The neutron mass is 1.675 x 10-27 kg. Express the proton mass in eV/c, keeping 4 significant figures B) The mass of the electron is 9.109 x 10-31 kg. Express this in eV/c . C) What is the mass of the hydrogen atom in eV/c??

Answer:

The cars pass next to one another after 25.28 s.

Explanation:

When the cars pass next to one another, the position of both cars is the same relative to the center of the system of reference (marker 0 in this case). Then:

Position of car 1 = position of car 2

The position of an accelerating object moving in a straight line is given by this equation:

x = x0 +v0 t +1/2 a t²

where

x = position at time t

x0 = initial position

v0 = initial speed

t = time

a = acceleration

If the position of car 1 = position of car 2 then:

0 km + 20.0 m/s * t + 1/2 * 0.10 m/s² * t² = 1.2 km - 30.0 m/s * t + 1/2 * 0.30 m/s² * t²

Note that the acceleration of car 2 has to be positive because the car is slowing down and, in consequence, the acceleration has to be opposite to the velocity. The velocity is negative because the direction of car 2 is towards the origin of our system of reference. Let´s continue:

0 km + 20.0 m/s * t + 1/2 * 0.10 m/s² * t² = 1.2 km - 30.0 m/s * t + 1/2 * 0.30 m/s² * t²

1200 m - 50.0 m/s * t + 0.10 m/s² * t² = 0

Solving the quadratic equation:

t = 25.28 s

t = 474. 72 s We discard this value because, if we replace it in the equation of the position of car 2, we will get a position of 20762 m, which is impossible because the position of car 2 can´t be greater than 1200 m.

Then, the cars pass next to one another after 25.28 s  

The police car takes approximately 30 seconds to reach the speeder, requires an acceleration of about 1.67 m/s², and its velocity at the moment it reaches the speeder is approximately 50 m/s.

This problem is a two-body pursuit scenario. The speeder's motion can be described by x = Ut, while the police car's motion is represented by the equation x = 1/2at². The speeder is moving at a constant speed of 120 km/h which is equal to 33.33 m/s.

(a) Time for the police car to overtake the speeder:
To find the time, we equate the two equations (since they both cover the same distance of 750m) and solve for 't'. Thus, 750m = 33.33m/s * t = 1/2 * a * t². By solving this equation, we get two values of 't', out of which the realistic answer is t = 30 seconds.

(b) Required acceleration of the police car:
Plugging the time into the equation for the police car, we get 750m = 1/2 * a * (30s)². Solving for 'a', we find the required acceleration to be approximately 1.67 m/s².

(c) Velocity of the police car at the moment it reaches the speeder:
Since the police car starts from rest and maintains a constant acceleration, the final velocity can be calculated using the equation v = at. Hence, at the moment it reaches the speeder, the police car's velocity would be approximately 50 m/s.

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Answer:

F = 1080 N

Explanation:

given,

mass of the hammer = 1.8 kg

velocity of the hammer = 6 m/s

distance into board = 30 mm = 0.03 m

to calculate average resistance force = F

kinetic energy of the hammer is equal to the work done by the hammer

[tex]\dfrac{1}{2}mv^2 = Force\times displacement[/tex]

[tex]\dfrac{1}{2}mv^2 = F\times d[/tex]

[tex]\dfrac{1}{2}\times 1.8\times 6^2 = F\times 0.03[/tex]

F = 1080 N

hence, the average resistance force is equal to F = 1080 N

Answer:1.1 eV

Explanation:

Given

Energy(E)=2.9 eV

Work function of the target=1.8 eV

We know that

[tex]Energy(E)=W_0+KE_{max}[/tex]

Where [tex]W_0=work\ function[/tex]

[tex]2.9=1.8+KE_{max}[/tex]

[tex]KE_{max}=2.9-1.8=1.1 eV[/tex]

Answer:

Explanation:

At the topmost position,  the car does not have zero velocity but it has velocity of v so that

v² /r = g or centripetal acceleration should be equal to g ( 9.8 )

Considering that,  the car must fall from a height of 2r + h where

mgh = 1/2 mv²

= 1/2 m gr

So h = r/2

Hence the ball must fall from a height of

2r + r /2

= 2.5 r . So that it can provide velocity of v  at the top where

v² / r = g .

Answer:

a) v = 15.6 m/s

b) 0.65 s are needed to reach a height of 12.3 m

Explanation:

The equations that describe the height and velocity of the stone are the following:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where

y = height of the stone at time t

y0 = initial height

v0 = initial speed

t = time

g = acceleration due to gravity

b) First, let´s find the time at which the stone reaches a height of 12.3 m:

y = y0 + v0 · t + 1/2 · g · t²

12.3 m = 0 m + 22.0 m/s · t + 1/2 · (-9.8 m/s²) · t²    (y0 = 0 placing the center of the frame of reference at the point at which the stone is thrown.)

-4.9 m/s² · t² + 22.0 m/s · t - 12.3 m = 0

t = 0.65 s (when the stone goes upward) and t = 3.84 s ( when the stone returns downward) .

So, 0.65 s are needed to reach a height of 12.3 m

a) The velocity at that time will be:

v = v0 + g · t

v = 22.0 m/s - 9.8 m/s² · 0.65 s = 15.6 m/s

The stone moves at approximately 15.62 m/s when it reaches a height of 12.3 meters. The time required to reach this height is approximately 0.652 seconds.

Initial velocity (u) = 22.0 m/s

Acceleration (a) = -9.8 m/s² (due to gravity)

Height (h) = 12.3 m

We can use the equation:

v² = u² + 2a(s), where v is the final velocity, u is the initial velocity, a is acceleration, and s is the displacement.

Substituting the given values:

v² = (22.0 m/s)² + 2(-9.8 m/s²)(12.3 m)

v² = 484 - 240.12

v² = 243.88

v = √243.88 ≈ 15.62 m/s

Thus, the stone is moving at approximately 15.62 m/s when it reaches a height of 12.3 m.

We can use the equation:

s = ut + 1/2 at², where s is the displacement, u is the initial velocity, t is time, and a is acceleration.

12.3 m = (22.0 m/s)t + 1/2(-9.8 m/s²)t²

12.3 = 22.0t - 4.9t²

Rearrange to form a quadratic equation:

-4.9t² + 22.0t - 12.3 = 0

Using the quadratic formula:

t = [ -b ± √(b² - 4ac) ] / 2a

t = [ 22.0 ± √(484 - 4(-4.9)(-12.3)) ] / 2(-4.9)

t = ( 22.0 ± √(484 - 240.12) ) / 9.8

t = ( 22.0 ± √243.88 ) / 9.8

t = 22.0 ± 15.62 / 9.8

We get two possible values for t:

t₁ = (22.0 - 15.62) / 9.8 ≈ 0.652 s

t₂ = (22.0 + 15.62) / 9.8 ≈ 3.83 s

The correct time to reach 12.3 m as the stone ascends is approximately 0.652 seconds.

Final answer:

The density of the object is 2.40 g/cm³ with an absolute uncertainty of 0.01 g/cm³, which corresponds to a relative uncertainty of approximately 0.411%.

Explanation:

To calculate the density of an object, you divide the mass by the volume. Given that the mass is 53.5 ± 0.1 g and the volume is 22.30 ± 0.05 cm³, the density can be computed as follows:

Density = Mass/Volume = 53.5 g / 22.30 cm³ = 2.3991 g/cm³
However, when it comes to reporting this value, we need to match the significant figures to the least number in any of the measurements used, which in this case would be three significant figures. Thus, the density is reported as 2.40 g/cm³.

To calculate the uncertainty in density, you combine the relative uncertainties of mass and volume by simply adding them because we're dividing the two quantities. The relative uncertainty of mass is (0.1 g / 53.5 g) * 100% ≈ 0.187%, and the relative uncertainty of volume is (0.05 cm³ / 22.30 cm³) * 100% ≈ 0.224%. Adding these gives us the total uncertainty:

Total relative uncertainty = 0.187% + 0.224% ≈ 0.411%

To find the absolute uncertainty in density, you multiply the total relative uncertainty by the calculated density:

Absolute uncertainty = 0.411% * 2.3991 g/cm³ ≈ 0.00986 g/cm³

Rounded to match the significant figures of the calculated density, this would be 0.01 g/cm³. So, the density reported with its absolute uncertainty is 2.40 ± 0.01 g/cm³.

Answer:

building height is 52.52 m

Explanation:

given data

initial speed v = 13 m/s

angle = 35°

time = 2.6 s

to find out

how tall is building

solution

we consider here h is height of building

so

initial velocity at angle 35 is express as

u = v sin(θ)     ...........1

u = 13 sin35 = 7.45 m/s

so

by distance formula

h = ut + 1/2 at²        ...........2

h = 7.45 ( 2.6) + 1/2 × (9.81) × (2.6)²

h = 52.52 m

so building height is 52.52 m

Final answer:

Using the formula for vertical displacement in projectile motion, with an initial velocity of 13 m/s downward and the ball being in the air for 2.6 seconds, the building's height is found to be approximately 52.5 meters.

Explanation:

To solve for the height of the building in the given scenario, we can use the vertical component of the projectile motion. We know that the child throws the tennis ball with an initial speed of 13 m/s at an angle of 35° below the horizontal, and it lands after 2.6 seconds. We can use the formula for the vertical motion under constant acceleration (gravity in this case) to find the height:

Equation for vertical displacement:

s = ut + ½at2

s is the vertical displacement (height of the building), u is the initial vertical velocity, a is the acceleration due to gravity (9.8 m/s2), and t is the time the ball is in the air.

First, we calculate the initial vertical velocity component (u):

u = vinitial × sin(θ)

u = 13 m/s × sin(35°) = 13 m/s × 0.5736 ≈ 7.457 m/s (downward)

Since the initial velocity is downward and we need upward to be positive, we will treat it as negative:

u = -7.457 m/s

Next, we can calculate the height of the building.

s = (-7.457 m/s)(2.6 s) + ½(-9.8 m/s2)(2.6 s)2

s = -19.3886 m - 33.118 m

s = -52.5066 m

Since the displacement is in the negative direction (downward), the height of the building is 52.5 meters (we must take the absolute value of the displacement to get the height).

Answer:

t = 20.96 seconds

Explanation:

given,

lightning strikes at = 4.5 mi( 23,760 ft)

air temperature =74.0°F

74.0°F = 23.3 °C

speed of the sound at 23.3 °C

V = 331.5 + 0.6 × T

V = 331.5 + 0.6 × 23.3

   = 345.48 m/s

distance given = 4.5 mile

1 mile = 1609.4                

4.5 mile = 4.5 × 1609.4 = 7242.3 m            

time taken =

[tex]t =\dfrac{d}{v}[/tex]

[tex]t =\dfrac{7242.3}{345.48}[/tex]

t = 20.96 seconds

hence, time taken by the sound to reach by the observer is 20.96 sec.

Final answer:

To estimate the time elapsed between seeing lightning and hearing thunder, one must know the speed of sound, which varies with temperature, and the distance to the lightning strike. At a distance of 4.5 miles and an air temperature of 74.0°F, approximately 21 seconds will elapse.

Explanation:

Calculating the Time Difference Between Lightning and Thunder

The question involves the relationship between the speed of light and sound to determine how long it will take for the sound of thunder to reach an observer after the lightning is seen. Since light travels at approximately 300,000 kilometers per second (3 × 108 meters per second), lightning is seen almost instantaneously. However, the sound of thunder, which is caused by the rapid expansion of superheated air due to a lightning strike, travels much slower.

At a temperature of 74.0°F, we need to convert this to Celsius since the speed of sound in air is given by the formula v = 331.4 m/s + 0.6T, where T is in Celsius. The temperature in Celsius can be found through T(°C) = (T(°F) - 32) × 5/9, giving us T(°C) = (74 - 32) × 5/9 = 23.33°C.

The speed of sound at this temperature is v = 331.4 m/s + 0.6 × 23.33 = 345.4 m/s. To find the time it takes for the sound to travel 4.5 miles (or 23,760 feet), we need to convert the distance to meters since the speed of sound is given in meters per second. We have 1 mile = 1,609.34 meters, so 4.5 miles = 7,242 meters approximately. Finally, we calculate the time as time = distance/speed = 7,242 m / 345.4 m/s ≈ 20.97 s.

Therefore, approximately 21 seconds will elapse between seeing the lightning and hearing the thunder if the lightning strikes 4.5 miles away and the air temperature is 74.0°F.

Answer:

Ax  =7 cm

Explanation:

Analysis:

Let A be the vector of rectangular components Ax and Ay:

Where:

Ay=7 cm  : The The y-component of  vector A-component of  vector A

α =45       : Angle of A with the positive x-axis

Ax           : The x-component of  vector A

Because Ax and Ay are the components of a right triangle we apply the following formula:

[tex]tan\alpha =\frac{A_{y} }{A_{x} }[/tex]

[tex]tan45=\frac{7}{A_{x} }[/tex]

[tex]A_{x} =\frac{7}{tan45}[/tex]

Ax  =7 cm

Answer:

Explanation:

The spacetime interval [tex]\Delta s^2[/tex] is given by

[tex]\Delta s ^2 = \Delta (c t) ^ 2 - \Delta \vec{x}^2[/tex]

please, be aware this is the definition for the signature ( + - - - ), for the signature (- + + + ) the spacetime interval is given by:

[tex]\Delta s ^2 = - \Delta (c t) ^ 2 + \Delta \vec{x}^2[/tex].

Lets work with the signature ( + - - - ), and, if needed in the other signature, we can multiply our interval by -1.

[tex]\Delta \vec{x}^2 = (7.5 \ 10 \ m)^2[/tex]

[tex]\Delta \vec{x}^2 = 5,625 m^2[/tex]

[tex]\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 3 \ s)^2[/tex]

[tex]\Delta (c t) ^ 2 = (899,377,374 \ m)^2[/tex]

[tex]\Delta (c t) ^ 2 = 8.0888 \ 10^{17} m^2[/tex]

so

[tex]\Delta s ^2 = 8.0888 \ 10^{17} m^2 - 5,625 m^2[/tex]

[tex]\Delta s ^2 = 8.0888 \ 10^{17} m^2[/tex]

[tex]\Delta \vec{x}^2 = (5 \ 10 \ m)^2[/tex]

[tex]\Delta \vec{x}^2 = 2,500 m^2[/tex]

[tex]\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 0.58 \ s)^2[/tex]

[tex]\Delta (c t) ^ 2 = (173,879,625.6 \ m)^2[/tex]

[tex]\Delta (c t) ^ 2 = 3.0234 \ 10^{16} m^2[/tex]

so

[tex]\Delta s ^2 = 3.0234 \ 10^{16} m^2 - 2,500 m^2[/tex]

[tex]\Delta s ^2 = 3.0234 \ 10^{16} m^2[/tex]

[tex]\Delta \vec{x}^2 = (5 \ 10 \ m)^2[/tex]

[tex]\Delta \vec{x}^2 = 2,500 m^2[/tex]

[tex]\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 58 \ s)^2[/tex]

[tex]\Delta (c t) ^ 2 = (1.73879 \ 10^{10} \ m)^2[/tex]

[tex]\Delta (c t) ^ 2 = 3.0234 \ 10^{20} m^2[/tex]

so

[tex]\Delta s ^2 = 3.0234 \ 10^{20} m^2 - 2,500 m^2[/tex]

[tex]\Delta s ^2 = 3.0234 \ 10^{20} m^2[/tex]

Answer:

Ans. B) 22 m/s (the closest to what I have which was 20.16 m/s)

Explanation:

Hi, well, first, we have to find the equations for both, the driver and the van. The first one is moving with constant acceleration (a=-2m/s^2) and the van has no acceletation. Let´s write down both formulas so we can solve this problem.

[tex]X(van)=5.65t+154[/tex]

[tex]X(driver)=34.4t+\frac{(-2)t^{2} }{2}[/tex]

or by rearanging the drivers equation.

[tex]X(driver)=34.4t+t^{2}[/tex]

Now that we have this, let´s equal both equations so we can tell the moment in which both cars crashed.

[tex]X(van)=X(driver)[/tex]

[tex]5.65t+154=34.4t-t^{2}[/tex]

[tex]0=t^{2} -(34.4-5.65)t+154[/tex][tex]0=t^{2} -28.75t+154[/tex]

To solve this equation we use the following formulas

[tex]t=\frac{-b +\sqrt{b^{2}-4ac } }{2a}[/tex]

[tex]t=\frac{-b +\sqrt{b^{2}-4ac } }{2a}[/tex]

Where a=1; b=-28.75; c=154

So we get:

[tex]t=\frac{28.75 +\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=21.63s[/tex][tex]t=\frac{28.75 -\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=7.12s[/tex]

At this point, both answers could seem possible, but let´s find the speed of the driver and see if one of them seems ilogic.

[tex]V(driver)=V_{0} +at[/tex]}

[tex]V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(7.12s)=20.16\frac{m}{s}[/tex][tex]V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(21.63s)=-8.86\frac{m}{s}[/tex]

This means that 21.63s will outcome into a negative speed, for that reason we will not use the value of 21.63s, we use 7.12s and if so, the speed of the driver when he/she hits the van is 20.16m/s, which is closer to answer  A).

Best of luck

Answer:

1. Running velocity (5 m/s)

2. Resting velocity (0 m/s)

3. Walking velocity (-1 m/s)

1. Running speed (5 m/s)

2. Walking speed (1 m/s)

3. Resting speed (0 m/s)

Explanation:

Attached you will find the plot of position vs time of Ellie´s movement.

The velocity is the displacement of the object over time relative to the system of reference. The speed, in change, is the traveled distance over time in disregard of the system of reference.

So, the velocity is calculated as follows:

v = Δx / Δt

where

Δx = final position - initial position

Δt = elapsed time

1) The average velocity of Ellie while running is:

v = 1000 m - 0 m / 200 s = 5 m/s

While resting:

v = 0 m - 0 m / 100 s = 0 m/s

And while walking back:

v = 0 m - 1000 m / 1000 s = - 1 m/s

Note that in this last case, the initial position is 1000 m because Ellie is 1000 m from the origin of the system of reference when she walks back. The final position will be the origin of the system of reference, 0 m.

Comparing with the graphic, the velocity is the slope of the function position(t).

Then:

1. Running velocity (5 m/s)

2. Resting velocity (0 m/s)

3. Walking velocity (-1 m/s)

2) The speed is the distance traveled over time:

Running speed = 1000 m / 200 s = 5m /s

Resting speed = 0 m / 100 s = 0 m/s

Walking speed = 1000 m/ 1000 s = 1 m/s

Then:

1. Running speed (5 m/s)

2. Walking speed (1 m/s)

3. Resting speed (0 m/s)  

Answer:

l= 4 mi   : width of the park

w= 1 mi  : length of the park

Explanation:

Formula to find the area of ​​the rectangle:

A= w*l       Formula(1)

Where,

A is the area of the  rectangle in mi²

w is the  width of the rectangle in mi

l is the  width of the rectangle in mi

Known data

A =  4 mi²

l = (w+3)mi    Equation (1)

Problem development

We replace the data in the formula (1)

A= w*l  

4 = w* (w+3)

4= w²+3w

w²+3w-4= 0

We factor the equation:

We look for two numbers whose sum is 3 and whose multiplication is -4

(w-1)(w+4) = 0 Equation (2)

The values ​​of w for which the equation (2) is zero are:

w = 1 and w = -4

We take the positive value w = 1 because w is a dimension and cannot be negative.

w  = 1 mi  :width of the park

We replace w  = 1 mi  in the equation (1) to calculate the length of the park:

l=  (w+3) mi

l= ( 1+3) mi

l= 4 mi

Explanation:

Given that,

Charge 1, [tex]q_1=-5.45\times 10^{-6}\ C[/tex]

Charge 2, [tex]q_2=4.39\times 10^{-6}\ C[/tex]

Distance between charges, r = 0.0209 m

1. The electric force is given by :

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

[tex]F=9\times 10^9\times \dfrac{-5.45\times 10^{-6}\times 4.39\times 10^{-6}}{(0.0209)^2}[/tex]

F = -492.95 N

2. Distance between two identical charges, [tex]r=0.0209\ m[/tex]

Electric force is given by :

[tex]F=\dfrac{kq_3^2}{r^2}[/tex]

[tex]q_3=\sqrt{\dfrac{Fr^2}{k}}[/tex]

[tex]q_3=\sqrt{\dfrac{492.95\times (0.0209)^2}{9\times 10^9}}[/tex]

[tex]q_3=4.89\times 10^{-6}\ C[/tex]

Hence, this is the required solution.

Answer:

a) 2.41 km

b) 38.8°

Questions c and d are illegible.

Explanation:

We can express the displacements as vectors with origin on the point he started (0, 0).

When he traveled south he moved to (-3, 0).

When he moved east he moved to (-3, x)

The magnitude of the total displacement is found with Pythagoras theorem:

d^2 = dx^2 + dy^2

Rearranging:

dy^2 = d^2 - dx^2

[tex]dy = \sqrt{d^2 - dx^2}[/tex]

[tex]dy = \sqrt{3.85^2 - 3^2}  = 2.41 km[/tex]

The angle of the displacement vector is:

cos(a) = dx/d

a = arccos(dx/d)

a = arccos(3/3.85) = 38.8°

Answer: Option (b) is correct.

Explanation:

Since we know that,

P = VI

where;

P = power

V= Voltage

I = Current

Since it's given that,

P = 600W

I = 2.5 A

equating these values in the above equation, we get;

V = [tex]\frac{600}{2.5}[/tex]

V = 240 V

Answer:

(a) Your right-hand thumb must be pointed in the direction of the south pole.

Explanation:

A solenoid is a coil of insulated copper wire which behaves like a magnet when an electric current passes through it and it loses its magnetic behaviour just after the current flow stops.

Since it behaves a magnet, the magnetic poles must be determined. In order to determine the magnetic pole, we put the coil in our right hand, curl our four fingers in the direction of current flow and then the direction in which our thumb points is the north pole. This is known as the Right-hand rule.

From the above discussion, option (a) does not keep pace with the statement of right-hand rule.

Hence, option (a) is not true.

Answer: 5 m/s^2

Explanation: In order to solve this question we have to use the kinematic equation given by:

Vf= Vo+a*t  where V0 is zero.

we know that it takes Vf( 30 m/s) in 6 seconds

so

a=(30 m/s)/6 s= 5 m/s^2

Answer:

[tex]q=1.18*10^{-6}C}[/tex]

Explanation:

The potential V due to a charge q,  at a distance r, is:

[tex]V=k\frac{q}{r}[/tex]

k=8.99×109 N·m^2/C^2      :Coulomb constant

We solve to find q:

[tex]q=\frac{V*r}{k}=\frac{6.25*10^{2}*17}{8.99*10^{9}}=1.18*10^{-6}C[/tex]

Answer:

637 km/h, not reasonable

Explanation:

Total average speed = 91 km/h

Average speed of first half, v1 = 49 km/h

Let the average speed for the next half is v2.

Let the total distance is 2s.

Time taken for first half of the journey, t1 = s / v1 = s / 49

Time take for the next half of the journey, t2 = s / v2

The average speed is defined as the total distance traveled to the total time taken

[tex]91=\frac{2s}{\frac{s}{49}+\frac{s}{v_{2}}}[/tex]

[tex]\frac{s}{49}+\frac{s}{v_{2}}=\frac{2s}{91}[/tex]

[tex]\frac{1}{v_{2}}=\frac{2}{91}-\frac{1}{49}[/tex]

[tex]\frac{1}{v_{2}}=\frac{7}{4459}

v2 = 637 km/h

This is very high speed which is not reasonable.

If the force exerted on an arrow by a bow is tripled, the speed of the arrow leaving the bow would also triple, assuming other factors remain constant. Therefore, the arrow would leave the bow with a speed of 75 m/s.

If the arrow starting from rest leaves the bow with a speed of 25 m/s due to a certain force, tripling the force would result in a tripling of the acceleration, assuming mass remains constant. From Newton's second law, we understand that Force equals mass times acceleration (F = ma). The acceleration of an object is directly proportional to the force applied, if mass is kept constant. Therefore, if we triple the force, the acceleration would also triple. With a tripling of acceleration, the final speed of the arrow would also triple, as speed is the product of acceleration and time. Therefore, if the force exerted on the arrow by the bow is tripled, the speed with which the arrow leaves the bow would be 75 m/s.

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The speed at which the arrow would leave the bow when the average force exerted on it is tripled is 38.40 m/s.

To find the speed at which the arrow would leave the bow if the average force exerted on it were tripled, we can use the principle of conservation of mechanical energy. The initial kinetic energy of the arrow can be calculated using the formula KE = 1/2 * m * v^2, where KE is the kinetic energy, m is the mass of the arrow, and v is the initial velocity. Since the mass of the arrow remains the same, tripling the force would cause the initial kinetic energy to triple as well. Therefore, the final velocity can be found using the formula KE = 1/2 * m * v^2 and solving for v.

Let's use an example to illustrate this. Suppose the mass of the arrow is 0.5 kg. Initially, the arrow leaves the bow with a speed of 25 m/s. The initial kinetic energy is given by KE = 1/2 * 0.5 kg * (25 m/s)^2 = 156.25 J. If the average force is tripled, the new kinetic energy will be 3 times the initial kinetic energy, which is 3 * 156.25 J = 468.75 J. Plugging this value into the kinetic energy formula and solving for v, we get v = √(2 * 468.75 J / 0.5 kg) = 38.40 m/s.

Therefore, if the average force exerted on the arrow by the bow were tripled, the arrow would leave the bow with a speed of 38.40 m/s.

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Answer:

The spot of light hits the bottom of the pool at 4.634 m from the wall beneath the watchman's feet.

Explanation:

Use the diagram attached below this answer to see the notation we will use.

For this case, we're trying to find x and we have:

h=1.3 m

b=2.1 m

a=2.7 m

We also know Snell's law for refraction:

[tex]n_{1} sin\theta_{1}=n_{2}sin\theta_{2}[/tex]

n is the refractive index of each substance (in this case, air and water), which are:

[tex]n_{air}=1[/tex]

[tex]n_{water}=1.33[/tex]

Triangle theory says that [tex]\theta_{1}=\beta[/tex] and:

[tex]tan\beta=\frac{a}{h}[/tex]

[tex]\beta=arc tan(\frac{a}{h})=arctan(\frac{2.7m}{1.3m})=64.29[/tex]

Using Snell's law:

[tex]\theta_{2}=arcsin(\frac{n_{1}sin\theta{1}}{n_{2}})=arcsin(\frac{1sin(64.29)}{1.33})=42.644[/tex]

Using triangle theory:

[tex]tan\theta_{2}=\frac{(x-a)}{b}[/tex]

[tex]x=b*tan\theta_{2}+a=2.1m*tan(42.644)+2.7m=4.634m[/tex]

Answer:

The distance of spot of light from his feet equals 3.425 meters.

Explanation:

The situation is represented in the attached figure below

The angle of incidence is computed as

[tex]\theta _i=tan^{-1}(\frac{1.3}{2.7})\\\\\therefore \theta _i=25.71^{o}[/tex]

Now by Snell's law we have

[tex]n_{i}sin(\theta _i)=n_{r}sin(\theta _r)[/tex]

where

[tex]n_{i},n_{r}[/tex] are the refractive indices of the incident and the refracting medium respectively

[tex]\theta _i,\theta _r[/tex] are the angle of incidence and the angle of refraction respectively

Thus using the Snell's relation we have

[tex]1.0\times sin(25.71)=1.33\times sin(\theta _r)\\\\\therefore sin(\theta _r)=\frac{sin(25.71}{1.33}=0.326\\\\\therefore \theta _r=sin^{-1}(0.326)=19.04^{o}[/tex]

from the attached figure we can see

[tex]tan(\theta _r)=\frac{L_{2}}{H}=\frac{L_{2}}{2.1}\\\\\therefore L_{2}=2.1\times tan(19.04)=0.725m[/tex]

Thus distance of spot on the pool bed from his feet equals [tex]2.7+0.725=3.425m[/tex]

Answer:

a) 0.2399 mi³

b) 440.8 × 10³ Pounds

Explanation:

Given:

Volume of cumulus cloud, V = 1 km³

Liquid water content = 0.2 g/m³

Now,

a) 1 km = [tex]\frac{\textup{1 miles}}{\textup{1.6093}}[/tex]

thus,

1 km³ = [tex](\frac{\textup{1 miles}}{\textup{1.6093}})^3[/tex]

1 km³ =  0.2399 mi³

Hence, volume of cloud in cubic miles is 0.2399 mi³

b)

Liquid water content = 0.2 g/m³

Now,

1 Km = 1000 m

thus,

1 km³ = 1000³ m³

Therefore,

Liquid water content in 1 Km³ of cloud = 0.2 g/m³ × 1000³ m³

= 200 × 10⁶ gram

or

= 200 × 10³ Kg

also,

1 kilogram =  2.204 pounds

Therefore,

200 × 10³ Kg = 200 × 10³ × 2.204 pounds = 440.8 × 10³ Pounds

The volume of a cumulus cloud occupying one cubic kilometer is equal to 0.386102 cubic miles. The water inside this cloud, with a liquid water content of 0.2 g/m³, weighs approximately 441 pounds.

The subject of this problem is to find the volume of a cloud in a different unit of measure and the weight of the water within it. Volume conversion from cubic kilometers to cubic miles and weight calculation of water in pounds are required.

To find the volume of the cloud in cubic miles, we use the conversion factor that 1 cubic kilometer is equal to about 0.386102 cubic miles. Hence, the volume of 1 cubic kilometer in cubic miles is:

1 cubic kilometer * 0.386102 cubic miles/cubic kilometer = 0.386102 cubic miles.

Next, we calculate the weight of the water in pounds, considering that the liquid water content is 0.2 grams per cubic meter and there are 1,000,000 cubic meters in a cubic kilometer:

0.2 g/m³ * 1,000,000 m³/km³ = 200,000 grams of water in the cloud.

Now, convert grams to pounds using the conversion factor 453.59237 grams per pound:

200,000 grams * (1 pound / 453.59237 grams) = 441 pounds.

Therefore, the water weight in the cloud is 441 pounds.

Answer:

The gravitational force is 3.509*10^17 times larger than the electrostatic force.

Explanation:

The Newton's law of universal gravitation and Coulombs law are:

[tex]F_{N}=G m_{1}m_{2}/r^{2}\\F_{C}=k q_{1}q_{2}/r^{2}[/tex]

Where:

G= 6.674×10^−11 N · (m/kg)2

k =  8.987×10^9 N·m2/C2

We can obtain the ratio of these forces dividing them:

[tex]\frac{F_{N}}{F_{C}}=\frac{Gm_{1}m_{2}}{kq_{1}q_{2}}=0.742\times10^{-20}\frac{C^{2}}{kg^{2}}\frac{m_{1}m_{2}}{q_{1}q_{2}}[/tex]   --- (1)

The mass of the moon is 7.347 × 10^22 kilograms

The mass of the earth is  5.972 × 10^24 kg

And q1=q2=Na*e=(6.022*10^23)*(1.6*10^-19)C=9.635*10^4 C

Replacing these values in eq1:

[tex]\frac{F_{N}}{F_{C}}}}=0.742\times10^{-20}\frac{C^{2}}{kg^{2}}\frac{7.347\times5.972\times10^{46}kg^{2}}{(9.635\times10^{4})^{2}}[/tex]

Therefore

[tex]\frac{F_{N}}{F_{C}}}}=3.509\times10^{17}[/tex]

This means that the gravitational force is 3.509*10^17 times larger than the electrostatic force, when comparing the earth-moon gravitational field vs 1mol electrons - 1mol protons electrostatic field

Answer:

The answer is [tex] 3.994 \times 10^{13}\ electrons[/tex]

Explanation:

The amount of negative charge that must be removed is

[tex]\Delta = Final\ charge - initial\ Charge = 2.6 - (-3.8) = 6.4 \ \mu C = 6.4 \times 10^{-6}\ C[/tex]

and the charge of one electron is

[tex]1 e = 1.60217662\times 10^{-19} \ C[/tex]

So the amount of electrons we need to remove is

[tex]x = \frac{6.4 \times 10^{-6}}{1.60217662\times 10^{-19}} \approx 3.994 \times 10^{13}\ electrons[/tex]

Answer:

Explanation:

The pressure of the gas can be found out as follows

Gas law formula is as follows

PV = nRT

P = nRT / V

= 32.4 X 8.31 X ( 273 + 459.38 ) / 51 X 10⁻⁶

3866.45 X 10⁶ Pa.

The first change is isobaric therefore

V₁ / T₁ = V₂ / T₂

V₂ = V₁ X T₂/ T₁

= 51 X 10⁻⁶ X ( 855.6 +273) / (459.38 +273)

= 78 X 10⁻⁶

78 cm³

After the first operation , the pressure of the gas remains at

= 3866.45 X 10⁶ Pa.

Now volume of the gas changes from 78 cm³ to 25.3 cm³ isothermally so

P₁V₁ = P₂V₂

P₂ = P₁V₁ / V₂

= 3866.45 X 10⁶ X 78 / 25.3

= 11920 X 10⁶ . Pa

Answer:

[tex]1.189eV[/tex]

Explanation:

The electric potential energy is the potential energy that results from the Coulomb force and is associated with the configuration of two or more charges. For an electron in the presence of an electric field produced by a proton, the electric potential energy is defined as:

[tex]U=\frac{kq_{e}q_{p}}{r}[/tex]

where  [tex]q_{e}[/tex] is the electron charge, [tex]q_{p}[/tex] is the proton charge, r is the separation distance between the charges and k is the coulomb constant.

Knowing this, we can calculate how much electric potential energy was lost:

[tex]\Delta U=U_{f}-U{i}\\\Delta U=\frac{kq_{e}q_{p}}{r_{f}}-\frac{kq_{e}q_{p}}{r_{i}}\\\Delta U=kq_{e}q_{p}(\frac{1}{r_{f}}-\frac{1}{r_{i}})\\\Delta U=(8.99*10^9\frac{Nm^2}{C^2})(-1.60*10^{-19}C)(1.60*10^{-19}C)(\frac{1}{0.105*10^{-9}m}-\frac{1}{0.115*10^{-9}m})\\\Delta U=1.90*10^{-19}J*\frac{6.2415*10^{18}eV}{1J}=1.189eV[/tex]

The change in electric potential energy between an electron and a proton with changing separation can be calculated using the Coulomb potential energy equation and can be expressed in electron volts (eV).

The question involves calculating the change in electric potential energy of a system consisting of an electron and a proton as they change separation distance during a reaction. The energy of the system is described by the Coulomb electrostatic potential energy equation Epot = -e2 / (4πε₀r), where e is the charge of the electron, ε₀ is the vacuum permittivity, and r is the separation distance between the electron and proton.

To find the amount of potential energy lost, we calculate the potential energy at the initial separation (0.115 nm) and the final separation (0.105 nm), then take the difference between these values. The energy will be expressed in electron volts (eV), a common unit used to describe energy at subatomic scales.

Answer:

a) 138.6 m/s

b) 762.3 m

c) 122.3 m/s

d) 24.47

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

[tex]v=u+at\\\Rightarrow v=0+12.6\times 11\\\Rightarrow v=138.6 \ m/s[/tex]

Velocity at the end of its upward acceleration is 138.6 m/s

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}\times 12.6\times 11^2\\\Rightarrow s=762.3\ m[/tex]

Maximum height the rocket reaches is 762.3 m

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as-u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 762.3-0^2}\\\Rightarrow v=122.3\ m/s[/tex]

The velocity with which the rocket crashes to the Earth is 122.3 m/s

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 762.3=0\times t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{762.3\times 2}{9.81}}\\\Rightarrow t=12.47\ s[/tex]

Total time from launch to crash is 12.47+11 = 24.47 seconds

Final answer:

The rocket's velocity at the end of its upward acceleration is 138.6 m/s. The maximum height it reaches is 1353.2 m. The rocket crashes to Earth with a velocity of -1.0 m/s.

Explanation:

(a)  To find the velocity at the end of the rocket's upward acceleration, we can use the formula:

v = u + at

Where:
u = initial velocity = 0 m/s (since the rocket starts from rest)
a = acceleration = 12.6 m/s2
t = time = 11.0 s

Substituting the values, we get:

v = 0 + 12.6 * 11.0 = 138.6 m/s

Therefore, the velocity at the end of its upward acceleration is 138.6 m/s.

(b)  To find the maximum height the rocket reaches, we can use the second equation of motion:

s = ut + 0.5at2

Where:
s = distance
u = initial velocity
a = acceleration
t = time

In this case, we need to consider the time taken for both the upward acceleration and free fall.

For the upward acceleration:

u = 0 m/s (since the rocket starts from rest)
a = 12.6 m/s2
t = 11.0 s

Substituting the values, we get:

s1 = 0 * 11.0 + 0.5 * 12.6 * (11.0)2 = 388.5 m

For the free fall:

u = 138.6 m/s (velocity at the end of the upward acceleration)
a = -9.8 m/s2 (acceleration due to gravity)
t = ?

To find the time for free fall, we can use the equation:

u = at

Substituting the values, we get:

138.6 = -9.8t

Solving for t, we get:

t = -14.1 s

However, time cannot be negative in this case. So, we take the absolute value of t:

t = 14.1 s

Substituting the values in the equation for free fall distance, we get:

s2 = 138.6 * 14.1 + 0.5 * (-9.8) * (14.1)2 = 964.7 m

The maximum height reached by the rocket is s1 + s2 = 388.5 m + 964.7 m = 1353.2 m.

(c)  To find the velocity at which the rocket crashes to Earth, we again consider the free fall phase. Using the equation:

v = u + at

Where:
u = 138.6 m/s (velocity at the end of the upward acceleration)
a = -9.8 m/s2 (acceleration due to gravity)
t = 14.1 s

Substituting the values, we get:

v = 138.6 - 9.8 * 14.1 = -1.0 m/s

The velocity at which the rocket crashes to Earth is -1.0 m/s. The negative sign indicates that the velocity is directed downward.

(d)  The total time from launch to crash is the sum of the time for upward acceleration (11.0 s) and the absolute value of the time for free fall (14.1 s). Therefore, the total time is 11.0 s + 14.1 s = 25.1 s.

The electron can make angles of 118° and 62° with the magnetic field.

An electron moving at a speed of 4.00 × 10³ m/s in a 1.25-T magnetic field experiences a magnetic force of 1.40 × 10-16 N. What angle does the velocity of the electron make with the magnetic field?

One possible angle is 118°, and the other possible angle is 62° with the magnetic field.

The correct answer is [tex]\(\boxed{89.9999999999999\°}\).[/tex]

To determine the angle between the electron's velocity and the magnetic field, we can use the formula for the magnitude of the magnetic force on a moving charge, which is given by the Lorentz force law:

[tex]\[ F = qvB \sin(\theta) \][/tex]

where:

- [tex]\( F \)[/tex] is the magnitude of the magnetic force,

- [tex]\( q \)[/tex] is the charge of the electron,

- [tex]\( v \)[/tex] is the speed of the electron,

- [tex]\( B \)[/tex] is the magnitude of the magnetic field, and

- [tex]\( \theta \)[/tex] is the angle between the electron's velocity and the magnetic field.

The charge of an electron is [tex]\( 1.60 \times 10^{-19} \)[/tex] Coulombs (C), and the mass of an electron is approximately [tex]\( 9.11 \times 10^{-31} \)[/tex] kilograms (kg). Using Newton's second law, [tex]\( F = ma \)[/tex], where [tex]\( m \)[/tex] is the mass of the electron and [tex]\( a \)[/tex] is its acceleration, we can equate the magnetic force to the mass times acceleration:

[tex]\[ qvB \sin(\theta) = ma \][/tex]

Given that the electron experiences only a magnetic force, we can solve for [tex]\( \sin(\theta) \)[/tex]:

[tex]\[ \sin(\theta) = \frac{ma}{qvB} \][/tex]

Plugging in the given values:

[tex]\[ \sin(\theta) = \frac{(9.11 \times 10^{-31} \text{ kg})(2.30 \times 10^{14} \text{ m/s}^2)}{(1.60 \times 10^{-19} \text{ C})(7.69 \times 10^{6} \text{ m/s})(9.21 \times 10^{-4} \text{ T})} \][/tex]

[tex]\[ \sin(\theta) = \frac{(9.11 \times 2.30)}{(1.60 \times 7.69 \times 9.21)} \times 10^{-31 + 14 - (-19) - 6 - (-4)} \][/tex]

[tex]\[ \sin(\theta) = \frac{(20.953)}{(119.9424)} \times 10^{-31 + 14 + 19 - 6 + 4} \][/tex]

[tex]\[ \sin(\theta) = 0.1747 \times 10^{-31 + 14 + 19 - 6 + 4} \][/tex]

[tex]\[ \sin(\theta) = 0.1747 \times 10^{-10} \][/tex]

[tex]\[ \sin(\theta) = 1.747 \times 10^{-11} \][/tex]

Now, we can find the angle [tex]\( \theta \)[/tex] by taking the inverse sine (arcsin) of [tex]\( \sin(\theta) \)[/tex]:

[tex]\[ \theta = \arcsin(1.747 \times 10^{-11}) \][/tex]

Since the value of [tex]\( \sin(\theta) \)[/tex] is extremely small, the angle [tex]\( \theta \)[/tex] will be very close to 0 degrees. However, because the electron is experiencing an acceleration, [tex]\( \theta \)[/tex] must be slightly greater than 0 degrees. Using a calculator, we find:

[tex]\[ \theta \approx \boxed{89.9999999999999\°} \][/tex]

This result indicates that the electron's velocity is nearly parallel to the magnetic field, with an angle that is almost 90 degrees but infinitesimally less.