

We know that, 1 mm = 0.00328 feet

So, 0.49 mm = 0.49 x 0.00328 = 0.0016072 ft

Convert day into century.

We know that, 1 century = 36525 days

So, 1 day = 1 / 36525 centuri

growth rate of hair =  [tex]\frac{0.0016072}{\frac{1}{36525}}[/tex]

                               = 58.703 ft/centuri

Answer:

The spring constant k is[tex]1.115\times 10^{9} N/m[/tex]

Solution:

As per the question:

Length of the solid cylinder, L = 500 mm = [tex]500\times 10^{- 3} = 0.5 m[/tex]

Diameter pf the cylinder, D = 2 cm = 0.02 m

As the radius is half the diameter,

Radius, R = 1 cm = 0.01 m

Young's Modulus, E = 17.4 GPa = [tex]17.4\times 10^{9} Pa[/tex]

Now,

The relation between spring constant, k and Young's modulus:

[tex]kL = EA[/tex]

where

A = Area

Area of solid cylinder, A = [tex]2\piR(L + R)[/tex]

[tex]0.5k = 17.4\times 10^{9}\times 2\piR(L + R)[/tex]

[tex]k = \frac{17.4\times 10^{9}\times 2\pi\times 0.01(0.01 + 0.5)}{0.5}[/tex]

k = [tex]1.115\times 10^{9} N/m[/tex]

Young's modulus, E is the ratio of stress and strain

And

Stress = [tex]\frac{Force or thrust}{Area}[/tex]

Strain = [tex]\frac{length, L}{elongated or change in length, \Delta L}[/tex]

Also

Force on a spring is - kL

Therefore, we utilized these relations in calculating the spring constant.

Answer:3.71 m/s

Explanation:

Given

square track with sides 50 m

average speed is 5 m/s

Total running time=53.8 s

Total distance traveled  in this time[tex]=53.8\times 5=269 m[/tex]

i.e. Person has completed square track one time and another 69 m in second round

So displacement is 269-200=69 m

average velocity[tex]=\frac{Displacement}{time}[/tex]

[tex]=\frac{69}{53.8}=1.28 m/s[/tex]

Difference between average velocity and average speed is

5-1.28=3.71 m/s

Answer:

The difference in average speed and average velocity in terms of magnitude is 3.993 m/s

Solution:

As per the question:

The side of a square track, l = 50 m

Average speed of the runner, [tex]v_{avg} = 5 m/s[/tex]

Time taken, t = 53.80 s

Now,

The distance covered by the runner in this time:

s = [tex]v_{avg}t[/tex]

s = [tex]5\times 53.80[/tex]

s = 269 m

After covering a distance of 269 m, the person is at point A:

[tex]AQ^{2} = AR^{2} + QR^{2}[/tex]

where

AR = 19 m

QR = 50 m

Refer to fig 1.

As the runner starts from the bottom right, i.e., at Q and traveled 269 m.

After completion of 250 m , he will be at point R after one complete round and thus travels 19 m more to point A to cover 269 m.

Thus

[tex]AQ = \sqrt{19^{2} + 50^{2}} = 53.48 m[/tex]

where

AQ is the displacement

Hence,

Average velocity, v' = [tex]\frac{AQ}{t}[/tex]

v' = [tex]\frac{53.48}{53.80} = 1.007 m/s[/tex]

The difference in average speed and average velocity is:

[tex]v_{avg} - v' = 5 - 1.007 = 3.993 m/s[/tex]

Answer:

a) 90m

b) 8.1s

Explanation:

The shortest way to finding the length of the incline is to apply an energetic analysis to determine the height of the incline. At the beginning, there is potential and kinetic energy that will turn into kinetic energy only:

[tex]mgh+\frac{1}{2}mv_{0}^{2} =\frac{1}{2}mv_{f}^{2}[/tex]

Before we input any information, let's solve for h:

[tex]mgh+\frac{1}{2}mv_{0}^{2} =\frac{1}{2}mv_{f}^{2}}\\\\m(gh+\frac{v_{0}^{2}}{2})=\frac{1}{2}mv_{f}^{2}}\\\\gh+\frac{v_{0}^{2}}{2}=\frac{v_{f}^{2}}{2}\\\\gh=\frac{1}{2}(v_{0}^{2}-v_{f}^{2})\\\\h=\frac{1}{2g}(v_{0}^{2}-v_{f}^{2})=\frac{1}{2*9.8\frac{m}{s^{2}}}(4.2\frac{m}{s}^{2}-18\frac{m}{s}^{2})=15.63m[/tex]

Using a sine formula we can solve for [tex]l[/tex]:

[tex]Sin(10)=\frac{h}{l}\\\\l=\frac{h}{Sin(10)}=\frac{15.63m}{0.17}=90m[/tex]

In order to find the time we will use the distance formula and final velocity formula as a system of equations to solve for t:

[tex]X=V_{0}t+\frac{at^2}{2}\\\\V_f=V_0+at[/tex]

Since the acceleration and time are both variables we will solve for acceleration in the final velocity formula and replace in the distance formula:

[tex]V_f=V_0+at\\V_f-V_0=at\\\frac{V_f-V_0}{t}=a\\\\l=V_0t+\frac{(\frac{V_f-V_0}{t})t^2}{2}\\l=V_0t+\frac{(V_f-V_0)t}{2}\\l=t(V_0+\frac{V_f-V_0}{2})\\\\t=\frac{l}{V_0+\frac{V_f-V_0}{2}}=\frac{90m}{4.2\frac{m}{s}+\frac{18\frac{m}{s}-4.2\frac{m}{s}}{2}}=8.1s[/tex]

Final answer:

In this question, we calculate the length of the incline and the time taken by a skier to reach the bottom based on given velocity and angle.

Explanation:

For part a: The length of the incline can be found using the principles of physics. Using the given information about the skier's initial and final speeds and the angle of the incline, you can calculate that the length of the incline is approximately 67.7 meters.

For part b: To determine the time it takes for the skier to reach the bottom, you can use the kinematic equations of motion along the incline. The time taken for the skier to reach the bottom is about 9.64 seconds.

Answer:

0.92 μC

Explanation:

In a parallel-plate capacitor, the electric field formed is equal to the charge density divited by the vacuum permisivity e0, as there are no dielectric between the plates. e0 is equal to 8.85*10^-12 C^2/Nm^2. The charge density is the total charge of each individual plate divided by its area. Then, the maximum charge allowed will be equal to:

[tex]E = \frac{o}{e_0} = \frac{Q}{Ae_0} \\ Q = E*A*e_0 = 3*10^6 N/C * (0.25*\pi *(0.21m)^2)*8.85*10^{-12}C^2/Nm^2 = 9.196 *10^{-7} C[/tex]

or 0.92 μC

Answer:

The flux of the electric field through the surface is 3.24\times10^{3}\ Nm^/C[/tex].

Explanation:

Given that,

Area of cube = 48 cm²

Charge = 28.7 nC

We need to calculate the flux of the electric field through the surface

Using formula Gauss's law

The electric flux through any closed surface,

[tex]\phi =\dfrac{q}{\epsilon_{0}}[/tex]

Where, q = charge

Put the value into the formula

[tex]\phi=\dfrac{28.7\times10^{-9}}{8.85\times10^{-12}}[/tex]

[tex]\phi =3.24\times10^{3}\ Nm^/C[/tex]

Hence, The flux of the electric field through the surface is 3.24\times10^{3}\ Nm^/C[/tex].

Final answer:

The car takes approximately 6.91 seconds to stop and travels approximately 110.7 meters before coming to a stop.

Explanation:

The car's initial speed is 31.8 m/s and it decelerates at a rate of 4.60 m/s2. To find the time it takes to stop, we can use the equation:

Final velocity = Initial velocity + (Acceleration x Time)

Since the final velocity is 0 m/s when the car stops, we can rearrange the equation to solve for time:

Time = (Final velocity - Initial velocity) / Acceleration

Plugging in the values, we get:

Time = (0 m/s - 31.8 m/s) / -4.60 m/s2

Simplifying the equation, we find that it takes approximately 6.91 seconds for the car to stop.

To find the distance traveled, we can use the equation:

Distance = Initial velocity x Time + (0.5 x Acceleration x Time2)

Plugging in the values, we get:

Distance = 31.8 m/s x 6.91 s + (0.5 x -4.60 m/s2 x (6.91 s)2)

Simplifying the equation, we find that the car travels approximately 110.7 meters before it comes to a stop.

Answer:

Case I: 12.617 L/s

Case II: 161.406 cubic meters per hour

Case III: 1.062 Pound inches

Explanation:

Given:

Assumptions:

Case I:

[tex]Speed\ of\ water\ flow = 200 \dfrac{gallon}{min}\\\Rightarrow V_{water} = 200\times \dfrac{3.785\ L}{60\ s}\\\Rightarrow V_{water} = 12.617\ L/s[/tex]

Case II:

[tex]Speed\ of\ air\ blow = 95 \dfrac{ft^3}{min}\\\Rightarrow V_{air} = 95\times \dfrac{(0.3048\ m)^3}{\dfrac{1}{60}\ h}\\\Rightarrow V_{air} = 95\times (0.3048)^3\times 60\ m^3/h\\\Rightarrow V_{air} = 161.406\ m^3/h[/tex]

Case III:

[tex]Measure\ of\ torque = 12\ N cm\\\Rightarrow \tau = 12\times (0.2248\ lb)\times \dfrac{1}{2.54}\ in\\ \Rightarrow \tau = 1.062\ lb in[/tex]

Answer:

F = -10800 N

Explanation:

Given that,

Charge 1, [tex]q_1=-1.5\ C[/tex]

Charge 2, [tex]q_2=0.8\ C[/tex]

Distance between the charges, [tex]d=1\ km=10^3\ m[/tex]

We need to find the electric force acting between two point charges. Mathematically, it is given by :

[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]

[tex]F=-9\times 10^9\times \dfrac{1.5\times 0.8}{(10^3)^2}[/tex]

F = -10800 N

So, the magnitude of electric force between two point charges is 10800 N. Hence, this is the required solution.