A thin metal bar, insulated along its sides, is composed of five different metal connected together. The left end bar is immersed in a heat bath at 100°C and right end in a heat bath at 0°C. Starting at the left end, the pieces and lenghts are steel(2cm), brass(3cm), copper(1cm), aluminum(5cm) and silver(1cm). What is the temperature of the steel/brass interface?

Answer:

[tex]acceleration=0.8 \frac{m}{s^{2} } \\distance=112.5 m[/tex]

Explanation:

We can find these answers following the equations of motion.

[tex]a=[/tex]Δ[tex]V/t[/tex]

Where Δ[tex]V[/tex] is the difference between the final speed and the initial speed. And [tex]t[/tex] is the time spent

We replace the terms:

[tex]a=\frac{18\frac{m}{s} -12\frac{m}{s} }{7.5s}[/tex]

We solve the difference:

[tex]a=\frac{6\frac{m}{s}}{7.5s}[/tex]

We divide the terms, so we can have the answer:

[tex]a=0.8 \frac{m}{s^{2}}[/tex]

2. To find the distance traveled by the truck, we use the equation:

[tex]x=V_0t+\frac{1}{2}at^{2}[/tex]

Where [tex]x[/tex] is the distance traveled, [tex]V_0[/tex] is the initial speed, [tex]a[/tex] is the acceleration and [tex]t[/tex] is the time.

We replace the terms:

[tex]x=(12\frac{m}{s}*7.5s)+\frac{1}{2}[0.8\frac{m}{s^{2} }*(7.5)^{2} ][/tex]

We multiply and solve the exponential:

[tex]x=90m+\frac{1}{2}(0.8\frac{m}{s^{2} }*56.25s^{2} )[/tex]

Then, we multiply the terms left:

[tex]x=90m+22.5m[/tex]

And add, so we can have the answer:

[tex]x=112.5m[/tex]

Answer:

E the electric field remains unchanged.

Explanation:

potential difference between the plates of a capacitor

V = Q / C  Where Q is charge on the capacitor and C is capacity of capacitor

Here Q is unchanged.

C the capacity has value equal to ε A / d

Here d is distance between plates. when  it is halved, capacitance C becomes double.

V = Q /C , When C becomes double V becomes half

E = V/d , E is electric field between plates having separation of d.

When V becomes half and d also becomes half , there is no change in the value of E.  

Hence E the electric field remains unchanged.

The magnitude of the electric field between two charged parallel plates remains constant when the plate separation is reduced from d to d/2, provided the charge density on the plates stays the same. Thus, the magnitude of the electric field halfway between the plates would still be E.

The student asked what the magnitude of the electric field would be if the separation between two large, flat, parallel, charged plates is reduced from d to d/2, assuming the charge on the plates remains constant.

According to Essential Knowledge 2.C.5, the electric field (E) between two oppositely charged parallel plates with uniformly distributed electric charge is constant in magnitude and direction at points far from the edges of the plates.

The magnitude of the electric field E between parallel plates is given by the equation E = σ/ε₀, where σ is the surface charge density and ε₀ is the permittivity of free space. Since the surface charge density (σ = Q/A, charge per area) does not change when the separation between the plates is altered and the permittivity of free space (ε₀) is a constant, the magnitude of the electric field between the plates remains the same even if the separation is reduced to d/2.

Therefore, the magnitude of the electric field halfway between the plates when the separation is reduced to d/2 would still be E.

The equations of motion are used to find the motion characteristics of an object with time

A) The time it takes the motorcycle to reach the car is approximately 5.39 seconds

B) The distance the motorcycle travel from the moment it starts to accelerate to the moment it catches up with the car is approximately 155 meters

The reason the above values are correct is as follows:

The given parameters are;

The initial speed of the car and the motorcycle, v₁ = 18.0 m/s

The initial distance between the motorcycle following the car = 58.0 m

The time after which the motorcycle starts to accelerate, t₁ = 5.00 secs

The acceleration of the motorcycle = 4.00 m/s²

The time at which the motorcycle catches up with the car = t₂

Required:

A) The time it takes the motorcycle to accelerate before it catches up with the car, which is to find t₂ - t₁

Method:

The distance moved by motorcycle to reach the car = The distance moved by the car in the same time + 58.0 meters

Solution;

The appropriate equation of motion to use are [tex]s = u \cdot t + \dfrac{1}{2} \cdot a\cdot t^2[/tex] and [tex]d = u\cdot t[/tex]

Where;

s = The distance between the motorcycle and the car = 58.0 m

u = The initial velocity of the motorcycle = 18.0 m/s

t = Δt = t₂ - t₁ = The time the motorcycle accelerates

a = The acceleration of the motorcycle = 4.00 m/s²

d = The distance moved by the car

By the method, we have;

[tex]s = d + 58.0[/tex]

[tex]u \cdot t + \dfrac{1}{2} \cdot a\cdot t^2 = u \cdot t[/tex]

Plugging in the values into the equation gives;

[tex](18.0 \times t) + \dfrac{1}{2} \times 4 \times t^2 = (18.0 \times t) + 58.0[/tex]

Cancelling the like term, (18.0 × t), on both sides of the equation gives;

[tex]\dfrac{1}{2} \times 4 \times t^2 = 58.0[/tex]

[tex]2\cdot t^2 = 58.0[/tex]

[tex]t = \Delta t = \sqrt{\dfrac{58.0}{2} } = \sqrt{29} \approx 5.39[/tex]

The time it takes the motorcycle to reach the car, t ≈ 5.39 seconds

The time it takes the motorcycle to reach the car is approximately 5.39 seconds

B) Required:

The distance the motorcycle travels from the moment it starts to accelerate, t₁ to the time it catches up with the car, t₂

Method:

The distance travelled during the time interval t₂ - t₁ should be calculated

Solution:

The distance, s, travelled during the time interval [tex]t_2 - t_1 = \Delta t = t[/tex] is given as follows;

[tex]s = u \cdot t + \dfrac{1}{2} \cdot a \cdot t^2[/tex]

From part (A), u = 18.0 m/s t = [tex]\sqrt{29}[/tex] s, and a = 4.00 m/s², therefore;

[tex]s = 18.0 \times \sqrt{29} + \dfrac{1}{2} \times 4.00 \times (\sqrt{29} )^2 \approx 154.933[/tex]

Rounded to three significant figures, the distance the motorcycle travel, s = 155 meters

Learn more about kinetic equation motion here:

https://brainly.com/question/16793944

To solve the question, we use the equations of motion to equate the distances traveled by the car and motorcycle. After solving the quadratic equation, we find the time the motorcycle catches up with the car and the distance it traveled.

The question involves a situation where a motorcycle is following a car traveling at a constant speed on a highway. Initially, both the car and the motorcycle are traveling at the same speed of 18.0 m/s, and there is a distance of 58.0 m between them. After 5.00 seconds, the motorcycle begins to accelerate at 4.00 m/s² until it catches up with the car. To find the time t2-t1, which is the duration the motorcycle catches up to the car, and the distance the motorcycle covers from the moment it starts to accelerate until it catches up with the car, we use the equations of motion.

We know that the car travels with a constant velocity, so the distance it covers in time t after the motorcycle starts accelerating can be given by:

d_car = v_car × t

For the motorcycle, which starts to accelerate at time t1, the distance it covers can be given by:

d_moto = d_initial + v_initial × (t - t1) + ½ × a × (t - t1)²

Given that the initial distance d_initial is 58.0 m, v_initial is 18.0 m/s, and a is 4.00 m/s², we can equate the distances for both the car and motorcycle when the motorcycle catches up:

18.0 × t = 58.0 + 18.0 × (t - 5.00) + ½ × 4.00 × (t - 5.00)²

By solving this quadratic equation, we can find the value of t, and then t2-t1 would simply be t - 5.00 seconds. Finally, to find out the distance the motorcycle traveled, we substitute the value of t into the distance equation for the motorcycle.

https://brainly.com/question/35709307

#SPJ3

Answer:197.504 N

Explanation:

Given

Two Charges with magnitude Q experience a  force of 12.344 N

at distance r

and we know Electrostatic force is given

[tex]F=\frac{kq_1q_2}{r^2}[/tex]

[tex]F=\frac{kQ\cdot Q}{r^2}[/tex]

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

Now the magnitude of charge is 2Q and is at a distance of [tex]\frac{r}{2}[/tex]

[tex]F'=\frac{k2Q\cdot 2Q}{\frac{r^2}{2^2}}[/tex]

F'=16F

F'=197.504 N

Final answer:

To change 350 g of ice at -20 degrees Celsius to water at 20 Celsius, you need to calculate the heat required for two processes: warming the ice and melting the ice. The heat needed to warm the ice can be calculated using the equation: Q = mcΔT, and the specific heat capacity of ice is 2.09 J/g°C. The heat needed to melt the ice can be calculated using the equation: Q = mL, where L is the heat of fusion, which is 334 J/g for ice.

Explanation:

To calculate the heat necessary to change 350 g of ice at -20 degrees Celsius to water at 20 degrees Celsius, we need to consider the energy required for two processes: warming the ice from -20°C to 0°C and then melting the ice at 0°C to form water at 0°C.

The heat needed to warm the ice can be calculated using the equation: Q = mcΔT, where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. The specific heat capacity of ice is 2.09 J/g°C.

The heat needed to melt the ice can be calculated using the equation: Q = mL, where L is the heat of fusion, which is 334 J/g for ice.

Let's calculate the heat needed for each process and add them together:

Finally, add both values to find the total heat required.

A) -0.15 s

First of all, we need to velocity of the swimmer in absence of current. This is given by:

[tex]v_0 = \frac{d}{t}[/tex]

where

d = 50.0 m

t = 25.0 s

Substituting,

[tex]v_0 = \frac{50.0}{25.0}=2.0 m/s[/tex]

In lane 1, the velocity of the current is

[tex]v_c = +1.2 cm/s = +0.012 m/s[/tex]

where the + sign means it is in the same direction as the swimmer. Therefore, the net velocity of the swimmer in lane 1 will be

[tex]v=v_0+v_c = 2.0 + 0.012 = = 2.012 m/s[/tex]

And so, the time the swimmer will take to cover the 50.0 m will be:

[tex]t=\frac{d}{v}=\frac{50.0}{2.012}=24.85 s[/tex]

So, the time would change by

[tex]\Delta t = 24.85 - 25.0 = -0.15 s[/tex]

which means that the swimmer will be 0.15 s faster.

B) +0.15 s

To solve this part, we just need to consider that the current goes in the opposite direction, so its velocity actually is:

[tex]v_c = -12 cm/s = -0.012 m/s[/tex]

where the negative sign indicates the opposite direction.

So, the net velocity of the swimmer in lane 8 is

[tex]v=v_0+v_c = 2.0 - 0.012 = = 1.988 m/s[/tex]

And so, the time the swimmer will take to cover the 50.0 m will be:

[tex]t=\frac{d}{v}=\frac{50.0}{1.988}=25.15 s[/tex]

So, the time would change by

[tex]\Delta t = 25.15 - 25.0 = +0.15 s[/tex]

which means that the swimmer will be 0.15 s slower.

Answer:

7.1 m

Explanation:

Given:

Distance traveled by the student in the first attempt = [tex](2.9 \pm 0.1)\ m[/tex]

Distance traveled by the student in the second attempt = [tex](3.9 \pm 0.2)\ m[/tex]

So, the maximum distance that the student could travel in this attempt = [tex](2.9+0.1)\ m = 3.0\ m[/tex]

So, the maximum distance that the student could travel in this attempt = [tex](3.9+0.2)\ m = 4.1\ m[/tex]

Since the student first moves straight in a particular direction, rests for a while and then moves some distance in the same direction.

So, the largest distance that the student could possibly be from the starting point would be the largest distance of the final position of the student from the starting point.

And this distance is equal to the sum of the maximum distance possible in the first attempt and the second attempt of walking which is 7.1 m.

Hence, the largest distance that the student could possibly be from the starting point is 7.1 m.

Answer:

The wavelength of the particle is [tex]\lambda=\frac{h}{\sqrt{2V\cdot e\cdot m}}[/tex]

Explanation:

We know that after accelerating across a potential 'V' the potential energy of the charge is converted into kinetic energy as

[tex]\frac{1}{2}mv^{2}=V\cdot e\\\\\therefore v=\sqrt{\frac{2V\cdot e}{m}}[/tex]

now according to De-Broglie theory the wavelength associated with a particle of mass 'm' moving with a speed of 'v' is given by

[tex]\lambda =\frac{h}{mv}[/tex]

where

'h' is planck's constant

'm' is the mass of particle

'v' is the velocity of the particle

Applying the values in the above equation we get

[tex]\lambda =\frac{h}{m\cdot \sqrt{\frac{2V\cdot e}{m}}}[/tex]

Thus

[tex]\lambda=\frac{h}{\sqrt{2V\cdot e\cdot m}}[/tex]

Answer:

Wavelength of the particle is [tex]\frac{h}{\sqrt{2meV}}[/tex]

Solution:

As per the question:

The particle with mass, m and charge, e accelerates through V (potential difference).

The momentum of the particle, [tex]p_{p}[/tex] if it travels with velocity, [tex]v_{p}[/tex]:

[tex]p_{p} = mv_{p}[/tex]

Now, squaring both sides and dividing by 2.

[tex]\frac{1}{2}p_{p}^{2} = \frac{1}{2}m^{2}v_{p}^{2}[/tex]

[tex]K.E =\frac{1}{2m}p_{p}^{2} = \frac{1}{2}mv_{p}^{2}[/tex]

[tex]K.E = \frac{1}{2}mv_{p}^{2}[/tex]

Also,

[tex]p_{p} = \sqrt{2mK.E}[/tex]          (1)

Now, we know that Kinetic energy of particle accelerated through V:

K.E = eV       (2)

where

e = electronic charge = [tex]1.6\times 10^{- 19} C[/tex]

From eqn (1) and (2):

[tex]p_{p} = \sqrt{2mK.E}[/tex]              (3)

From eqn (2) and (3):

[tex]p_{p} = \sqrt{2meV}[/tex]

From the de-Broglie relation:

[tex]\lambda_{p} = \frac{h}{p_{p}}[/tex]      (4)

where

[tex]\lambda_{p}[/tex] = wavelength of particle

h = Planck's constant

From eqn (3) and (4):

[tex]\lambda_{p} = \frac{h}{\sqrt{2meV}}[/tex]

Final answer:

The final momentum of the car pushed by two men with a net force of 690 N for 6.6 seconds is 4554 N·s. This is calculated using the formula for impulse, which is the product of the net force and the time over which the force is applied.

Explanation:

The question is asking to calculate the final momentum of a car after two men have pushed it with a net force of 690 N for 6.6 seconds. According to Newton's second law of motion, the change in momentum (also known as impulse) is equal to the net force multiplied by the time over which the force is applied. The formula for impulse is:

Impulse (J) = Force (F) × Time (t)

By applying this formula, we can find the final momentum:

Therefore, the impulse is:

J = 690 N × 6.6 s = 4554 N·s

Since the car was initially at rest and assuming there is no external resistance, the final momentum of the car will be equal to the impulse given to it:

Final momentum = 4554 N·s

The final momentum of the car, with an applied force of +690 N for 6.6 seconds, is 4554 Ns.  The initial momentum is zero since the car starts from rest.

To determine the final momentum of the car, we will utilize the relationship between force, time, and momentum. According to the impulse-momentum theorem, the impulse applied to an object is equal to the change in its momentum.

Impulse is given by the formula:

Impulse (J) = Force (F) × Time (t)

Given:

Now, calculate the impulse:

J = 690 N × 6.6 s = 4554 Ns

In this scenario, the car starts from rest, which means its initial momentum (p_initial) is 0. Hence, the final momentum ([tex]p_{final}[/tex]) after 6.6 seconds is equal to the impulse:

[tex]p_{final}[/tex] = 4554 Ns

The final momentum of the car is 4554 Ns.

Answer: a) 1.6 * 10 ^-7 N/C (upward) ; b) -2.5*10^-6N (downward)

Explanation: In order to solve this proble we have to use the Coulomb law given by:

F=q*E from this expression we have

E=F/q=4.8*10^-6/30 C= 1.6 * 10 ^-7 N/C

The force on the particle charge by -1.6 X10 C place intead of the initial charge we have

F=q*E= -16 C* 1.6 * 10 ^-7 N/C= -2.5*10^-6N

Answer:

a)X=70.688m

b)v=18.8m/S

Explanation:

the truck has constant speed while the car moves in uniformly accelerated motion

A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t  (1)\\\\

\frac{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\

X=Xo+ VoT+0.5at^{2}    (3)\\

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve .

For the automobile

we use the ecuation number 3

Xo=0

Vo=0

a=2.5

X=0.5(2.5)t^2

X=1.25t^2

for the truck

x=Vt

X=9.4t

the distance is the same we can match the previous equations

1.25t^2=9.4t

t=9.4/1.25=7.52s

using the ecuation for the truck

x=9.4*7.52=70.688m

b) for this point we can use the ecuation number 1 for automobile

Vf=Vo+at

Vf=0+2.5(7.52)

Vf=18.8m/S

Answer:

The possible value of power fluctuation is 3.9811 W

Solution:

As per the question:

Power output, P = 10 W

Power fluctuation is from +3 dB to -3 dB

Therefore,

Power fluctuation in dB = 6 dB

Now,

P (in dB) = [tex]10log_{10}P'[/tex]

6 dB = [tex]10log_{10}P'[/tex]

[tex]0.6 = log_{10}P'[/tex]

[tex]P' = 10^{0.6} W = 3.9811 W[/tex]

Answer:

θ = 3.38 degrees

r = 0.48mi

Explanation:

a) Start by converting 1.7 mi into feet.

1mile = 5280 feet,

so 1.7 mi = 8976 ft

[tex]y=rsin\theta,[/tex]

so [tex]\theta = sin^{1] {530}{8976}[/tex]

and θ = 3.38 degrees

b )Now to gain another 150 ft of elevation, again use the

equation y=rsin(θ)

150 = rsin(3.38)

r = 2544.187 ft.

Convert r = 2544.187 ft into miles by dividing by 5280,

and r = 0.48mi

Answer:

[tex]y_{o}=v_{o}*sin(\alpha )*t + 1/2*gt^{2}[/tex]

Explanation:

Kinematics equation, with gravity as deceleration

[tex]y =y_{o}-v_{o}*sin(\alpha )*t - 1/2*gt^{2}[/tex]

If the base of the building is taken to be the origin of the coordinates, then :

[tex]y_{o} [/tex] is the initial coordinates of the ball

for t=5.00, the ball strikes the ground ⇒ y=0m

then:

[tex]y_{o}=v_{o}*sin(\alpha )*t + 1/2*gt^{2}[/tex]

Final answer:

The initial coordinates of the ball below the horizontal are (0, y0), with y0 representing the height from which the ball was tossed and 0 representing the horizontal origin.

Explanation:

The student's question asks for the initial coordinates of a ball tossed from a window with a given initial velocity and angle relative to the horizontal. Since we're dealing with projectile motion, we need to resolve the initial velocity into its horizontal (x-direction) and vertical (y-direction) components.

The initial horizontal velocity (vx0) can be found using trigonometry: vx0 = v0 × cos(θ), where v0 is the initial velocity of 8.40 m/s and θ is the launch angle of 15.0°. However, since the angle is below the horizontal, the vertical component of the initial velocity (vy0) will be negative: vy0 = v0 × sin(θ), pointing downward.

Given the vertical direction is positive upwards, and taking the base of the building to be at the origin (0,0), the initial coordinates of the ball will be (0, y0), where y0 represents the height of the window from which the ball was tossed.

Answer:

[tex]Q=-8.85*10^{-12}*0.36 C =-3.186*10^{-12} C[/tex]

(negative charge)

Explanation:

Hi!

To solve this problem we use Gauss Law for electric fields, which relates the flux of electric field E through a closed surface S with the charge Q enclosed by that surface:

[tex]\int_{S} \vec{E}\cdot \vec{n} \;dA = \frac{Q}{\epsilon_0}[/tex]

[tex]\vec{n} = \text{outwards surface normal}\\\epsilon_0 = 8.85*10^{-12}\frac{C^2}{Nm^2}[/tex]

In this problem S is a cube of side length 2cm. The integral is easy because the electric field is uniform in each face, and normal to the face. The total integral is the sum of the integrals in each of the six faces.

[tex]\int_{S} \vec{E}\cdot \vec{n} \;dA = \frac{Q}{\epsilon_0} = 3(-500\frac{N}{C} (2cm)^2) + 3(200\frac{N}{C} (2cm)^2)[/tex]

[tex](-1500*4*10^{-4}\frac{Nm^2}{C} + 600*4*10^{-4}\frac{Nm^2}{C} )=-0.36\frac{Nm^2}{C}[/tex]

[tex]Q=-8.85*10^{-12}*0.36 C =-3.186*10^{-12} C[/tex]

Answer:

The value of resistance will be 42.08 ohm    

Explanation:

We have given capacitance [tex]C=10\mu F=10\times 10^{-6}F[/tex]

Frequency f = 1 kHz = 1000 Hz

Impedance Z = 45 ohm

Capacitive reactance [tex]X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}=\frac{1}{2\times 3.14\times 1000\times 10^{-5}}=15.92ohm[/tex]

We know that impedance is given by [tex]Z=\sqrt{R^2+X_C^2}[/tex]

[tex]45=\sqrt{R^2+15.92^2}[/tex]

Squaring both side [tex]2025=R^2+253.4464[/tex]

R = 42.08 ohm

Answer:

The rise from A to B is 0.887

Solution:

As per the question:

The following reading of an inverted staff is given as:

A = 2.915

B = -2.028

Here, for inverted staff, the greater reading shows greater elevation and lesser reading shows lower elevation.

Thus

The rise from A to B is given as:

A - B = 2.915 - 2.028 = 0.887

Answer:

a) 10.2 eV

b) 122 nm

Explanation:

a) First we must obtain the energy for each of the states, which is given by the following formula:

[tex]E_{n}=\frac{-13.6 eV}{n^2}[/tex]

So, we have:

[tex]E_{1}=\frac{-13.6 eV}{1^2}=-13.6 eV\\E_{2}=\frac{-13.6 eV}{2^2}=-3.4 eV[/tex]

Now we find the energy that the electron loses when it falls from state 2 to state 1, this is the energy carried away by the emitted photon.

[tex]E_{2}-E_{1}=-3.4eV-(-13.6eV)=10.2eV[/tex]

b) Using the Planck – Einstein relation, we can calculate the wavelength of the photon:

[tex]E=h\nu[/tex]

Where E is the photon energy, h the Planck constant and  [tex]\nu[/tex] the frequency.

Recall that [tex]\nu=\frac{c}{\lambda}[/tex], Rewriting for [tex]\lambda[/tex]:

[tex]E=\frac{hc}{\lambda}\\\lambda=\frac{hc}{E}\\\lambda=\frac{(4.13*10^{-15}eV)(3*10^8\frac{m}{s})}{10.2eV}=1.22*10^{-7}m[/tex]

Recall that [tex]1 m=10^9nm[/tex], So:

[tex]1.22*10^{-7}m*\frac{10^9nm}{1m}=122nm[/tex]

Answer:

75 m /s

Explanation:

Acceleration of ball a = change in velocity / time

=( 90 - 60)/ 3 = 10 m s⁻²

Initial velocity u = 60 m/s

a = 10 m s⁻²

time t = 3

distance traveled s = ut + 1/2 at²

= 60 x 3 + .5 x 10 x 3 x 3

= 225 m

Average velocity = total distance traveled /time

= 225 / 3 = 75 m/s

Answer:

net electrostatic force = 46.76 N

and direction is vertical downward

Explanation:

given data

charge =  +5μC

corners identical charges = -6 μC

length of side = 0.10 m

to find out

What is the net electrostatic force

solution

here apex is the vertex where the two sides of equal length meet

so upper point A have charge +5μC and lower both point B and C have charge -6 μC

so

force between +5μC and -6 μC is express as

force  = [tex]k \frac{q1q2}{r^2}[/tex]       ..............1

put here electrostatic constant  k = 9 × [tex]10^{9}[/tex] Nm²/C² and q1 q2 is charge given and r is distance 0.10 m

so

force  = [tex]9*10^{9} \frac{5*6*10^(-12)}{0.10^2}[/tex]

force = 27 N

so net force is vector addition of both force

force = [tex]\sqrt{x^{2}+x^{2}+2x^{2}cos60}[/tex]

here x is force 27 N

force = [tex]\sqrt{27^{2}+27^{2}+2(27)^{2}cos60}[/tex]

net electrostatic force = 46.76 N

and direction is vertical downward

Answer:

2.97m

Explanation:

Hi!

To solve this problem we must consider the directions of the electric fields, since both charges are positive each field will point outwards form the sources meaning that the field will point yo the -x direction formpoints along the x-axis to the left of the charges and the field will point to the +x direction for points along the x-axis to the rigth of the sources, having said this, the only place where the fields point to oposite ditections is between them:

In the following diagram X is the 17nC charge and O the 8nC charge, the arrows represent the direction of their respective electric fields and d is an arbitrary distance measured from the origin.

|-------d------|-----5-d------|

X-----------> | <------------O

|---------------5--------------|

At d the electric fields of the charges are:

[tex]E_{X}=k\frac{17nC}{d^{2}} \\E_{O}=-k\frac{8nC}{(5-d)^{2}}[/tex]

And the total field is the sum of both. Since we are looking for a position d at which the total or net electric field is zero we must solve the following equation:

[tex]0=k(\frac{17}{d^{2}} - \frac{8}{(5-d)^2} )[/tex]

Reorganizing terms:

17*(5-d)^2-8*d^2=0

17*25-170d+9d^2=0

Solving for d, we find two solutions

  d1 = 2.9655     d2=15.923

The second solution implies a distance outside the region between the charges for which the equations are no longer valid, since the direction of the 8nC field would be wrong.

So the solution is

 d=2.97 m

Answer:

Acceleration, a = [tex]97.52 m/s^{2}[/tex]

t = 0.08 s

Solution:

As per the question:

The velocity of the player, v = 7.9 m/s

distance, d = 0.32 m

Now, consider the direction of the initial velocity of the player as positive and the acceleration to be constant:

Also, the final velocity of the player, v' = 0 m/s as he finally stops

Using the third eqn of motion:

[tex]v'^{2} = v^{2} - 2ad[/tex]

[tex]0 = 7.9^{2} - 2a\times 0.32[/tex]

a = [tex]97.52 m/s^{2}[/tex]

Also, From eqn (1) of motion:

v' = v - at

0 = 7.9 - 97.52t

t = 0.08 s

Answer:

The ball will be at 700 m above the ground.

Explanation:

We can use the following kinematic equation

[tex]y(t) = \ y_0 \ + \ v_0 \ t \ + \frac{1}{2} \ a \ t^2[/tex].

where y(t) represent the height from the ground. For our problem, the initial height will be:

[tex]y_0 \ = \ 1000 m[/tex].

The initial velocity:

[tex]v_0 = - 20 \frac{m}{s}[/tex],

take into consideration the minus sign, that appears cause the ball its thrown down.  The same minus appears for the acceleration:

[tex]a=-10\frac{m}{s}[/tex]

So, the equation for our problem its:

[tex]y(t) = \ 1000 m \ - \ 20 \ \frac{m}{s} \ t \ - \frac{1}{2} \ 10 \frac{m}{s^2} \ t^2[/tex].

Taking t=6 s:

[tex]y(6 \ s) = \ 1000 m \ - \ 20 \ \frac{m}{s} \ * \ 6 \ s \ - \frac{1}{2} \ 10 \frac{m}{s^2} \ * \ (6 s)^2[/tex].

[tex]y(6 \ s) = \ 1000 m \ - 120 m - \frac{1}{2} \ 10 \frac{m}{s^2} \ * \ 36 s^2[/tex].

[tex]y(6 \ s) = \ 1000 m \ - 120 m - 180 m[/tex].

[tex]y(6 \ s) = \ 1000 m \ - 300 m[/tex].

[tex]y(6 \ s) = \ 700 m [/tex].

So this its the height of the ball 6 seconds after being thrown.

Given a ball thrown straight downward from a 1000 m high cliff with an initial speed of 20 m/s and considering an acceleration due to gravity of -10 m/s², it can be calculated using a kinematics equation that the ball will be 760 m above the ground after 6 seconds.

The subject of this question is in the field of Physics, specifically kinematics. The problem is set up to have an object that is thrown downward with an initial speed and is then under the influence of gravity, causing it to accelerate downward. To calculate how far above from the ground the ball will be at a certain time, we use the equation for position in uniformly accelerated motion which is:

y = y₀ + v₀×t + 0.5×a×t²

Here, y₀ represents the initial height, v₀ is the initial speed, a is the acceleration due to gravity, and t is the time. Since the boy is standing at the edge of a 1000 m cliff, y₀ = 1000 m. The ball is thrown downwards with an initial speed of 20 m/s, so v₀=-20 m/s (we take down as the negative direction). On earth, the acceleration due to gravity is approximately -10 m/s² and a = -10 m/s². The time t = 6 seconds is given in the problem.

Substituting these values into our equation gives y = 1000 + (-20×6) + 0.5×(-10) (6²), which simplifies to y = 760 m. Therefore, the ball is 760 meters above the ground 6 seconds after it was thrown.

https://brainly.com/question/7590442

#SPJ3

Final answer:

To estimate the surface temperature of a star modeled as an ideal blackbody, we can use the Stefan-Boltzmann law. The law states that the power radiated by a blackbody is proportional to its surface area and temperature to the fourth power. We can rearrange the equation to solve for temperature: T = (I/(σ*A))^1/4.

Explanation:

To estimate the surface temperature of a star modeled as an ideal blackbody, we can use the Stefan-Boltzmann law. The law states that the power radiated by a blackbody is proportional to its surface area and temperature to the fourth power. We can rearrange the equation to solve for temperature: T = (I/(σ*A))^1/4, where T is the temperature, I is the intensity of radiation, σ is the Stefan-Boltzmann constant, and A is the surface area.

Plugging in the given values, we have T = (0.055W/m² / (5.670 x 10^-8 W/(m²K^4) * 4π(5.0 x 10^8m)^2))^1/4.

Answer:

displacement is 481.3 m  with 40.88° north east

Explanation:

given data

walk 1 = 300 m north = 300 j

walk 2 = 400 m northwest =400 cos(45) i + sin(45) j

walk 3 = 700 m east southeast  = 700 cos(22.5) i- sin(22.5) j

to find out

displacement and direction

solution

we consider here direction x as east and direction y as north

so

displacement = distance 1 + distance 2 + distance 3

displacement = 300 j + 400 cos(45) i + sin(45) j + 700 cos(22.5) i- sin(22.5) j  

we get here

displacement = 363.9 i + 315 j

so magnitude

displacement = [tex]\sqrt{363.9^{2}+315^{2}  }[/tex]

displacement = 481.3 m

and angle will be = arctan(315/369.9)

angle = 0.713494059 rad

angle is 40.88 degree

so displacement is 481.3 m  with 40.88° north east

Answer:

Magnitude of displacement = 481.24 m

Direction = 40.88 degrees north east.

Explanation:

Let east is along real axis and north is along imaginary axis. So,

First walk = d1 = J300 m

Second walk = d2 = -400Cos(45) + J400Sin(45) = (-282.84 + J282.84) m  

Third walk = d3 = 700Cos(22.5) – J700Sin(22.5) = (646.71 – J267.88) m

Total displacement = d = d1 + d2 + d2 = (363.86 + J314.96)m

Magnitude = √((363.86)^2+(314.96)^2) = 481.24 m

Direction = arctan(314.96/363.86) = 40.88 degrees north east.

Answer:

The person should use a convex lens of power -0.625 D.

Explanation:

Given that, the far point of the person, with myopic eye, is 160 cm.

It means that the image of an object, which is at infinity, is formed at this point 160 cm distance away from the person's eye.

If the object is infinity, then the object distance, u = -[tex]\infty[/tex].

Image distance, v = -160 cm.

u and v are taken to be negative because the object and the image of the object are on the side of the eye from where the light is coming.

Let the focal length of the lens which is to be used for the correction of this myopic eye be f, then using lens equation,

[tex]\rm \dfrac 1f = \dfrac 1v-\dfrac 1u=\dfrac 1{-160 }-\dfrac1{-\infty}=\dfrac 1{-160}-0=\dfrac 1{-160}.\\\Rightarrow f=-160\ cm.[/tex]

The negative focal length indicates that the lens should be convex.

The power of the lens is given by

[tex]\rm P=\dfrac{1}{f}=\dfrac{1}{-160\ cm }=\dfrac{1}{-1.6\ m}=-0.625\ D.[/tex]

So, the person should use a convex lens of power -0.625 D.

Answer:

- 1.428 D

Explanation:

A myopic person is able to see the nearby objects clearly but cannot see the far off objects very clearly. It can be cured by using concave lens of suitable power.

far point = 70 cm

v = - 70 cm (position of image from the lens)

u = ∞ (position of object from lens)

Let f be the focal length of the lens

Use lens equation

[tex]\frac{1}{f}=\frac{1}{v}-\frac{1}{u}[/tex]

[tex]\frac{1}{f}=\frac{1}{-70}-\frac{1}{∞}[/tex]

f = - 70 cm

[tex]P=\frac{1}{f}[/tex]

Where, P is the power of lens

P = - 1.428 Dioptre

Thus, the power of lens used is - 1.428 D.

Answer:

The value of acceleration equals [tex]0.0556m/s^{2}[/tex]

Explanation:

The  distances covered in each of the three phases are calculated as under

1) Distance covered in accelerating phase for a period of 5 minutes or 300 seconds ( 1 minute = 60 seconds)

Using second equation of kinematics

[tex]s=ut+\frac{1}{2}at^{2}\\\\s_{1}=0\times 300+\frac{1}{2}\times a\times (300)^{2}\\\\s_{1}=0.5\times a\times (300)^{2}[/tex]

2) Distance covered in 5 minutes while travelling at a constant speed which is The speed after 5 minutes of travel is obtained by first equation of kinematics as

[tex]v=u+at\\\\v=0+a\times 300\\\\v=300a[/tex]

Thus distance traveled equals

[tex]s_{2}=300a\times 300\\\\s_{2}=(300)^{2}a[/tex]

3)

The phase when the car stops the distance it covers during this phase can be obtained using third equation of kinematics as

[tex]v^{2}=u^{2}+2as\\\\\therefore s=\frac{v^{2}-u^{2}}{2a}\\\\s_{3}=\frac{0-(300a)^{2}}{2\times -a}\\\\s_{3}=\frac{300^{2}a}{2}[/tex]

Now the sum of [tex]s_{1}+s_{2}+s_{3}[/tex] equals 10 kilometers or 10000 meters.

Thus we get

[tex]0.5\times a\times 300^{2}+300^{2}a+300^{2}\times \frac{a}{2}=10000\\\\a(0.5\times 300^{2}+300^{2}+0.5\times 300^{2})=10000\\\\\therefore a=\frac{10000}{(0.5\times 300^{2}+300^{2}+0.5\times 300^{2})}=0.0556m/s^{2}[/tex]

To solve for the acceleration, we can use the constant acceleration equation and find the distance traveled during each phase.

The constant acceleration equation can be used to solve this problem. The total distance traveled by the train is equal to the sum of the distances traveled during each phase.

In the first phase, the train starts at rest and accelerates for 5 minutes. The distance traveled during this phase can be found using the equation s = 0.5at^2, where s is the distance, a is the acceleration, and t is the time. Since the initial velocity is 0, we can simplify the equation to s = 0.5at^2.

In the second phase, the train travels at a constant speed for 5 minutes. The distance traveled during this phase is equal to the product of the speed and the time.

In the third phase, the train decelerates with a negative acceleration. The distance traveled during this phase can be found using the equation s = ut + 0.5at^2, where u is the initial velocity, a is the acceleration, and t is the time. Since the final velocity is 0, we can simplify the equation to s = ut.

The total distance traveled is given as 10 km. By plugging in the values for the distances and using the given times for each phase, we can solve for the acceleration, a, and find that a = 0.014 km/s^2.

https://brainly.com/question/32817533

#SPJ11

That statement is false.

Angular velocity is the rate at which that angle changes. As long as the reference line doesn't move, it doesn't matter where it is. The rate at which the angle changes is still the same, constant number.

The angular velocity of a rigid body that undergoes plane motion depends on the reference line from which its angle of rotation is measured is False.

When anything is moving, its velocity tells us how quickly that something's location is changing from a certain vantage point and as measured by a particular unit of time.

When a point travels down a path and completes a certain distance in a predetermined amount of time, its average speed over that time is equal to the distance covered divided by the travel time. A train traveling 100 kilometers in two hours, for instance, is doing it at an average speed of 50 km/h.

The pace at which that angle change is known as the angular velocity. It doesn't matter where the reference line is, as long as it stays put. The rate of change of the angle remains constant and the same.

Therefore, the angular velocity of a rigid body that undergoes plane motion depends on the reference line from which its angle of rotation is measured is False.

To know more about velocity:

https://brainly.com/question/19979064

#SPJ2

Answer:

(a) 1.73 s

(b) 14.75 m

(c) 3.36 s

(d) double

(e) 63.32 m

Explanation:

Vertical component of initial velocity, uy = 17 m/s

Horizontal component of initial velocity, ux = 18.3 m/s

(A) At highest point of trajectory, the vertical component of velocity is zero. Let the time taken is t.

Use first equation of motion in vertical direction

vy = uy - gt

0 = 17 - 9.8 t

t = 1.73 seconds

(B) Let the highest point is at height h.

Use III equation of motion in vertical direction

[tex]v^{2}=u^{2}-2gh[/tex]

0 = 17 x 17 - 2 x 9.8 x h

h = 14.75 m

(C) The time taken by the ball to return to original level is T.

Use second equation of motion i vertical direction.

[tex]h = ut + 0.5at^2[/tex]

h = 0 , u = 17 m/s

0 = 17 t - 0.5 x 9.8 t^2

t = 3.46 second

(D) It is the double of time calculated in part A

(E) Horizontal distance = horizontal velocity x total time

d = 18.3 x 3.46 = 63.32 m

Answer:

a) 0.1 lb/ft³

b) 20 lb

c) 88.788 N

Explanation:

Given:

Volume of the object = 200 ft³

Specific volume = 10 ft³/ lb

Now,

Density = [tex]\frac{\textup{1}}{\textup{Specific volume}}[/tex]

or

Density = [tex]\frac{1}{10}[/tex] = 0.1 lb/ft³

also,

Specific volume = [tex]\frac{\textup{Volume}}{\textup{Mass}}[/tex]

or

10 = [tex]\frac{200}{\textup{Mass}}[/tex]

or

Mass = 20 lb

And,

Weight =  Mass in kg × Acceleration due to gravity

Mass = 20 lb

and 1 lb = 0.453 kg

thus,

20 lb = 20 × 0.453 = 9.06 kg

Therefore,

weight = 9.06 × 9.8 = 88.788 N

For the object,

Specific volume of the object is inverse to the density of the object.

Given information-

The volume of the object is 200 ft cubed.

The specific volume of the object is 10 ft cubed.

The density of a object is the inverse of its specific volume, thus density of the given object is,

[tex]\rho=\dfrac{1}{\rm specific volume} \\\rho=\dfrac{1}{10} \\\rho=0.1 \rm ib/ft^3[/tex]

The mass of a object is the ratio of the volume to the specific volume, thus mass of the given object is,

[tex]m=\dfrac{\rm volume}{\rm specific volume} \\m=\dfrac{200}{10} \\m=20 \rm ib[/tex]

As 1 Ib is equal to the 0.453 kg.

The weight of a object is the product of the mass of the object to the acceleration due to gravity, thus weight of the given object is,

[tex]W=0.453\times20\times9.81\\W=88.788\rm N[/tex]

Thus for the object,

Learn more about the specific density here;

https://brainly.com/question/24799156