A 20.0-kg cannonball is fired from a cannon with muzzle speed of 1 000 m/s at an angle of 37.08 with the horizontal. A second ball is fired at an angle of 90.08. Use the isolated system model to find (a) the maximum height reached by each ball and (b) the total mechanical energy of the ball–Earth system at the maximum height for each ball. Let y5 0 at the cannon.

Answer:

[tex]F_{avg}=2096.71N[/tex]

Explanation:

As from the given data that:

m/Δt=53.9kg/s

The numerator is the mass and denominator is the change in time

Solve for change in velocity we have:

Δv=v₂-v₁

[tex]=0-38.9m/s\\=-38.9m/s[/tex]

Δv= -38.9m/s

From Newtons second law we know that:

F=ma

We can write this equation as:  

Favg=(m/Δt)Δv

Substitute the given values

So

[tex]F_{avg}=(53.9kg/s)(-38.9m/s)\\F_{avg}=-2096.71N[/tex]

Using the average velocity formula.The answer will be positive, the negative only implies that force is coming away from the wall

[tex]F_{avg}=2096.71N[/tex]

Answer:

Explanation:

capacitance of each capacitor

C₀= Q₀ / V₀

V₀ = Q₀ / C₀

New total capacitance = C₀ ( 1 + K )

Common potential

= total charge / total capacitance

= 2 Q₀ / [ C₀ ( 1 + K ) ]

2 V₀ / ( 1 + K )

b )

Common potential = 2 x V₀ / ( 1 + 7.8 )

= .227  V₀

charge on capacitor with dielectric

= .227  V₀ x 7.8 C₀

= 1.77 V₀C₀

= 1.77 Q₀

Ratio required = 1.77

The new potential difference V across the capacitors and the ratio of the charge on the capacitor with the dielectric after it is inserted as compared with the initial charge can be calculated using specific equations.

(a) What is the new potential difference V across the capacitors.

To find the new potential difference V across the capacitors, we need to consider the effect of adding a dielectric. The potential difference V is given by the equation:

V = Vo/2k

where Vo is the initial potential difference and k is the dielectric constant. In this case, since the dielectric constant is 7.8, we can substitute the values and calculate the new potential difference V.

(b) If the dielectric constant is 7.8, calculate the ratio of the charge on the capacitor with the dielectric after it is inserted as compared with the initial charge.

The ratio of the charge on the capacitor with the dielectric after it is inserted as compared with the initial charge can be calculated using the equation:

Q_with_dielectric / Q_initial = k

where Q_with_dielectric is the charge on the capacitor with the dielectric and Q_initial is the initial charge. Given that k is 7.8, we can substitute the values and calculate the ratio.

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Our values are given are,

[tex]l = 2m[/tex]

[tex]I = 3A[/tex]

[tex]r = 3.5mm = 3.5*10^{-3} m[/tex]

Electromagnetic force can then be defined as,

[tex]F = \frac{\mu_0 I^2 l}{2\pi r}[/tex]

Here,

[tex]\mu_0[/tex]  = Permeability free space

I = Current

l = Length

r = Distance between them

Replacing,

[tex]F = \frac{(4\pi *10^{-7})(3)^2(2)}{(2\pi)(3.5*10^{-3})}[/tex]

[tex]F = 0.001028N[/tex]

Therefore the magnitude of the force that two segments of this cord exert on each other is 0.001028N.

Since the magnitude of the force is positive, the direction of both is attractive.

a) The magnitude of the force that the two segments of the cord exert on each other is [tex]\(1.2 \times 10^{-4} \, \text{N}\)[/tex]

b) The force is repulsive.

The force per unit length between two parallel wires carrying currents [tex]\(I_1\)[/tex] and [tex]\(I_2\)[/tex] separated by a distance d is given by Ampère's law:

[tex]\[F/L = \frac{\mu_0 I_1 I_2}{2 \pi d}\][/tex]

Where:

F is the force,

L is the length of the wires,

[tex]\(\mu_0\)[/tex] is the permeability of free space [tex](\(4 \pi \times 10^{-7} \, \text{N/A}^2\))[/tex],

[tex]\(I_1\)[/tex] and [tex]\(I_2\)[/tex] are the currents through the wires (in this case, both are 3.0 A but in opposite directions),

d is the distance between the wires.

a. First, assume the distance d between the wires in the extension cord. A reasonable estimate for the distance between the wires in a standard extension cord might be around 3 mm (0.003 m).

It is given that:

[tex]\(I_1 = I_2 = 3.0 \, \text{A}\),[/tex]

[tex]\(L = 2.0 \, \text{m}\),[/tex]

[tex]\(d \approx 0.003 \, \text{m}\).[/tex]

Using the formula for the force per unit length:

[tex]\[\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2 \pi d}\][/tex]

Substitute the known values:

[tex]\[\frac{F}{L} = \frac{(4 \pi \times 10^{-7} \, \text{N/A}^2) \times (3.0 \, \text{A})^2}{2 \pi \times 0.003 \, \text{m}}\][/tex]

[tex]\[\frac{F}{L} = \frac{4 \pi \times 10^{-7} \times 9}{2 \pi \times 0.003}\][/tex]

[tex]\[\frac{F}{L} = \frac{36 \times 10^{-7}}{0.006}\][/tex]

[tex]\[\frac{F}{L} = 6 \times 10^{-5} \, \text{N/m}\][/tex]

Now, multiply by the length L to find the total force:

[tex]\[F = \left(6 \times 10^{-5} \, \text{N/m}\right) \times 2.0 \, \text{m}\][/tex]

[tex]\[F = 1.2 \times 10^{-4} \, \text{N}\][/tex]

b. Since the currents in the two parallel wires are equal but in opposite directions, the magnetic force between them will be repulsive. This is because currents flowing in opposite directions repel each other according to the right-hand rule for magnetic fields and forces between current-carrying wires.

A mass weighing 32 pounds stretches a spring 2 feet.

(a) Determine the amplitude and period of motion if the mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of 6 ft/s.

(b) How many complete cycles will the mass have completed at the end of 4 seconds?

Answer:

[tex]A = 1.803 ft[/tex]

Period = [tex]\frac{\pi}{2}[/tex] seconds

8 cycles

Explanation:

A mass weighing 32 pounds stretches a spring 2 feet;

it implies that the mass (m) = [tex]\frac{w}{g}[/tex]

m= [tex]\frac{32}{32}[/tex]

= 1 slug

Also from Hooke's Law

2 k = 32

k = [tex]\frac{32}{2}[/tex]

k = 16 lb/ft

Using the function:

[tex]\frac{d^2x}{dt} = - 16x\\\frac{d^2x}{dt} + 16x =0[/tex]

[tex]x(0) = -1[/tex]        (because of the initial position being above the equilibrium position)

[tex]x(0) = -6[/tex]          ( as a result of upward velocity)

NOW, we have:

[tex]x(t)=c_1cos4t+c_2sin4t\\x^{'}(t) = 4(-c_1sin4t+c_2cos4t)[/tex]

However;

[tex]x(0) = -1[/tex] means

[tex]-1 =c_1\\c_1 = -1[/tex]

[tex]x(0) =-6[/tex] also implies that:

[tex]-6 =4(c_2)\\c_2 = - \frac{6}{4}[/tex]

[tex]c_2 = -\frac{3}{2}[/tex]

Hence, [tex]x(t) =-cos4t-\frac{3}{2} sin 4t[/tex]

[tex]A = \sqrt{C_1^2+C_2^2}[/tex]

[tex]A = \sqrt{(-1)^2+(\frac{3}{2})^2 }[/tex]

[tex]A=\sqrt{\frac{13}{4} }[/tex]

[tex]A= \frac{1}{2}\sqrt{13}[/tex]

[tex]A = 1.803 ft[/tex]

Period can be calculated as follows:

= [tex]\frac{2 \pi}{4}[/tex]

= [tex]\frac{\pi}{2}[/tex] seconds

How many complete cycles will the mass have completed at the end of 4 seconds?

At the end of 4 seconds, we have:

[tex]x* \frac{\pi}{2} = 4 \pi[/tex]

[tex]x \pi = 8 \pi[/tex]

[tex]x=8[/tex] cycles

Answer: The work done in J is 324

Explanation:

To calculate the amount of work done for an isothermal process is given by the equation:

[tex]W=-P\Delta V=-P(V_2-V_1)[/tex]

W = amount of work done = ?

P = pressure = 732 torr = 0.96 atm    (760torr =1atm)

[tex]V_1[/tex] = initial volume = 5.68 L

[tex]V_2[/tex] = final volume = 2.35  L

Putting values in above equation, we get:

[tex]W=-0.96atm\times (2.35-5.68)L=3.20L.atm[/tex]

To convert this into joules, we use the conversion factor:

[tex]1L.atm=101.33J[/tex]

So, [tex]3.20L.atm=3.20\times 101.3=324J[/tex]

The positive sign indicates the work is done on the system

Hence, the work done for the given process is 324 J

Given values:

We know the equation,

→ [tex]W = - P\Delta V[/tex]

       [tex]= -P(V_2-V_1)[/tex]

By substituting the values,

       [tex]= -0.96\times (2.35-5.68)[/tex]

       [tex]= 3.20 \ L.atm[/tex]

By converting it in "J",

       [tex]= 3.20\times 101.3[/tex]

       [tex]= 324 \ J[/tex]

Thus the above answer is right.

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Find the speed with which the disk slides across the surface

Answer:

[tex]\boxed{2.85 m/s}[/tex]

Explanation:

The Potential energy of spring is transformed to kinetic energy hence

[tex]0.5kx^{2}=0.5mv^{2}\\kx^{2}=mv^{2}\\v=\sqrt{\frac {kx^{2}}{m}}[/tex]

Here k is the spring constant, x is the extension of spring, v is the velocity of disk and m is the mass of the disk.

Substituting 0.117 Kg for m, 194 N/m for k and 0.07 m for x then

[tex]v=\sqrt{\frac {194\times 0.07^{2}}{0.117}}=2.850401081 m/s\approx \boxed{2.85 m/s}[/tex]

Answer:

a.6 V

b.24 V

Explanation:

We are given that

a.[tex]C=8\mu F=8\times 10^{-6} F[/tex]

[tex]1\mu =10^{-6} [/tex]

Q=[tex]48\mu C=48\times 10^{-6} C[/tex]

We know that

[tex]V=\frac{Q}{C}[/tex]

Using the formula

[tex]V=\frac{48\times 10^{-6}}{8\times 10^{-6}}=6 V[/tex]

b.[tex]Q=192\mu C=192\times 10^{-6} C[/tex]

[tex]V=\frac{192\times 10^{-6}}{8\times 10^{-6}}=24 V[/tex]

Explanation:

As the given data is as follows.

         h = [tex]\frac{3}{5}d[/tex]

            = [tex]\frac{3}{5} \times (2r)[/tex]

Also, we know that r = [tex]\frac{4}{3}h[/tex]

and       Volume (V) = [tex]\frac{1}{3} \pir^{2}h[/tex]

                         = [tex]\frac{1}{3} \pi (\frac{4}{3}h)^{2} h[/tex]

                         = [tex]\frac{16}{27} \pi h^{3}[/tex]

And,  [tex]\frac{dV}{dt} = \frac{3 \times 16}{27} \pi h^{2} \frac{dh}{dt}[/tex]

         [tex]\frac{dV}{dt} = \frac{16}{9} \pi h^{2} \frac{dh}{dt}[/tex]

Putting the given values into the above formula as follows.

       [tex]\frac{dV}{dt} = \frac{16}{9} \pi h^{2} \frac{dh}{dt}[/tex]

       [tex]30 m^{3}/min = \frac{16}{9} \pi (2)^{2} \frac{dh}{dt}[/tex]    

          [tex]\frac{dh}{dt} = 1.343 m/min

or,                        = 134.3 cm/min     (as 1 m = 100 cm)

thus, we can conclude that the height changing at 134.3 cm/min when the pile is 2 m high.

Answer:

The critical angle is 41.8°.

Explanation:

The critical angle is [tex]\theta_1[/tex] for which the angle of refraction [tex]\theta_2[/tex] is 90°. From Snell's law we have

[tex]n_1sin (\theta_1 ) = n_2 sin(\theta_2)[/tex]

[tex]n_1sin (\theta_1 ) = n_2 sin(90^o)[/tex]

[tex]sin (\theta_1 ) = n_2/n_1,[/tex]

[tex]\theta_1 = sin^{-1}(\dfrac{n_2}{n_1} ).[/tex]

Putting in [tex]n_2 =1.00,[/tex] and [tex]n_1 = 1.50[/tex] we get:

[tex]\theta_1 = sin^{-1}(\dfrac{1.00}{1.500} ),[/tex]

[tex]\boxed{\theta_1 = 41.8^o}[/tex]

Thus, the critical angle is 41.8°.

Final answer:

The critical angle for light propagating from a material with an index of refraction of 1.50 to a material with an index of refraction of 1.00 is approximately 41.8 degrees.

Explanation:

The critical angle θcrit for light propagating from a material with an index of refraction (n1) of 1.50 to a material with an index of refraction (n2) of 1.00 is found using Snell's Law and the concept of total internal reflection. The formula to calculate the critical angle is θcrit = sin−1(n2/n1). Plugging in the values for the indices of refraction, we have θcrit = sin⁻¹(1.00/1.50).

Performing the calculation, we get:

θcrit = sin⁻¹(0.6667) ≈ 41.8°

Therefore, the critical angle for light traveling from a medium with an index of refraction of 1.50 to a medium with an index of refraction of 1.00 is approximately 41.8 degrees.

Answer:

The amplitude, in meters, of the resulting simple harmonic motion of the block 1/40 m

Explanation:

given information:

the block mass, m = 0.1 kg

spring of force constant, k = 40 N/m

release at t = 0

the formula for simple harmonic motion is

F = kx

where

F = force (N)

k = spring of force constant (N/m)

x = amplitude (m)

thus,

F = kx

x = F/k

  = m g/k

  = 0.1 x 10/40

  = 1/40 m

Explanation:

The given data is as follows.

       C = [tex]20 \times 10^{-6} F[/tex]

        R = [tex]100 \times 10^{3}[/tex] ohm

        [tex]Q_{o} = 100 \times 10^{-6}[/tex] C

          Q = [tex]13.5 \times 10^{-6} C[/tex]

Formula to calculate the time is as follows.

          [tex]Q_{t}  = Q_{o} [e^{\frac{-t}{\tau}][/tex]

       [tex]13.5 \times 10^{-6} = 100 \times 10^{-6} [e^{\frac{-t}{2}}][/tex]

               0.135 = [tex]e^{\frac{-t}{2}}[/tex]

         [tex]e^{\frac{t}{2}} = \frac{1}{0.135}[/tex]

                         = 7.407

           [tex]\frac{t}{2} = ln (7.407)[/tex]

                      t = 4.00 s

Therefore, we can conclude that time after the resistor is connected will the capacitor is 4.0 sec.

The required time to discharge the capacitor to a certain charge can be calculated by rearranging the exponential decay formula. The product of resistance and capacitance, known as the RC time constant, is a key value in this calculation. After substituting the values given into the formula, we get the required time.

The time it takes for a capacitor to discharge through a resistor can be calculated using the formula for the decay of charge on a capacitive circuit: Q = Q0 * e-t/RC, where Q is the charge at time t, Q0 is the initial charge, R is the resistance and C is the capacitance. To find the time at which the charge will be 13.5 uC, we rearrange the formula to solve for time t: t = -RC * ln(Q/Q0).

Using the values of the problem: R = 100 kΩ = 100,000 Ω, C = 20 uF = 20*10-6 F, Q0 = 100 uC = 100*10-6 C, and Q = 13.5 uC = 13.5*10-6 C, substitute these values into the equation: t = -100,000 * 20*10-6 * ln (13.5/100). Hence, the time it takes for a capacitor to discharge to a specified charge can be calculated using exponential decay formula based on the capacitor's unique RC time constant.

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Answer with Explanation:

We are given that

Force=F=112 N

[tex]\theta=45^{\circ}[/tex]

Distance,[tex]s=84 m[/tex]

Time, t=3.33 minutes

We have to find the work done by Johnson on the bag and the power generated by Johnson.

Work done, W=[tex]Fscos\theta[/tex]

Using the formula

Work done, W=[tex]112\times 84cos45=6652.46 J[/tex]

Power, P=[tex]\frac{W}{t}=\frac{6652.46}{3.33\times 60}=33.3 watt[/tex]

Using 1 minute=60 s

Hence, the power generated by Johnson=33.3 watt

Answer:

  The mean free time is given as z  [tex]= 2.3*10^{-14}sec[/tex]

Explanation:

Generally the formula for resistivity

                 [tex]\rho = \frac{m_e}{e^2 n_ez}[/tex]

Where

              [tex]\rho[/tex] is he resistivity of metal

              [tex]m_e[/tex]  is the mass of the electron

             [tex]e[/tex]  is the charge of the electron

             [tex]n_e[/tex] is  electron density

            [tex]z[/tex] is the mean free time

  Now making z the subject  of the formula

              =>  [tex]z = \frac{m_e}{e^2n_e\rho}[/tex]

    Substituting the given values

           [tex]z = \frac{9*10^{-31}}{(1.6*10^{-19}(16.1*10^{28}(2.5*10^{-8})))}[/tex]

                [tex]= 2.3*10^{-14}sec[/tex]

Answer:

[tex]W_f=-62460\ J[/tex]

Explanation:

Given that

mass of the car ,m = 120 kg

Radius ,R= 12 m

Speed at the bottom , u = 25 m/s

Speed at top ,v= 8 m/s

We know that

Work done by all the forces = Change in the kinetic energy

Work done by gravity +  Work done by friction =Change in the kinetic energy

By taking point A as reference

[tex]m g \times (2R) + W_f=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2[/tex]

Now by putting the values in the above equation we get

[tex]120\times 10\times 2\times 12+ W_f=\dfrac{1}{2}\times 120\times 8^2-\dfrac{1}{2}\times 120\times 25^2[/tex]

[tex]W_f=\dfrac{1}{2}\times 120\times 8^2-\dfrac{1}{2}\times 120\times 25^2-120\times 10\times 2\times 12\ J[/tex]

[tex]W_f=-62460\ J[/tex]

Therefore the work done by friction force will be -62460 J.

Answer:

(A) t = 4s

(B) H = 40m

(C) g = 5m/s²

(D) V = -20m/s

(E) t = 8s

The detailed solution to this problem requires the knowledge of costant linear acceleration motion.xplanation:

The detailed solution to this problem requires the knowledge of costant linear acceleration motion.

Explanation:

The full solution can be found in the attachment below.

To answer part A, we have been given some values of velocities bounding the time jnterval of 1s from which we can calculate the acceleration due to gravity and then the time.

Part B

Requires just using the acceleration due to gravity and the time taken in the equation

H = ut - 1/2gt²

Part C

Has already been calculated in part A

Part D

V = -20m/s because the tennis ball is coming down as the upward direction was assumed positive.

Part E

When the ball retirns to its original position it is the same as it never left and so H = 0m

The calculation can be found in the attachment below.

Final answer:

Ball reaches peak at 4s, rises 80m, experiences -5m/s² gravity. Returns at -20m/s and spends 8s in the air (4s up + 4s down).

Explanation:

Here are the answers to your questions about the tennis ball on the atmosphere-free planet:

A) Time to reach maximum height: Between the beginning and one second later, the ball's velocity decreases by 5.0 m/s. Since acceleration is constant (no atmosphere), this decrease is due to gravity acting against the initial velocity. We can use the equation:

v = u + at

where:

v = final velocity (15.0 m/s)

u = initial velocity (20.0 m/s)

a = acceleration due to gravity (unknown, negative direction)

t = time (1 second)

Solving for a:

a = (v - u) / t = (15.0 m/s - 20.0 m/s) / 1 s = -5.0 m/s²

Now, to find the time to reach maximum height, we know that at the peak, the velocity is 0 m/s. We can again use the same equation:

0 = u + at

Solving for t:

t = -u / a = -20.0 m/s / (-5.0 m/s²) = 4.0 seconds (positive result since time cannot be negative)

Therefore, it takes 4.0 seconds for the ball to reach its maximum height.

B) Maximum height:

We can use the kinematic equation to find the maximum height (h):

h = ut + 1/2 × at²

Substituting the values:

h = 20.0 m/s × 4.0 s + 1/2 × (-5.0 m/s²) × (4.0 s)² = 40.0 m + 40.0 m = 80.0 meters

C) Acceleration due to gravity:

We already calculated the acceleration due to gravity in part A: -5.0 m/s² (the negative sign indicates downward direction).

D) Velocity upon returning to the original position:

Since the motion is symmetrical (same initial and final positions), the velocity when the ball returns will be the negative of its initial velocity: -20.0 m/s.

E) Time spent in the air:

The total time in the air is the sum of the time to reach the maximum height and the time to fall back to the original position. Since the motion is symmetrical, the time to fall back is also 4.0 seconds. Therefore, the total time in the air is:

4.0 seconds (upward) + 4.0 seconds (downward) = 8.0 seconds

Explanation:

Below is an attachment containing the solution.

The average power supplied by the force of kinetic friction acting on the block if the block moves 4 m before coming to stop is  -31.25 W.

In physics, power is the amount of energy that is transferred or transformed in a given amount of time. The International System of Units uses the watt, or one joule per second, as the unit of power. Ancient literature frequently used the term "activity" to describe power. Powers are scalar variables.

According to the question, the given values are :

Block mass, m = 4 kg,

The initial speed of the block at the horizontal surface, v₁ = 5 m/s,

The final speed of the block will be, v₂ = 0 m/s.

Distance covered, d = 4 m

Let t be the time when the box is going to stop.

d = [(v₁+v₂)/2] × t

4 = (5/2) t

t = 1.6 sec.

The change in kinetic energy of the block will be ΔE.

ΔE = mv₂²/2 - mv₁²/2

ΔE = 0 - (4)(5)²/2

ΔE = -50 J

Average power supplied by the force of kinetic friction acting on the block = P(avg)

P(avg) = ΔE / t

P(avg) = -50 / 1.6

P(avg) = -31.26 W.

So, the average power supplied by the force of kinetic friction on the block will be -31.26 W.

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Answer: The coefficient of static friction is 3.85 and  The coefficient of kinetic friction is 2.8

Explanation:

in the attachment

Answer:

minimum speed v=[tex]\sqrt({Fr}/m)}[/tex]

Explanation:

Recall the formula for centripetal force;

Centripetal force is the force that is required to keep an object moving in circular part

     [tex]F=mv^{2} /r[/tex]

where;

F=centripetal force

m=mass of object

r=radius of curvature

v= minimum speed

To find minimum speed make v the subject of formula;

   v=[tex]\sqrt({Fr}/m)}[/tex]

The question is incomplete.

The complete question is:

A student with a mass of m rides a roller coaster with a loop with a radius of curvature of r. What is the minimum speed the rollercoaster can maintain and still make it all the way around the loop? g= 9.8m/s

r=5.5 m= 55kg

Answer: 7.3m/s

Explanation:

Centripetal force is the force acting on a body causing circular motion towards the center which allows to keep the body on track or right path. Any combination of forces in nature can cause centripetal acceleration in various systems.

Where:

V is the tangential velocity

W is the angular velocity

M is the Mass in kg

g is the gravitational force

r is the circular radius

Apply Newton's second law of motion.

The minimum speed is also known as the critical speed is the point where the tension, frictional or normal force is zero and the only thing keeping the object in circular motion is the force of gravity.

This is explained as follows:

N + mg = mw^2r

mg = mv^2/r

Therefore,the minimum speed is:

v^2 = gr

v = square root(gr)

= sqrt(9.8N/kg)(5.5m)

=(9.8N/kg)(5.5m) ^1/2

v = 7.3m/s

Answer: the total velocity of the air is 67.69km/h to the north and 35.4km/h to the east.

Explanation: The initial velocity of the plane is 200km/h south (supose that south is our positive x-axis here and east is the positive y-axis)

In one hour, the plane is located 137km away from the initial position, and the position in X is equal to 137km*cos(15°) = 132.33, this means that the velocity in the x axis is equal to 132.33 km/h, knowing that the initial velocity of the plane was 200km in the x-axis, this means that the velocity of the air must be:

132.33km/h - 200km/h = -67.69km/h

km and the position in "y" is equal to 137km*sin(15°) = 35.4km

This means that the velocity of the air in the y-axis is 35.4km/h

So the total velocity of the air is 67.69km/h to the north and 35.4km/h to the east.

Answer:

Δy=70.3885 downwards

Explanation:

First we get the acceleration on moon by substituting the values we have(v=0m/s at t=0s and v=-3.40m/s at t=2.0 s) into equation of simple motion:

[tex]v_{f}-v_{i}=at\\-3.40m/s-0=(2s)a\\a=-1.7m/s^{2}[/tex]

As we know that

Δy=vit+(1/2)at²

Substitute the values of acceleration and time t=9.10s

So

Δy=vit+(1/2)at²

[tex]=(0)(9.10s)+(1/2)(-1.7m/s^{2})(9.10s)^{2}\\ =-70.3885m[/tex]  

Δy=70.3885 downwards

The camera has fallen approximately 402.5 meters after 9.10 seconds.

To find the distance the camera has fallen after 9.10 s, we can use the equation of motion:

s = ut + (1/2)at^2

Where:
s = distance fallen
u = initial velocity (0 m/s)
t = time (9.10 s)
a = acceleration (due to gravity, -9.8 m/s^2)

Plugging the values into the equation, we get:

s = 0 + (1/2)(-9.8)(9.10)^2

Simplifying, we find that the camera has fallen approximately 402.5 meters after 9.10 seconds.

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Answer:

your question is incomplete as the options are not given. I guess following is the complete question.

Which of these atoms is most likely to share electrons with other atoms?

a) chlorine (7 valence electrons)

b) calcium (2 valence electrons)

c) argon (8 valence electrons)

d) carbon (4 valence electrons)

e) potassium (1 valence electron)

The correct option is d) carbon (4 valence electrons)

Explanation:

Carbon has four electrons in its valence shell. In order to complete the 8 electrons in its valence shell carbon has to make four covalent bonds by sharing its four electrons with the other atom. Carbon atom will neither gain the electrons nor it losses the electrons to follow the octet rule. So in the above mentioned options carbon is the atom that will share maximum electrons.

→ Chlorine has 7 electrons, it will gain 1 electron. It will not do the sharing.

→ Calcium has 2 electrons, it will lost these 2 electrons to complete its shell.

→ Argon has already a completed shell. It will not react with other atom.

→ Potassium has only 1 valence electron which it will lose to complete its shell.

Answer: F

Explanation:

When pushing the body it is necessary to break the frictional force generated by the floor. Once this frictional force is overcome, the body will begin to move. Ideally, if a constant velocity is maintained or close to this value, the acceleration that will be exerted will tend to be zero and therefore, by Newton's second law the value of the Force will also tend to minimum values.

Remember that this law tells us that

[tex]F= ma[/tex]

[tex]F= m \frac{\Delta v}{t}[/tex]

Therefore the best strategy is A. keep pushing the box forward at a steady speed

To minimize the average force applied to a box being pushed across a rough floor, you should keep pushing the box forward at a steady speed. This way, you're only balancing the frictional force, and there's no need for an extra force to accelerate the box.

To minimize the average force applied to the box on a rough floor, you would opt for method a. Keep pushing the box forward at a steady speed. This option will ensure constant velocity, meaning that the net force on the box is zero. In this case, the force you're applying is just balancing the frictional force the box experiences due to the rough floor.

Contrastingly, methods b and c involve changing the box's velocity, which necessitates an acceleration. According to Newton's second law (F=ma), a force is required for acceleration. Thus, these methods will require a greater average force compared to method a.

Learn more about Force and Motion

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Answer:

981.41 secs

Explanation:

Parameters given:

Voltage, V = 1.5V

Power, P = 0.204W

Number of electrons, n = 8.33 * 10^20

First, we calculate the current:

P = I*V

I = P/V

I = 0.204/1.5 = 0.136A

The total charge of 8.33 * 10^20 electrons is:

Q = 8.33 * 10^20 * 1.6023 * 10^(-19)

Q = 133.47 C

Current, I, is given as:

I = Q/t

=> t = Q/I

t = 133.47/0.136 = 981.41 secs

In physics, power is calculated using the formula P = IV, where P is power, I is current, and V is voltage. By applying this formula, along with the relationship between charge, current, and time, the time the flashlight was used can be calculated.

Power is determined by the rate at which energy is transferred, calculated as P = IV where P is power, I is current, and V is voltage. In this scenario, the power supplied is 0.204 W from a 1.50-V battery. The formula P = IV can be rearranged to find the current flowing, which is 0.136 A.

To determine the time the flashlight was used, we can utilize the relationship between charge, current, and time: Q = It. Given 8.33 x 10^20 electrons moved, we need to convert this to Coulombs by recognizing that 1 electron has a charge of approximately 1.6 x 10^-19 C. By dividing the total charge by the current, we find the time t to be approximately 1.23 x 10^19 seconds.

Answer:

The temperature is 749.8 K

Explanation:

Final temperature (T2) = (distance apart/thermal coefficient of expansion×length) + initial temperature

distance apart = 1.5 mm = 1.5/1000 = 0.0015 m

thermal coefficient of expansion for copper = 16.6×10^-6/K

Length of copper = 20 cm = 20/100 = 0.2 m

Initial temperature = 25 °C = 25 + 273 = 298 K

T2 = (0.0015/16.6×10^-6×0.2) + 298 = 451.8 + 298 = 749.8 K

Answer:

The mass of the ice block is equal to 70.15 kg

Explanation:

The data for this exercise are as follows:

F=90 N

insignificant friction force

x=13 m

t=4.5 s

m=?

applying the equation of rectilinear motion we have:

x = xo + vot + at^2/2

where xo = initial distance =0

vo=initial velocity = 0

a is the acceleration

therefore the equation is:

x = at^2/2

Clearing a:

a=2x/t^2=(2x13)/(4.5^2)=1.283 m/s^2

we use Newton's second law to calculate the mass of the ice block:

F=ma

m=F/a = 90/1.283=70.15 kg

70.31kg

Step I: Consider Newton's second law of motion which states that;

∑F = m x a;

Where;

∑F = net force acting on a body

m = the mass of the body

a = acceleration due to the force on the body.

Step II: Now to the question;

Since frictional force is negligible and the only force acting on the block of ice is the applied force by the dockworker, the net force on the body (block of ice) is the constant horizontal force. i.e

∑F = 90.0N

Also;

the block starts from rest and moves a distance (s) of 13.0m in a time (t) of 4.50s. Here, we can get the acceleration in that duration of time using one

the equations of motion as follows;

s = ut + [tex]\frac{1}{2}[/tex]at²            ------------------------------(ii)

Where;

s = distance covered = 13.0m

u = initial velocity = 0      [since the block starts from rest]

t = time taken to cover the distance = 4.50s

a = acceleration of the body.

Substitute these values into equation (ii) as follows;

13.0 = 0(4.5) + [tex]\frac{1}{2}[/tex](a)(4.50)²

13.0 = 0 + [tex]\frac{1}{2}[/tex](a)(20.25)

13.0 = [tex]\frac{1}{2}[/tex](a)(20.25)

13.0 = 10.125a

Solve for a;

a = [tex]\frac{13.0}{10.125}[/tex]

a = 1.28m/s²

Step III: Now substitute the values of a = 1.28m/s² and ∑F = 90.0N into equation (i) as follows;

90.0 = m x 1.28

m = [tex]\frac{90.0}{1.28}[/tex]

m = 70.31

Therefore, the mass of the block of ice is 70.31kg

Answer:

Explanation:

component of force along the ramp

= 370 cos θ where θ is the slope of the ramp with respect to ground.

sinθ = 1.4 / 5

θ = 16 degree

370 cos16

= 355.67 N

Work done

= component of force along the ramp x length of the ramp

= 355.67 x 5

= 1778.35 J

Answer:

The distance between the two slits is 1.2mm.    

Explanation:

The physicist Thomas Young establishes, through its double slit experiment, a relationship between the interference (constructive or destructive) of a wave, the separation between the slits, the distance between the two slits to the screen and the wavelength.

[tex]\Lambda x = L\frac{\lambda}{d} [/tex]  (1)

Where [tex]\Lambda x[/tex] is the distance between two adjacent maxima, L is the distance of the screen from the slits, [tex]\lambda[/tex] is the wavelength and d is the separation between the slits.  

If light pass through two slits a diffraction pattern in a screen will be gotten, at which each bright region corresponds to a crest, a dark region to a trough, as consequence of constructive interference and destructive interference in different points of its propagation to the screen.  

Therefore, d can be isolated from equation 1.

[tex]d = L\frac{\lambda}{\Lambda x} [/tex]  (2)

Notice that it is necessary to express L and [tex]\lambda[/tex] in units of millimeters.

[tex]L = 4m \cdot \frac{1000mm}{1m}[/tex] ⇒ [tex]4000mm[/tex]

[tex]\lambda = 600nm \cdot \frac{1mm}{1x10^{6}nm}[/tex] ⇒ [tex]0.0006mm[/tex]

[tex]d = (4000mm)\frac{0.0006mm}{2mm} [/tex]

[tex]d = 1.2mm[/tex]

Hence, the distance between the two slits is 1.2mm.

Answer:

Explanation:

Analysis of structure gives

a=gsinθ−μkgcosθ

Notice that all the expression are right but we want to know of we can simplify the expression further.

We want to analyse if we can still further simplify the expression,

Inspecting the Right hand side of the equation, we notice that the acceleration due to gravity is common to both side, so we can bring it out i.e.

So option a is wrong because the expression can be simplified further to

a=g(sinθ−μkcosθ)

Option b is right and the best option.

Since we are given that, g=9.8m/s²

We can as well substitute that to option a

So we will have

a=9.8metre/second²(sinθ−μkcosθ)

Also option C is correct but it is not best inserting the values of g directly without simplifying the expression first

So it will have been the best option if it was written as

a=9.8metre/second²(sinθ−μkcosθ)

So the best option is B.

Answer:

Explanation:

The the observer must not be stationed between speakers . Let its position be d cm from one of the speaker. When sound has maximum intensity , the the two sounds reaching his ear must have constructive interference and during zero intensity , they must have destructive interference. Distance between sources will be equal to path difference.

For constructive interference

n λ = path diff

= 19 cm

For destructive interference

(2n+1) λ /2= path diff

= 29 cm

n λ +  λ /2 = 29

19 +  λ /2 = 29

λ  = 20 cm

B )

The path difference of 19 cm corresponds to constructive interference or maximum sound intensity . Wave length is 20 cm . Therefore , next path difference that can create condition of constructive interference will be 19 + 20 = 39 cm . Present path difference is 29 cm. So the distance between the speaker must be increased to 39 cm , so that path difference becomes 39 cm and constructive interference takes place.

ANS 39 cm .

Answer:

595391.482946 m/s

[tex]3.21875\times 10^{6}[/tex]

Explanation:

E = Energy = 1.85 keV

I = Current = 5.15 mA

e = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]

t = Time taken = 1 second

m = Mass of proton = [tex]1.67\times 10^{-27}\ kg[/tex]

Velocity of proton is given by

[tex]v=\sqrt{\dfrac{2E}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 1.85\times 10^3\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}}\\\Rightarrow v=595391.482946\ m/s[/tex]

The speed of the proton is 595391.482946 m/s

Current is given by

[tex]I=\dfrac{\Delta Q}{t}\\\Rightarrow \Delta Q=It\\\Rightarrow \Delta Q=5.15\times 10^{-3}\times (1\ sec)\\\Rightarrow Q=5.15\times 10^{-3}\ C[/tex]

Number of protons is

[tex]n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{5.15\times 10^{-3}}{1.6\times 10^{-19}}\\\Rightarrow n=3.21875\times 10^{6}\ protons[/tex]

The number of protons is [tex]3.21875\times 10^{6}[/tex]

The speed of the protons accelerated by the Van de Graaff generator is approximately 3.19 million meters per second. Additionally, the generator produces roughly 3.21 x 10^16 protons every second.

The Van de Graaff generator is a particle accelerator that can be used to accelerate charged particles. In the given scenario, a beam of protons with an energy of 1.85 keV and a beam current of 5.15 mA is being generated.

(a) The energy of a proton (kinetic energy = 1/2 mv²) is given by the equation E = mv²/2, where m is the mass of the proton (1.67262192369 × 10⁻²⁷ kg) and v is the speed of the proton. Solving the equation v = sqrt((2*E)/m), where E is the energy in joules (1.85 keV = 1.85 * 10^-16 joules), we find that v ≈ 3.19 x 10^6 m/s. This implies that the speed of the protons is approximately 3.19 x 10^6 meters per second.

(b) The number of protons produced each second, i.e., the beam current, can be calculated using the formula I = qN/t, where I is the current, q is the charge of a proton (1.602 x 10^-19 Coulombs), N is the number of protons, and t is time. Rearranging the formula, we find that N = It/q. Substituting the values I=5.15 mA = 5.15 x 10^-3 A and t=1s, we get N ≈ 3.21 x 10^16 protons. Therefore, approximately 3.21 x 10^16 protons are produced every second under the mentioned conditions.

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