A 238 92U nucleus is moving in the x-direction at 5.0 × 105 m/s when it decays into an alpha particle 4 2He and a 234 90Th. If the alpha particle moves off at 25.4° above the x axis with a speed of 1.4 × 107 m/s, what is the recoil velocity of the thorium nucleus? Assume the uranium-thorium-alpha system is isolated; you may also assume the particles are pointlike.

A. To find the orientation angles of the dipole for which the torque is zero, we need to consider the torque equation for an electric dipole in an external electric field.

The torque ([tex]\(\tau\)[/tex]) acting on an electric dipole (p) in an electric field (E) is given by:

[tex]\[ \tau = p \cdot E \cdot \sin(\theta) \][/tex]

Where:

p is the magnitude of the electric dipole moment,

E is the magnitude of the electric field,

[tex]\(\theta\)[/tex] is the angle between the dipole moment (p) and the electric field (E).

For the torque to be zero, [tex]\(\sin(\theta)\)[/tex] must be zero. This happens when [tex]\(\theta = 0\)[/tex] or [tex]\(\theta = \pi\) (180 degrees)[/tex], as [tex]\(\sin(0) = 0\)[/tex] and [tex]\(\sin(\pi) = 0\)[/tex].

So, the two possible orientation angles for which the torque is zero are:

[tex]\(\theta = 0\)[/tex] degrees (aligned with the electric field)

[tex]\(\theta = 180\)[/tex] degrees (opposite to the electric field)

B. Now, let's analyze the stability of these orientations:

1. [tex]\(\theta = 0\)[/tex] degrees (aligned with the electric field):

In this orientation, the dipole moment is aligned with the electric field.

Any slight deviation from this orientation will result in a torque that tends to restore the dipole to its original position. Therefore, this orientation is stable.

2. [tex]\(\theta = 180\)[/tex] degrees (opposite to the electric field):

In this orientation, the dipole moment is opposite to the electric field. Similar to the previous case, any slight deviation from this orientation will result in a torque that tends to restore the dipole to its original position. Therefore, this orientation is also stable.

Both orientations are stable because they represent energy minima. Any deviation from these orientations results in a restoring torque that tries to bring the dipole back to its stable position.

Thus, both orientations ([tex]\(\theta = 0\) degrees and \(\theta = 180\)[/tex] degrees) are stable orientations for the electric dipole in a uniform electric field.

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Final answer:

The torque on an electric dipole in a uniform electric field is zero when the dipole is parallel to the field. This orientation is stable.

Explanation:

In general, the torque vector on an electric dipole, p, from an electric field, E, is given by the equation T = p x E, where the cross product represents the direction of the torque. To find all the orientation angles of the dipole for which the torque is zero, we set the cross product equal to zero:

p x E = 0

The torque is zero when the dipole and electric field vectors are parallel. Therefore, the dipole will experience a torque that tends to align it with the electric field vector. This means that the dipole is in a stable equilibrium when it is parallel to the electric field.

Therefore, the orientation angles that result in a zero torque are all angles that make the dipole parallel to the electric field.

a) Time at which velocity is +20.0 m/s: 2.04 s

b) Time at which velocity is -20.0 m/s: 6.12 s

c) Time at which the displacement is zero: t = 0 and t = 8.16 s

d) Time at which the velocity is zero: t = 4.08 s

e) i) ii) iii) The acceleration of the boulder is always [tex]9.8 m/s^2[/tex] downward

f) See graphs in attachment

Explanation:

a)

The motion of the boulder is a uniformly accelerated motion, with constant acceleration

[tex]a=g=-9.8 m/s^2[/tex]

downward (acceleration due to gravity). So, we can use the following suvat equation:

[tex]v=u+at[/tex]

where:

v is the velocity at time t

u = 40.0 m/s is the initial velocity

a=g=-9.8 m/s^2 is the acceleration

We want to find the time t at which the velocity is

v = 20.0 m/s

Therefore,

[tex]t=\frac{v-u}{a}=\frac{20-40}{-9.8}=2.04 s[/tex]

b)

In this case, we want to find the time t at which the boulder is moving at 20.0 m/s downward, so when

v = -20.0 m/s

(the negative sign means downward)

We use again the suvat equation

[tex]v=u+at[/tex]

And substituting

u = +40.0 m/s

a=g=-9.8 m/s^2

We find the corresponding time t:

[tex]t=\frac{v-u}{a}=\frac{-20-(+40)}{-9.8}=6.12 s[/tex]

c)

To solve this part, we can use the following suvat equation:

[tex]s=ut+\frac{1}{2}at^2[/tex]

where

s is the displacement

u = +40.0 m/s is the initial velocity

[tex]a=g=-9.8 m/s^2[/tex] is the acceleration

t is the time

We want to find the time t at which the displacement is zero, so when

s = 0

SUbstituting into the equation and solving for t,

[tex]0=ut+\frac{1}{2}at^2\\t(u+\frac{1}{2}a)=0[/tex]

which gives two solutions:

t = 0 (initial instant)

[tex]u+\frac{1}{2}at=0\\t=-\frac{2u}{a}=-\frac{2(40)}{-9.8}=8.16 s[/tex]

which is the instant at which the boulder passes again through the initial position, but moving downward.

d)

To solve this part, we can use again the suvat equation

[tex]v=u+at[/tex]

where

u = +40.0 m/s is the initial velocity

[tex]a=g=-9.8 m/s^2[/tex] is the acceleration

We want to find the time t at which the velocity is zero, so when

v = 0

Substituting and solving for t, we find:

[tex]t=\frac{v-u}{a}=\frac{0-(40)}{-9.8}=4.08 s[/tex]

e)

In order to evaluate the acceleration of the boulder, let's consider the forces acting on it.

If we neglect air resistance, there is only one force acting on the boulder: the force of gravity, acting downward, with magnitude

[tex]F=mg[/tex]

where m is the mass of the boulder and [tex]g[/tex] the acceleration of gravity.

According to Newton's second law, the net force on the boulder is equal to the product between its mass and its acceleration:

[tex]F=ma[/tex]

Combining the two equations, we get

[tex]ma=mg\\a=g[/tex]

So, the acceleration of the boulder is [tex]g=9.8 m/s^2[/tex] downward at any point of the motion, no matter where the boulder is (because the force of gravity is constant during the motion).

f)

Find the three graphs in attachment:

- Position-time graph: the position of the boulder initially increases as the boulder goes upward; however, the slope of the curve decreases as the boulder goes higher (because the velocity decreases). The boulder reaches its maximum height at t = 4.08 s (when velocity is zero), then it starts going downward, until reaching its initial position at t = 8.16 s

- Velocity-time graph: the initial velocity is +40 m/s; then it decreases linearly (because the acceleration is constant), and becomes zero when t = 4.08 s. Then the velocity becomes negative (because the boulder is now moving downward) and its magnitude increases.

- Acceleration-time graph: the acceleration is constant and it is [tex]-9.8 m/s^2[/tex], so this graph is a straight horizontal line.

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Answer:

(b) they are not detectable with current technology

Explanation:

The improvement in exoplanet detection methods in recent years, thanks to spaces telescopes such as the Kepler, has led to a revolution in the field of astronomy. The findings have gone from focusing on Jupiters hot to super-earth and, ultimately, to Earth-like planets.

Answer:

a) 15.864s

b) 392.78m

c) 29.52 m/s

Explanation:

The total distance (relative to the truck) that the (front bumper of the) car travels from 24m behind the truck's rear bumper to in front of the car is

distance from car's front bumper to the truck's rear bumper + distance from truck's rear bumper to truck's front bumper (truck's length) + distance from truck's front bumper to car's rear bumper's + distance from the car's rear bumper to the car's front bumper (car's length)

= 24 + 21 + 26 + 4.5 = 75.5 m

As they start at the same speed, we can draw the following equation of motion for the car distance relative to the truck

[tex]s = at^2/2[/tex]

[tex]75.5 = 0.6t^2/2[/tex]

[tex]t^2 = 251.67[/tex]

[tex]t = \sqrt{251.67} = 15.864s[/tex]

b) The actual distance relative to Earth that the car has traveled during this time is the distance car traveled relative to the truck plus distance truck traveled relative to Earth within this time

= 75.5 + 20*15.864 = 392.78 m

c) final speed of the car is the initial speed plus the change in speed

[tex]v = v_0 + \Delta v = v_0 + at = 20 + 15.864*0.6 = 29.52 m/s[/tex]

To pass the truck, it takes the car 33.3 seconds to accelerate and overtake the truck. The car travels a distance of 333 meters during this time. The final speed of the car is 1.80 m/s.

u is the initial velocity of the car and a is the acceleration. Since the car is initially traveling at the same speed as the truck (20.0 m/s) and accelerates at a constant rate of 0.600 m/s², the equation becomes: t = (0 - 20) / -0.600. Solving for t gives us t = 33.3 seconds. To find the distance traveled by the car during this time, we can use the equation: s = ut + (1/2)at², where s is the distance, u is the initial velocity, t is the time, and a is the acceleration. Plugging in the values, we get: s = 20(33.3) + (1/2)(-0.600)(33.3)². Solving for s gives us s = 333 meters. To find the final speed of the car, we can use the equation: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Plugging in the values, we get: v = 20 + (-0.600)(33.3). Solving for v gives us v = 1.80 m/s.

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Answer:

E = -3.6859 x 10∧6 N/C

Explanation:

Let q1 = 3.5 μF = 3.5 x 10∧-6 F and q2 =-7.6 μF = -7.6 x 10∧-6 F

are the charges placed at two corner of equilateral triangle. Electric Field Magnitude "E" on the third corner will be equal to the sum of Electric Filed Magnitude generated by q1 and q2.

E = E1 + E2

E = Ke × q1 / ([tex]d^{2}[/tex]) + Ke × q2 / ([tex]d^{2}[/tex])

E = Ke/[tex]d^{2}[/tex] (q1 + q2)       (taking common)

(Ke 8.99 × 10∧9 N[tex]m^{2}[/tex]/[tex]C^{2}[/tex] and distance d= 0.1 m)  

E =   8.99 × 10∧9 N[tex]m^{2}[/tex]/[tex]C^{2}[/tex] /[tex](0.1 m)^{2}[/tex] (3.5 x 10∧-6 F  -  7.6 x 10∧-6 F)

E = -3685.9 x 10∧3 N/C

E = -3.6859 x 10∧6 N/C

To solve this problem we will define the order of magnitude of both points, then we will obtain the radius and obtain the conclusion of the order of magnitude.

A nanosecond is one billionth of a second while and a millisecond is one millionth of a second

[tex]\frac{\text{millisec}}{\text{nanosec}} = \frac{10^{-3}}{10^{-9}} = 10^6[/tex]

Therefore something that runs in nanoseconds is six times faster than something that runs in milliseconds

Word magnitude means the extent of something. It is the property that determines the object is larger or smaller than the other. The object's magnitude can be arranged into class.

The correct answer is 100,00,00.

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Answer:

1402.73 m

Explanation:

Mass of Azurite=3.25 lb

Percent of copper in AZurite mineral=55.1%

Diameter of  copper wire,d=0.0113  in

Radius of copper wire=[tex]r=\frac{d}{2}=\frac{0.0113}{2}=0.00565 in=\frac{565}{100000}=\frac{565}{100}\times \frac{1}{1000}=5.65\times 10^{-3}in[/tex]

[tex]\frac{1}{1000}=10^{-3}[/tex]

Density  of copper=[tex]\rho=8.96g/cm^3[/tex]

1 lb=454 g

3.25 lb=[tex]3.25\times 454=1475.5 g[/tex]

Mass of Azurite=[tex]1475.5 g[/tex]

Mass of copper=[tex]\frac{55.1}{100}\times 1475.5=813 g[/tex]

Density=[tex]\frac{Mass}{volume}[/tex]

Using the formula

[tex]8.96=\frac{813}{volume\;of\;copper}[/tex]

Volume of copper wire=[tex]\frac{813}{8.96}=90.7cm^3[/tex]

Radius of copper wire=[tex]5.65\times 10^{-3}\times 2.54=14.35\times 10^{-3} cm[/tex]

1 in=2.54 cm

Volume of copper wire=[tex]\pi r^2 h[/tex]

[tex]\pi=3.14[/tex]

Using the formula

[tex]90.7=3.14\times (14.35\times 10^{-3})^2\times h[/tex]

[tex]h=\frac{90.7}{3.14\times (14.35\times 10^{-3})^2}[/tex]

[tex]h=140273 cm[/tex]

1 m=100 cm

[tex]h=\frac{140273}{100}=1402.73 m[/tex]

Hence, the length of copper wire required=1402.73 m

Answer:

[tex]E=158.19\frac{kN}{m}[/tex]

Explanation:

Gauss's theorem states that the flux of the electric field through a closed surface is equal to the the charge enclosed by the surface divided by the vacuum permittivity:

[tex]\int\limits{\vec{E}\cdot \vec{dS}} \,=\frac{q}{\epsilon_0}[/tex]

The direction of the electric field ([tex]\vec{E}[/tex]) just outside of a conductor is parallel to its surface ([tex]\vec{S}[/tex]):

[tex]\int\limits{\vec{E}\cdot \vec{dS}} \,=\int\limits{EdScos(0^\circ)} \,=E\int\limits{dS} \,=ES[/tex]

Recall that the surface charge density is defined as:

[tex]\sigma=\frac{q}{S}[/tex]

Now, we get the electric field strength:

[tex]ES=\frac{q}{\epsilon_0}\\E=\frac{q}{S\epsilon_0}\\E=\frac{\sigma}{\epsilon_0}\\E=\frac{1.4*10^{-6}\frac{C}{m^2}}{8.85*10^{-12\frac{C^2}{N\cdot m^2}}}\\\\E=158192.09\frac{N}{C}=158.19\frac{kN}{C}[/tex]

The electric field strength just outside the surface of a conducting sphere with surface charge density of 1.4 μC/m^2 can be calculated using Gauss' Law. This gives an electric field strength of approximately 1.58 x 10^5 N/C.

The electric field strength just outside the surface of a conducting sphere, with a surface charge density of 1.4 μC/m2, can be calculated using Gauss' Law, which states that the electric field is equal to the surface charge density divided by the permittivity of free space, ε0.

It is given by the formula:

E = σ/ε0

Where E is the electric field strength, σ is the charge density, and ε0 is the permittivity of free space. Using the given value for σ (1.4 x 10-6 C/m2) and the known value for ε0 (8.85 x 10-12 C2/N.m2), we find that:

E = (1.4 x 10-6) / (8.85 x 10-12)

This simplifies to approximately E = 1.58 x 105 N/C.

Therefore, the electric field strength just outside the surface of the conducting sphere is approximately 1.58 x 105 N/C.

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Answer:

a)2.46 %

b)For 1 :101.52 %

For 2 : 99.08 %

c)100..4 %

Explanation:

Given that

g₁ = 9.96 m/s²

g₂ = 9.72 m/s²

The actual value of  g = 9.8 m/s²

a)

The difference Δ g =  9.96 -9.72 =0.24  m/s²

[tex]The\ percentage\ difference=\dfrac{0.24}{9.72}\times 100=2.46\ percentage\\[/tex]

b)

For first one :

[tex]Error\ in\ the\ percentage =\dfrac{9.96}{9.81}\times 100 =101.52\ perncetage[/tex]

For second  :

[tex]Error\ in\ the\ percentage =\dfrac{9.72}{9.81}\times 100 =99.08\ perncetage[/tex]

c)

The mean g(mean )

[tex]g(mean )=\dfrac{9.96+9.72}{2}\ m/s^2\\g(mean)=9.84\ m/s^2[/tex]

[tex]The\ percentage=\dfrac{9.84}{9.8}\times 100=100.40\ percentage[/tex]

a)2.46 %

b)For 1 :101.52 %

For 2 : 99.08 %

c)100..4 %

The percent difference between the two measurements is 2.44%. The percent error for the first measurement is 1.63%, and for the second measurement is 0.82%. The percent error of their mean is 0.41%.

In physics, the percent difference is calculated by subtracting the two values, taking the absolute value, dividing by the average of the two values, and then multiplying by 100. Therefore, the percent difference between the two measurements 9.96m/s² and 9.72m/s² is:

|(9.96-9.72)|/((9.96+9.72)/2)*100 = 2.44%

The percent error involves taking the absolute difference between the experimental value and the accepted value, divided by the accepted value, then multiplied by 100. So, the percent error for the two measurements with accepted value of 9.80m/s² are:

For 9.96m/s²: |(9.96-9.80)|/9.80*100 = 1.63%

For 9.72m/s²: |(9.72-9.8)|/9.8*100 = 0.82%

The percent error of the mean involves doing the above but using the mean of the experimental measurements:

|(Mean of measurements - Accepted value)|/Accepted value * 100 |(9.96+9.72)/2-9.8|/9.8*100 = 0.41%

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Answer:

Explanation:

Option a is correct  

If puck and pick constitute a system then the momentum of the system is conserved but not this may not be valid for the puck .

Option e is correct

If puck and pick is the system then momentum is conserved but because of the presence of friction, mechanical energy is not conserved.

Friction will cause the energy to dissipate in heat.

The final velocity is [tex]v=35.75\ ft/s[/tex] and the final elevation is [tex]45ft[/tex].

The potential energy is the energy possessed by virtue of configuration and the kinetic energy is the energy possessed due to motion.

Given:

Initial velocity, [tex]u=40 ft/s[/tex]

Initial elevation, [tex]h_i=30 ft[/tex]

Weight of the object, [tex]w=100lbf[/tex]

Increase in potential energy, [tex]\Delta PE=1500 ft{-}lbf[/tex]

Decrease in kinetic energy, [tex]\Delta KE=-500 ft{-}lbf[/tex]

(a)

The final velocity of the object is computed as:

[tex]\frac{1}{2} \frac{w}{g}v^2-\frac{1}{2} \frac{w}{g}u^2= \Delta KE\\v^2=40^2 + 2 \times \frac{32.17}{100} \times (-500)\\v= \sqrt{1278.3}\\v=35.75 \ ft/s[/tex]

(b)

The final elevation of the object is computed as:

[tex]mgh_f-mgh_i= \Delta PE\\h_f= h_i+ \frac {\Delta PE}{mg}\\h_f= 30 + \frac{1500}{100}\\h_f=45 ft[/tex]

Therefore, the final velocity is [tex]v=35.75\ ft/s[/tex] and the final elevation is [tex]45ft[/tex].

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A) Horizontal range: 16.34 m

B) Horizontal range: 16.38 m

C) Horizontal range: 16.34 m

D) Horizontal range: 16.07 m

E) The angle that gives the maximum range is [tex]41.9^{\circ}[/tex]

Explanation:

A)

The motion of the shot is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion is a uniformly accelerated motion, so we can use the following suvat equation to find the time of flight:

[tex]s=u_y t + \frac{1}{2}at^2[/tex] (1)

where

s = -1.80 m is the vertical displacement of the shot to reach the ground (negative = downward)

[tex]u_y = u sin \theta[/tex] is the initial vertical velocity, where

u = 12.0 m/s is the initial speed

[tex]\theta=40.0^{\circ}[/tex] is the angle of projection

So

[tex]u_y=(12.0)(sin 40.0^{\circ})=7.7 m/s[/tex]

[tex]a=g=-9.8 m/s^2[/tex] is the acceleration due to gravity (downward)

Substituting the numbers, we get

[tex]-1.80 = 7.7t -4.9t^2\\4.9t^2-7.7t-1.80=0[/tex]

which has two solutions:

t = -0.21 s (negative, we ignore it)

t = 1.778 s (this is the time of flight)

The horizontal motion is instead uniform, so the horizontal range is given by

[tex]d=u_x t[/tex]

where

[tex]u_x = u cos \theta=(12.0)(cos 40^{\circ})=9.19 m/s[/tex] is the horizontal velocity

t = 1.778 s is the time of flight

Solving, we find

[tex]d=(9.19)(1.778)=16.34 m[/tex]

B)

In this second case,

[tex]\theta=42.5^{\circ}[/tex]

So the vertical velocity is

[tex]u_y = u sin \theta = (12.0)(sin 42.5^{\circ})=8.1 m/s[/tex]

So the equation for the vertical motion becomes

[tex]4.9t^2-8.1t-1.80=0[/tex]

Solving for t, we find that the time of flight is

t = 1.851 s

The horizontal velocity is

[tex]u_x = u cos \theta = (12.0)(cos 42.5^{\circ})=8.85 m/s[/tex]

So, the range of the shot is

[tex]d=u_x t = (8.85)(1.851)=16.38 m[/tex]

C)

In this third case,

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

So the vertical velocity is

[tex]u_y = u sin \theta = (12.0)(sin 45^{\circ})=8.5 m/s[/tex]

So the equation for the vertical motion becomes

[tex]4.9t^2-8.5t-1.80=0[/tex]

Solving for t, we find that the time of flight is

t = 1.925 s

The horizontal velocity is

[tex]u_x = u cos \theta = (12.0)(cos 45^{\circ})=8.49 m/s[/tex]

So, the range of the shot is

[tex]d=u_x t = (8.49)(1.925)=16.34[/tex] m

D)

In this 4th case,

[tex]\theta=47.5^{\circ}[/tex]

So the vertical velocity is

[tex]u_y = u sin \theta = (12.0)(sin 47.5^{\circ})=8.8 m/s[/tex]

So the equation for the vertical motion becomes

[tex]4.9t^2-8.8t-1.80=0[/tex]

Solving for t, we find that the time of flight is

t = 1.981 s

The horizontal velocity is

[tex]u_x = u cos \theta = (12.0)(cos 47.5^{\circ})=8.11 m/s[/tex]

So, the range of the shot is

[tex]d=u_x t = (8.11)(1.981)=16.07 m[/tex]

E)

From the previous parts, we see that the maximum range is obtained when the angle of releases is [tex]\theta=42.5^{\circ}[/tex].

The actual angle of release which corresponds to the maximum range can be obtained as follows:

The equation for the vertical motion can be rewritten as

[tex]s-u sin \theta t + \frac{1}{2}gt^2=0[/tex]

The solutions of this quadratic equation are

[tex]t=\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g}[/tex]

This is the time of flight: so, the horizontal range is

[tex]d=u_x t = u cos \theta (\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g})=\\=\frac{u^2}{-2g}(1+\sqrt{1+\frac{2gs}{u^2 sin^2 \theta}})sin 2\theta[/tex]

It can be found that the maximum of this function is obtained when the angle is

[tex]\theta=cos^{-1}(\sqrt{\frac{2gs+u^2}{2gs+2u^2}})[/tex]

Therefore in this problem, the angle which leads to the maximum range is

[tex]\theta=cos^{-1}(\sqrt{\frac{2(-9.8)(-1.80)+(12.0)^2}{2(-9.8)(-1.80)+2(12.0)^2}})=41.9^{\circ}[/tex]

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Answer:

The muzzle speed u =200 m/s

Explanation:

Let u be the muzzle speed.

maximum height s= 500 m

Assuming angle of elevation to be 30°

then initial vertical speed  = usin30° = u/2

moreover, when the bullet reaches the maximum height its vertical velocity will be zero.

then using

[tex]v^2=u^2-2as[/tex]

a= 10 m/s^2

u= u/2

v= 0

s=500

we get

[tex]0^2=u^2/4-2\times10\times500[/tex]

u= 200 m/s.

Therefore, muzzle speed = 200 m/s

Answer:

The muzzle speed is 197.9 m/s.

Explanation:

Given that,

Maximum height = 500 m

Suppose the angle is 30°.

We need to calculate the muzzle speed

Using formula of maximum height

[tex]h_{max}=\dfrac{u^2}{2g}[/tex]

Where, h = maximum height

g = acceleration due to gravity

u = initial vertical velocity

Put the value into the formula

[tex]500=\dfrac{u^2\sin^{2}30}{2\times9.8}[/tex]

[tex]u^2=8\times500\times9.8[/tex]

[tex]u=\sqrt{4\times500\times9.8}[/tex]

[tex]u=197.9\ m/s[/tex]

Hence, The muzzle speed is 197.9 m/s.

Answer:

0.528m

Explanation:

a)58.7 cm = 0.587 m

Let g = 9.8m/s2. When the frog jumps from ground to the highest point its kinetic energy is converted to potential energy:

[tex]E_p = E_k[/tex]

[tex]mgh = mv^2/2[/tex]

where m is the frog mass and h is the vertical distance traveled, v is the frog velocity at take-off

[tex]v^2 = 2gh = 2*9.8*0.587 = 11.5[/tex]

[tex]v = \sqrt{11.5} = 3.4 m/s[/tex]

b) Vertical and horizontal components of the velocity are

[tex]v_v = vsin(\alpha) = 3.4sin(58^0) = 2.877 m/s[/tex]

[tex]v_h = vcos(\alpha) = 3.4cos(58^0) = 1.8 m/s[/tex]

The time it takes for the vertical speed to reach 0 (highest point) under gravitational acceleration g = -9.8m/s2 is

[tex]\Delta t = \Delta v / g = \frac{0 - 2.877}{-9.8} = 0.293s[/tex]

This is also the time it takes to travel horizontally, we can multiply this with the horizontal speed to get the horizontal distance it travels

[tex]s_h = v_ht = 1.8*0.293 = 0.528 m[/tex]

Answer:

Bulb A has a greater resistance.

Explanation:

Electric power (P) = V²/R

P = V²/R................ Equation 1

Where P = power, V = Voltage, R = Resistance.

Make R the subject of the equation

R = V²/P ................ Equation 2

For Bulb A,

Given: V = 120 V, P = 60 W.

Substitute into equation 2

R = 120²/60

R = 240 Ω

For bulb B

Given: V =120 V, P = 100 W.

Substitute into equation 2

R = 120²/100

R = 14400/100

R = 144 Ω

Hence Bulb A has a greater resistance.

bulb A has the greater filament resistance.

This can be defined as the ability of a conducting material to oppose the flow of electric current.

To know the bulb with the greatest filament resistance, we use the formula below.

Formula:

making R the subject of the equation

Where:

For Bulb A,

Given:

Substitute these values into equation 1

For Bulb B,

Given:

Substitute these values into equation 1

Hence, From the above, it can be seen that bulb A has the greater filament resistance.

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Answer:

3.72 m/s²

Explanation:

Horizontal component of the force applied = Fcosθ = 86 cos 30 = 74.478 N

Force = mass × acceleration = 74.478 N

ma = 74.478 N

a = 74.478 / 20 since friction can be neglected = 3.72 m/s²

Answer:

[tex]3.7m/s^{2}[/tex]

Explanation:

The forces acting on the cart is displayed in the attached file below.

Since the motion is only along the horizontal, we can conclude that the force bringing about that motion is the horizontal force.And from second newton law of motion, we deduce that

F=ma,

F=force, m=mass and a=acceleration

from the given question,

F=86N, mass,m=20kg,

Hence we can write the horizontal component of the force as

[tex]F_{X}=Fcos\alpha \\F_{x}=86cos30\\F_{x}=74.48\\Hence \\a=\frac{F_{x}}{m} \\a=\frac{74.48}{20}\\ a=3.7m/s^{2}[/tex]

Answer:

A. [tex]|\vec v_t|=52.42\ m/s[/tex]

B. [tex]\theta=256.74^o[/tex]

Explanation:

Velocity Vector

The velocity vector has two components. Depending on the reference system they could be magnitude and direction in the polar coordinates or x-component and y-component in the rectangular coordinates system.

We are given two velocities in the form of magnitude-direction. The plane's velocity goes south at 39 m/s. The zero reference for angles is pointed East, so the south direction has a 270° angle respect to the reference. If the polar coordinates are known, the rectangular coordinates are computed as

[tex]v_{xp}=|v_p|cos\alpha_p[/tex]

[tex]v_{yp}=|v_p|sin\alpha_p[/tex]

[tex]Since\ |v_p|=39 m/s,\ \alpha_p=270^o,[/tex]

[tex]v_{xp}=39cos\ 270^o=0[/tex]

[tex]v_{yp}=39sin\ 270^o=-39[/tex]

Thus, the velocity of the plane is

[tex]\vec v_p=0\hat i-39\hat j[/tex]

The wind is blowing toward the southwest. It means its angle is 225° (3rd quadrant):

[tex]v_{xw}=|v_w|cos\alpha_w[/tex]

[tex]v_{yw}=|v_w|sin\alpha_w[/tex]

[tex]v_{xw}=17cos\ 215^o=-12.02[/tex]

[tex]v_{yw}=17sin\ 215^o=-12.02[/tex]

Thus, the velocity of the wind is

[tex]\vec v_w=-12.02\hat i-12.02\hat j[/tex]

Now we perform the vector addition to compute the plane's final speed

[tex]\vec v_t=\vec v_p+\vec v_w[/tex]

[tex]\vec v_t=0\hat i-39\hat j-12.02\hat i-12.02\hat j[/tex]

[tex]\vec v_t=-12.02\hat i-51.02\hat j[/tex]

A) The magnitude of the total velocity is

[tex]|\vec v_t|=\sqrt{(-12.02)^2+(-51.02)^2}[/tex]

[tex]\boxed{|\vec v_t|=52.42\ m/s}[/tex]

B) The direction angle is given by

[tex]\displaystyle \tan\theta=\frac{-51.02}{-12.02}=4.24[/tex]

[tex]\theta=arctan\ 4.24[/tex]

[tex]\theta=76.74^o[/tex]

This angle is west of south, we must add 180° to express it in due reference, thus

[tex]\boxed{\theta=256.74^o}[/tex]

Answer

given,

Side of copper plate, L = 55 cm

Electric field, E = 82 kN/C

a) Charge density,σ = ?

  using expression of charge density

 σ = E x ε₀

ε₀ is Permittivity of free space = 8.85 x 10⁻¹² C²/Nm²

now,

 σ = 82 x 10³ x 8.85 x 10⁻¹²

 σ = 725.7 x 10⁻⁹ C/m²

 σ = 725.7 nC/m²

change density on the plates are 725.7 nC/m² and -725.7 nC/m²

b) Total change on each faces

   Q = σ  A

   Q = 725.7 x 10⁻⁹ x 0.55²

   Q = 219.52 nC

Hence, charges on the faces of the plate are 219.52 nC and -219.52 nC

Answer:

solution:

when the engine are fired the rocket has a linear constant acceleration motion:

[tex]V_{1} ^{2} =v_{0} ^{2} +2a_{1} (y_{1} -y_{0} )t=(0)^2+2(16)(y_{1}-0)\\V_{1} ^{2}=32y_{1}................. eq(1)[/tex]

[tex]V_{1}[/tex] is the final velocity of the rocket

when the engines are fired it become equal to the initial velocity of the rocket,

when the engines are shut off

[tex]V_{2} ^{2} =v_{1} ^{2} +2a_{2} (y_{2} -y_{1} )t=>v_{1} ^{2}-2(9.8)(960-y_{1})\\V_{2} ^{2} =v_{1} ^{2}-18816+19.6y_{1}...................eq(2)[/tex]

solve eq(1) and eq(2) we find

[tex](0)^2=(32y_{1} )-18816+19.6y_{1}[/tex]

solving for [tex]y_{1}[/tex]=364.65 m

Where [tex]y_{1}[/tex] is the distance travelled by the rockets for shutting off the engine

when the engines are fired:

[tex]y_{1} =y_{o} + v_{0}t_{1} +\frac{1}{2}at^{2} =>0+(0)T+\frac{1}{2}(16)t^{2\\\\\\364.65=8T^{2} -->T=6.75s[/tex]

NOTE:

DIAGRAM IS ATTACHED

Answer:

V₂  =541.94 m L.

Explanation:

Given that

V₁ = 500 mL  

T₁ = 25°C  = 273 + 25 = 298 K

T₂ = 5°C = 273+50 =323  K

The final volume = V₂  

We know that ,the ideal gas equation  

If the pressure of the gas is constant ,then we can say that

[tex]\dfrac{V_2}{V_1}=\dfrac{T_2}{T_1}[/tex]

[tex]\dfrac{V_2}{V_1}=\dfrac{T_2}{T_1}[/tex]

Now by putting the values in the above equation we get

[tex]V_2=500\times \dfrac{323}{298}\ mL\\V_2=541.94\ mL[/tex]

The final volume of the balloon will be 541.94 m L.

V₂  =541.94 m L.

The final volume of the balloon will be "541.94 mL".

Given:

Volume,

Temperature,

            [tex]= 273+25[/tex]

            [tex]= 298 \ K[/tex]

             [tex]=273+50[/tex]

             [tex]=323 \ K[/tex]

By using the Ideal gas equation, we get

→ [tex]\frac{V_2}{V_1} = \frac{T_2}{T_1}[/tex]

or,

→ [tex]V_2 = \frac{T_2\times V_1}{T_1}[/tex]

       [tex]= \frac{500\times 323}{298}[/tex]

       [tex]= 541.94 \ mL[/tex]

Thus the above approach is correct.  

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Answer:

66.02m/s

Explanation:

the equation describing the distance covered in the horizontal direction is

[tex]x=ucos\alpha t-(1/2)gt^{2}[/tex] but the acceleration in the horizontal path is zero, hence we have

[tex]x=ucos\alpha t[/tex]

Since the horizontal distance covered is 155m at 7.6secs, we have [tex]ucos\alpha =\frac{155}{7.6} \\ucos\alpha =20.38.............equation 1[/tex]

Also from the vertical path, the distance covered is expressed as

[tex]y=usin\alpha t-(1/2)gt^{2}[/tex]

since the horizontal distance covered in 7.6secs is 195m, then we have

[tex]y=usin\alpha t-(1/2)gt^{2}\\y=7.6usin\alpha -4.9(7.6)^{2}\\478.02=7.6usin\alpha \\usin\alpha =62.9...........equation 2[/tex]

Hence if we divide both equation 1 and 2 we arrive at

[tex]\frac{usin\alpha }{ucos\alpha } =\frac{62.9}{20.38} \\tan\alpha =3.08\\\alpha =tan^{-1}(3.08)\\\alpha =72.02^{0}\\[/tex]

Hence if we substitute the angle into the equation 1 we have

[tex]ucos72.02=20.38\\u=66.02m/s[/tex]

Hence the initial velocity is 66.02m/s

Answer:

[tex]t=5057.9167\ s[/tex]

Explanation:

Given:

Now the amount of heat required to heat the water by 50 K:

[tex]Q_w=m_w.c_w.\Delta T[/tex]

[tex]Q_w=0.051\times 4180\times 50[/tex]

[tex]Q_w=10659\ J[/tex]

Now the amount of heat required to heat the resistor by 50 K:

[tex]Q_r=m_r.c_r.\Delta T[/tex]

[tex]Q_r=0.008\times 3700\times 50[/tex]

[tex]Q_r=1480\ J[/tex]

Now the total heat to converted from the electrical energy:

[tex]Q=Q_w+Q_r[/tex]

[tex]Q=12139\ J[/tex]

Now Using Joule's law of heating:

[tex]Q=V.I.t[/tex]

[tex]12139=24\times 0.1\times t[/tex]

[tex]t=5057.9167\ s[/tex]

Final answer:

To determine the time needed to heat water by 5°K, calculate the total energy needed using the specific heat capacity of water, then use the power equation P = V × I to find the time. It will take approximately 444.125 seconds for the water to reach the desired temperature.

Explanation:

To calculate the time it will take for the water to heat up by 5°K using the provided values, first determine the total energy required to raise the temperature of the water by using the equation E = mw × cw × ΔT. Here, E is the energy in joules (J), mw is the mass of water, cw is the specific heat capacity of water, and ΔT is the change in temperature.

Using the given figures:
E = 51 g × 4.18 J/g°K × 5°K = 1065.9 J

Since power P is the rate at which energy is used and P = V × I where V is the voltage and I is the current, we can rearrange the equation to find time: t = E / P.

Substitute the power using the given voltage and current:
P = 24 V × 0.1 A = 2.4 W

The time in seconds to heat the water 5°K is thus:
t = 1065.9 J / 2.4 W = 444.125 seconds

Therefore, it will take approximately 444.125 seconds to heat up the water by 5°K.

Answer:

Charge enter a 0.100 mm length of the axon is [tex]8.98\times 10^{-12} C[/tex]

Explanation:

Electric field E at a point due to a point charge is given by

[tex]E=k \frac{q}{r^2}[/tex]

where [tex]k[/tex] is the constant =[tex]9.0 \times 10^9 Nm^2 / C^2[/tex]

[tex]q[/tex] is the magnitude of point charge and [tex]r[/tex] is the distance from the point charge

Charges entering one meter of axon is [tex]5.\times 10^{11} \times (+e)[/tex]

Charges entering 0.100 mm of axon is [tex]5.\times 10^{11} \times (+e) \times (0.1 \times 10^{-3}[/tex]

substituting the value of [tex]+e=1.6\times 10^{-19} C[/tex] in above equation, we get charge enter a 0.100 mm length of the axon is

[tex]q=5.\times 10^{11} \times1.6\times 10^{-19} \times (0.1 \times 10^{-3}\\q=8.98\times 10^{-12} C[/tex]

Answer:

Q = 3.363 x 10⁻² C

Explanation:

given,

Voltage, V= 169 V

Capacitance of the capacitor, C = 199 μF

Charge in the capacitor = ?

We know,

Q = CV

Q = 169 x 199 x 10⁻⁶

Q = 3.363 x 10⁻² C

Hence, the Charge stored in the capacitor is equal to Q = 3.363 x 10⁻² C

Complete question:

Electrons are ejected from a metallic surface with speeds ranging up to 4.60 x10⁵ m/s when light with a wavelength of 625nm is used. (a) What is the work function of the surface? (b) What is the cutoff frequency for this surface?

Answer:

Part(a) The work function of the surface is 22.177 x 10⁻²⁰ J = 1.384 eV

Part(b) The cutoff frequency for this surface is 3.347 x 10¹⁴ Hz

Explanation:

The kinetic energy (KE) of the emitted photon:

KE = 0.5mv²

m is mass of electron = 9.1 X 10⁻³¹ kg

KE = 0.5 * 9.1 X 10⁻³¹  * (460000)² = 9.628 X 10⁻²⁰ J

in eV = 9.628 X 10⁻²⁰ J x 6.242 X 10¹⁸ ev = 0.601 eV

The photon energy of the incoming radiation:

E = hf = hc/λ

c is speed of light (photon) = 3 x 10⁸

h is Planck's constant = 6.626 × 10⁻³⁴ J.s

E = (6.626 × 10⁻³⁴ *3 x 10⁸) /(625 X 10⁻⁹)

E = 31.805 X 10⁻²⁰ J

in eV = 31.805 X 10⁻²⁰ J x  6.242 X 10¹⁸ ev = 1.985 eV

Part (a) the work function of the surface

KE = hf - W

where;

W is work function

W = hf - KE

W =  31.805 X 10⁻²⁰ J - 9.628 X 10⁻²⁰ J = 22.177 x 10⁻²⁰ J

in eV = 22.177 x 10⁻²⁰ J x 6.242 X 10¹⁸ ev = 1.384 eV

Part(b) the cutoff frequency for this surface

W =hf

f = W/h

f = (22.177 x 10⁻²⁰ J)/(6.626 × 10⁻³⁴ J.s)

f = 3.347 x 10¹⁴ Hz

The work function is the minimum energy required to eject an electron from a metallic surface, which can be calculated using the maximum kinetic energy of ejected electrons and the energy of the incident light. The cutoff frequency is the minimum frequency of light required to eject an electron.

The question is asking to calculate the work function of the surface and the cutoff frequency in a context related to the photoelectric effect. The photoelectric effect is a phenomenon in which electrons are ejected from a metallic surface when it is exposed to light of a certain frequency, in this case, light with a wavelength of 625 nm.

The energy of the incident light is used to expel electrons from the surface of the metal. Any remaining energy contributes toward the ejected electrons' kinetic energy. This can be modeled by the equation KE_maximum = hf - Φ, where KE_maximum is the electron's maximum kinetic energy, h is Planck's constant, f is the frequency of the light, and Φ is the work function of the material. The work function Φ is the minimum amount of energy required to eject an electron from the material's surface.

The cutoff frequency or threshold frequency is the minimum frequency of the incident light required to eject electrons. If the frequency of the light is less than this value, no electrons are ejected

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Answer:

6.71 × 10^8 mi/hr

Explanation:

Light is usually defined as an electromagnetic wave that is comprised of a definite wavelength. It is of both types, visible and invisible. The light emitted from a source usually travels at a speed of about 3 × 10^8 meter/sec. This speed of light is commonly represented by the letter 'C'.

To write it in the metric system, it has to be converted into miles/hour.

We know that,

1 minute = 60 seconds

60 minutes = 1 hour

1 kilometer = 1000 meter

1 miles = 1.6 kilometer

Now,

= [tex]\frac{3 \times\ 10^8 meter \times\ 60 sec \times\ 60 min}{1 sec \times\ 1 min \times\ 1 hr}[/tex]

= 1.08 × 10^12 m/ hr (meter/hour)

= [tex]\frac{1.08 \times\ 10^{12} meter \times\ 1 km \times\ 1 miles}{1 hr \times\ 1000 meter \times\ 1.6 km}[/tex]

= 6.71 × 10^8 mi/hr (miles/hour)

Thus, the value for speed of light (C) in metric unit is 6.71 × 10^8 mi/hr.

Final answer:

The speed of light in a vacuum is precisely known and has a fixed value of c = 2.99792458 × 10^8 m/s, roughly equal to 3.00 × 10^8 m/s. It's a fundamental constant in physics but is slower in materials, as defined by their index of refraction.

Explanation:

The speed of light in a vacuum is represented by the symbol c and has a fixed value of c = 2.99792458 × 108 m/s, which is essentially 3.00 × 108 m/s when rounded to three significant digits. This constant speed is a fundamental physical quantity and is crucial in many areas of physics including relativity and electromagnetism. It's important to note that the speed of light reduces when it passes through matter, which is characterized by the index of refraction n of the material.

Answer:

because of the raindrop velocity relative of the car has a vertical and horizontal component  

Explanation:

Final answer:

The diver takes approximately 2.18 seconds to reach the water after leaving the edge of the cliff. The diver travels approximately 3.21 meters horizontally from the cliff face before hitting the water.

Explanation:

To find the time it takes for the diver to reach the water after leaving the edge of the cliff, we can use the equation of motion:[tex]h = 1/2 * g * t^2,[/tex]where h is the height of the cliff and g is the acceleration due to gravity. Rearranging the equation to solve for t, we get t = sqrt(2h/g). Plugging in the values given, we have t = sqrt(2*30/9.8) ≈ 2.18 s.

To find the horizontal distance the diver travels, we can use the equation s = v * t, where s is the distance, v is the horizontal velocity, and t is the time. Rearranging the equation to solve for v, we get v = s/t. Plugging in the values given, we have v = 7/2.18 ≈ 3.21 m/s.

Answer:

Q = 5267J

Explanation:

Specific heat capacity of copper (S) = 0.377 J/g·°C.

Q = MSΔT

ΔT = T2 - T1

ΔT=49.8 - 22.3 = 27.5C

Q = change in energy = ?

M = mass of substance =508g

Q = (508g) * (0.377 J/g·°C) * (27.5C)

Q= 5266.69J

Approximately, Q = 5267J

Final answer:

To raise the temperature of 508 g of copper from 22.3°C to 49.8°C, 5096.925 J of heat needs to be added, using the specific heat capacity of copper.

Explanation:

To calculate the amount of heat (Q) needed to raise the temperature of a substance, we use the formula Q = mcΔT, where m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature. For 508 g of copper with a specific heat of 0.377 J/g°C, needing to increase from 22.3°C to 49.8°C, the change in temperature (ΔT) is 49.8°C - 22.3°C = 27.5°C.
Therefore, Q = (508 g)(0.377 J/g°C)(27.5°C) = 5096.925 J. Hence, 5096.925 J of heat needs to be added to the copper to achieve the desired temperature increase.