Answer:
Part A:
[tex]d_{min} = \frac{v_0^2}{2g\mu_s}[/tex]
Part B:
[tex]d_{min} = 6.37~m[/tex]
Explanation:
Part A:
We should determine the free-body diagram of the small box.
For the first box, the only force exerted to the box is the static friction force in the direction of the motion.
(The direction of the static friction is always confusing to the students. The wrong idea is that the static friction is in the opposite direction with the motion. However, if you look at the Newton's Second Law, it states that the net force acting on an object is equal to the mass times acceleration. And in this case acceleration of the total system is equal to that of the small box, since it sits on the larger box.)
We can use the equations of kinematics to find the minimum distance to stop without the small box slipping.
[tex]v^2 = v_0^2 + 2a(\Delta x)\\0 = v_0^2 + 2ad_{min}[/tex]
The acceleration can be found by Newton's Second Law:
[tex]F = ma\\mg\mu_s = m(-a)\\a = -g\mu_s[/tex]
The negative sign comes from the fact that in order for the boxes to stop they have to apply a negative acceleration.
Now, we can combine the two equations to find the distance x:
[tex]0 = v_0^2 + 2ad_{min} = v_0^2 + 2(-g\mu_s)d_{min}\\d_{min} = \frac{v_0^2}{2g\mu_s}[/tex]
Part B:
We can apply the above formula to the truck and file cabinet.
[tex]d_{min} = \frac{v_0^2}{2g\mu_s} = \frac{10^2}{2(9.8)(0.80)} = 6.37~m[/tex]
The expression for the shortest distance which the large box can stop without the small box slipping is [tex]d_{min} = \frac{v_0^2 }{2\mu_s g}[/tex]
The shortest distance in which the pickup can stop without the file cabinet sliding is 6.38 m.
The given parameters;
mass of the bigger box, = Mspeed of the bigger box, = v0mass of the small box, = mcoefficient of static friction, = μscoefficient of kinetic friction, = μkThe expression for the shortest distance which the large box can stop without the small box slipping is calculated as follows;
Apply work-energy theorem;
[tex]\mu_s F \times d_{min} = \frac{1}{2} Mv_0^2\\\\\mu_s (Mg)d_{min} = \frac{1}{2} Mv_0^2\\\\\mu_s gd_{min} = \frac{v_0^2}{2} \\\\d_{min} = \frac{v_0^2}{2\mu_s g}\\\\[/tex]
At the given speed and coefficient of static friction, the shortest distance in which the pickup can stop without the file cabinet sliding is calculated as;
[tex]d_{min} = \frac{v^2}{2\mu_s g} = \frac{10^2 }{2\times 0.8 \times 9.8} = 6.38 \ m[/tex]
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Honeybees acquire a charge while flying due to friction with the air. A 120 mg bee with a charge of + 23 pC experiences an electric force in the earth's electric field, which is typically 100 N/C, directed downward.What is the ratio of the electric force on the bee to the bee's weight?F/W = ______What electric field strength would allow the bee to hang suspended in the air?E= _______What electric field direction would allow the bee to hang suspended in the air?Upward? or downward?
Answer:
[tex] \frac{F}{W}=1.95\times10^{-6}[/tex][tex] 51304447 \frac{N}{C} [/tex]UpwardExplanation:
The weight of the bee is:
[tex] W=mg=(120\times10^{-6}kg)(9.81\frac{m}{s^{2}})=1.18\times10^{-3}N[/tex]
with m the mass and g the gravity acceleration.
Electric force of the bee is related with the electric field of earth by:
[tex]F_{e}=qE=(23\times10^{-12}C)(100\frac{N}{C})=2.3\times10^{-9} [/tex]
with q the charge, E the electric field and Fe the electric force.
So:
[tex] \frac{F}{W}=\frac{2.3\times10^{-9}}{1.18\times10^{-3}}=1.95\times10^{-6}[/tex]
Because Newton's first law we should make the net force on it equals cero:
[tex] F+F_e+W=0[/tex]
[tex]F=-(F_e+W)=-(2.3\times10^{-9} +1.18\times10^{-3})=-1.1800023\times10^{-3} [/tex]
with W the weight, Fe the electric force on the bee due earth's electric field and F the force.
So, the applied electric field should be:
[tex]E_a=\frac{-1.1800023\times10^{-3}}{23\times10^{-12}}=-51304447 \frac{N}{C} [/tex]
The negative sign indicates that the electric field should be opposite to earth's electric field, so it should be upward.
Final answer:
The ratio of the electric force on the bee to the bee's weight is approximately 0.002. An electric field strength of about 51.1 N/C directed upward would allow the bee to hang suspended in the air.
Explanation:
To find the ratio of the electric force on the bee to the bee's weight (F/W), we first need to calculate both forces. The electric force (F) can be calculated using the equation F = qE, where q is the charge in coulombs (C) and E is the electric field in newtons per coulomb (N/C). The weight (W) of the bee can be calculated using W = mg, where m is the mass in kilograms (kg) and g is the acceleration due to gravity, approximately 9.8 m/s².
Given, q = 23 pC (23 × 10⁻¹² C) and E = 100 N/C. Thus, F = 23 × 10⁻¹² C × 100 N/C = 2.3 × 10⁻¹⁴ N. The mass of the bee is 120 mg (0.120 g or 0.000120 kg), so the weight W = 0.000120 kg × 9.8 m/s² = 1.176 × 10⁻³ N. Hence, the ratio F/W = (2.3 × 10⁻¹⁴ N) / (1.176 × 10⁻³ N) ≈ 0.002.
To allow the bee to hang suspended in the air, the upward electric force must equal the downward force of gravity. Therefore, the required electric field strength (E) can be found by rearranging F = qE to E = W/q. Substituting the values gives E = (1.176 × 10⁻³ N) / (23 × 10⁻¹² C) ≈ 51.1 N/C directed upward.
Cathode ray tubes (CRTs) used in old-style televisions have been replaced by modern LCD and LED screens. Part of the CRT included a set of accelerating plates separated by a distance of about 1.54 cm. If the potential difference across the plates was 27.0 kV, find the magnitude of the electric field (in V/m) in the region between the plates. HINT
Answer:
1753246.75325 V/m
Explanation:
d = Distance of separation = 1.54 cm
V = Potential difference = 27 kV
When the voltage is divided by the distance between the plates we get the electric field.
Electric field is given by
[tex]E=\dfrac{V}{d}\\\Rightarrow E=\dfrac{27\times 10^3}{1.54\times 10^{-2}}\\\Rightarrow E=1753246.75325\ V/m[/tex]
The magnitude of the electric field in the region between the plates is 1753246.75325 V/m
A juggler throws a bowling pin straight up with an initial speed of 8.20 m/s. How much time elapses until the bowling pin returns to the juggler’s hand?
The time it takes for the bowling pin to return to the juggler's hand is approximately [tex]\( 1.13 \, \text{s} \)[/tex].
To find the time it takes for the bowling pin to return to the juggler's hand, you can use the kinematic equation for vertical motion under constant acceleration. The equation is:
[tex]\[ h = v_0 t - \frac{1}{2}gt^2 \][/tex]
Where:
- [tex]\( h \)[/tex] is the displacement (in this case, the height the bowling pin reaches, which is zero when it returns to the hand),
- [tex]\( v_0 \)[/tex] is the initial velocity,
- [tex]\( t \)[/tex] is the time,
- [tex]\( g \)[/tex] is the acceleration due to gravity (approximately [tex]\( 9.8 \, \text{m/s}^2 \))[/tex].
In this case, the final height [tex](\( h \))[/tex] is zero because the bowling pin returns to the juggler's hand. The initial velocity [tex](\( v_0 \))[/tex] is given as [tex]\( 8.20 \, \text{m/s} \)[/tex], and [tex]\( g \) is \( 9.8 \, \text{m/s}^2 \)[/tex].
Plugging in these values, the equation becomes:
[tex]\[ 0 = (8.20 \, \text{m/s}) \cdot t - \frac{1}{2}(9.8 \, \text{m/s}^2) \cdot t^2 \][/tex]
Now, you can solve this quadratic equation for [tex]\( t \)[/tex]. The general form of a quadratic equation is [tex]\( at^2 + bt + c = 0 \)[/tex], where [tex]\( a = -\frac{1}{2}(9.8 \, \text{m/s}^2) \), \( b = 8.20 \, \text{m/s} \), and \( c = 0 \)[/tex]. The solutions to this equation give you the times when the bowling pin is at the initial and final heights.
You can use the quadratic formula to solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = -\frac{1}{2}(9.8 \, \text{m/s}^2) \), \( b = 8.20 \, \text{m/s} \), and \( c = 0 \)[/tex].
[tex]\[ t = \frac{-8.20 \, \text{m/s} \pm \sqrt{(8.20 \, \text{m/s})^2 - 4 \cdot \left(-\frac{1}{2}(9.8 \, \text{m/s}^2)\right) \cdot 0}}{2 \cdot \left(-\frac{1}{2}(9.8 \, \text{m/s}^2)\right)} \][/tex]
Simplifying further:
[tex]\[ t = \frac{-8.20 \, \text{m/s} \pm \sqrt{67.24}}{-9.8} \][/tex]
Now, calculate the two possible values for [tex]\( t \)[/tex] using both the plus and minus signs:
[tex]\[ t_1 = \frac{-8.20 + \sqrt{67.24}}{-9.8} \][/tex]
[tex]\[ t_2 = \frac{-8.20 - \sqrt{67.24}}{-9.8} \][/tex]
Calculating these values:
[tex]\[ t_1 \approx 1.13 \, \text{s} \][/tex]
[tex]\[ t_2 \approx -0.58 \, \text{s} \][/tex]
Since time cannot be negative in this context, we discard the negative solution. Therefore, the time it takes for the bowling pin to return to the juggler's hand is approximately [tex]\( 1.13 \, \text{s} \)[/tex].
A gas had an initial pressure of 4.80atm in a 5.50L container. After transfering it to a 9.60L container, the gas was found to have a pressure of 2.10atm and a temperature of 25.00∘C. What was the initial temperature in degrees Celsius?
To solve this problem we will apply the concepts related to the ideal gas equations. Which defines us that the relationship between pressure, temperature and volume in the first state must be equivalent in the second state of matter. In mathematical terms this is
[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]
If we rearrange the equation to find the Temperature at state 1 we have that
[tex]T_1 = \frac{P_1V_1T_2}{P_2V_2}[/tex]
Replacing our values we have that
[tex]T_1 = \frac{(4.8*5.5*298.15)}{(2.1*9.6)}[/tex]
[tex]T_1 = 390.435K[/tex]
Therefore the temperature is 390.435K
Answer:
117 ∘C
Explanation:
Use the combined gas law.
P1V1/T1 = P2V2/T2
Let the subscript 2 represent the 9.60L of gas at 25.0∘C and the subscript 1 represent the gas at the initial volume of 5.50L.
Remember to covert the temperature from degrees Celsius to Kelvin by adding 273.15.
Therefore, we have that T2=298.15K, P2=2.10atm, V2=9.60L, P1=4.80atm, V1=5.50L, and T1 is unknown.
Rearrange the equation for T1 and substitute in the known values to solve for the initial temperature.
T1T1T1=P1V1T2P2V2=(4.80atm)(5.50L)(298.15K)(2.10atm)(9.60L)=390.434K
Now, convert this temperature from Kelvin to degrees Celsius.
T1=390.434K−273.15 = 117.28∘C
Therefore, after rounding this value to three significant figures, we find that the initial temperature is 117∘C.
In a rainstorm with a strong wind, what determines the best position in which to hold an umbrella?
Explanation:
The best way to hold a strong umbrella in rainstorm with a strong wind will be against the direction of the wind. This can provide with the maximum protection in rain. Moreover, it should be placed slightly upward also, at an angle. This will again call for maximum protection.
The best position to hold an umbrella in a rainstorm with wind is determined by the direction of the wind. You should hold your umbrella facing towards the wind's direction, including both horizontal and vertical direction, to best protect from the rain.
Explanation:In a rainstorm with a strong wind, the best position in which to hold an umbrella is largely determined by the direction of the wind. Since rain in a storm tends to fall diagonally due to wind rather than vertically, you should position your umbrella in such a way that it faces the direction from which the wind and rain are coming.
This includes both the horizontal direction (north, south, east, or west) and the vertical direction (upwards or downwards), as wind and rain can also come from above or below. For example, if the wind is blowing from the north, you should hold your umbrella to the north.
If it's also blowing downwards, you should tilt your umbrella accordingly. By doing so, you can protect yourself from the rain to the greatest extent possible.
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A light rope is attached to a block with mass 4.10 kg that rests on a frictionless, horizontal surface. The horizontal rope passes over a frictionless, massless pulley, and a block with mass m is suspended from the other end. When the blocks are released, the tension in the rope is 14.7 N.
a. Draw two free-body diagrams: one for each block.
b. What is the acceleration of either block?
c. Find m.
d. How does the tension compare to the weight of the hanging block?
Answer:
a) please find the attachment
(b) 3.65 m/s^2
c) 2.5 kg
d) 0.617 W
T<weight of the hanging block
Explanation:
a) please find the attachment
(b) Let +x be to the right and +y be upward.
The magnitude of acceleration is the same for the two blocks.
In order to calculate the acceleration for the block that is resting on the horizontal surface, we will use Newton's second law:
∑Fx=ma_x
T=m1a_x
14.7=4.10a_x
a_x= 3.65 m/s^2
c) in order to calculate m we will apply newton second law on the hanging
block
∑F=ma_y
T-W= -ma_y
T-mg= -ma_y
T=mg-ma_y
T=m(g-a_y)
a_x=a_y
14.7=m(9.8-3.65)
m = 2.5 kg
the sign of ay is -ve cause ay is in the -ve y direction and it has the same magnitude of ax
d) calculate the weight of the hanging block :
W=mg
W=2.5*9.8
=25 N
T=14.7/25
=0.617 W
T<weight of the hanging block
A bus leaves New York City, takes a non-direct route and arrives in St. Louis, Missouri 23 hours, 16 minutes later. If the distance between the two cities is 1250 km, what is the magnitude of the bus' average velocity?
Answer:
vavg = 53.7 km/h
Explanation:
In order to find the magnitude of the bus'average velocity, we need just to apply the definition of average velocity, as follows:
[tex]vavg =\frac{xf-xo}{t-to}[/tex]
where xf - xo = total displacement = 1250 Km
If we choose t₀ = 0, ⇒ t = 23h 16'= 23h + 0.27 h = 23.27 h
⇒ [tex]vavg =\frac{1250 km}{23.27h} = 53.7 Km/h[/tex]
The average velocity of the bus, calculated by dividing the total displacement (1250 km) by the total time (23.27 hours), is approximately 53.69 km/h. This refers to the magnitude of the average velocity, not the direction.
Explanation:The subject of the question is average velocity, which is defined in physics as the total displacement divided by the total time taken. The displacement in this scenario is the straight-line distance between New York City and St. Louis, Missouri, which is 1250 km. The time taken for the bus to travel this route is 23 hours and 16 minutes, or approximatively 23.27 hours when converted into decimal hours to facilitate calculation. Using the formula for average velocity (v = d / t), we substitute the given values to find the average velocity of the bus:
v = 1250 km / 23.27 hours
Upon calculating this, we find that the average velocity of the bus is approximately 53.69 km/h. please note that this is the magnitude of the average velocity, indicating the speed rather than the direction of the bus's travel.
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Two point charges are placed on the x axis. The first charge, q1 = 8.00 nC, is placed a distance 16.0 m from the origin along the positive x axis; the second charge, q2 = 6.00 nC, is placed a distance 9.00 m from the origin along the negative x axis.Calculate the electric field at point A, located at coordinates (0 mm, 12.0 mm ). Give the x and y components of the electric field as an ordered pair. Express your answer in newtons per coulomb to three significant figures. EAx, EAy =
The x and y components of the electric field at point A are [tex]EAx = 3.11 * 10^4 N/C~ and~ EAy = 0 N/C.[/tex]
The net electric field at point A due to two point charges can be found by calculating the electric field contributed by each charge independently and then summing the components to find the net electric field as an ordered pair.
Finally, we sum the x-components and sum the y-components of the electric fields from both charges to get the net electric field at point A as an ordered pair (EAx, EAy).
For q1:
r1x = 16.0 m
r1y = 0.012 m (converting 12.0 mm to meters)
[tex]r1 = sqrt(r1x^2 + r1y^2)\\r_1 = sqrt(16.0^2 + 0.012^2) \\r_1 = 16.0001 m[/tex]
For q2:
[tex]r2x = -9.0 m\\r2y = 0.012 m (same as for q1)\\r2 = sqrt(r2x^2 + r2y^2) \\r2 = sqrt((-9.0)^2 + 0.012^2)\\ r2 = 9.0001 m[/tex]
Now, we can calculate the electric field contributions from each charge:
[tex]E1x = (8.99 * 10^9) * (8 * 10^-9) / (16.0001)^2 \\E1x = 1.94 * 10^4 N/C\\E1y = 0 E2y = (8.99 * 10^9) * (6 * 10^-9) / (9.0001)^2 \\E2y= 1.17 * 10^4 N/C\\E2y = 0 E1x[/tex]
Finally, we add the x-components of the electric fields vectorially:
[tex]EAx = E1x + E2x \\= 1.94 * 10^4 + 1.17 * 10^4 \\= 3.11 * 10^4 N/C[/tex]
The y-component of the electric field, EAy, is 0 since both charges are on the x-axis.
So, the x and y components of the electric field at point A are [tex]EAx = 3.11 * 10^4 N/C~ and~ EAy = 0 N/C.[/tex]
Some region of space contains uniform electric field directed towards south with magnitude 100 V/m. At point A electric potential is 400 V. What is electric potential at point B which is 2 meters directly towards the north from A
The value of electric potential at point B which is 2 meters directly towards the north from point A is 600 V.
What is electric field?The electric field is the field, which is surrounded by the electric charged. The electric field is the electric force per unit charge.
In terms of potential difference, the electric field can be given as,
[tex]E=\dfrac{\Delta V}{d}[/tex]
Here, (ΔV) is the potential difference and (d) is the distance.
Some region of space contains uniform electric field directed towards south with magnitude 100 V/m.
At point A electric potential is 400 V. The distance of point B directly towards the north from point A is 2 meters. Thus, by the above formula,
[tex]100=\dfrac{ V_b-400}{2}\\V_b=600\rm\; V[/tex]
Hence, the value of electric potential at point B which is 2 meters directly towards the north from point A is 600 V.
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Final answer:
The electric potential at point B, which is 2 meters directly north of point A within a uniform electric field pointing south with a magnitude of 100 V/m and an initial potential at A of 400 V, is 600 V.
Explanation:
The student asks about the electric potential at a point located 2 meters to the north of another point A within a uniform electric field pointing south, with an initial potential at point A of 400 V. Since the direction towards point B is opposite the direction of the electric field, the potential will increase as we move from A to B. The electric field has a magnitude of 100 V/m, which means that for each meter we move against the electric field, the potential increases by 100 V. As point B is 2 meters north of point A, we simply calculate the potential at B as follows:
VB = VA + E × d
VB = 400 V + (100 V/m) × 2 m = 600 V
So the electric potential at point B is 600 V.
flat sheet is in the shape of a rectangle with sides of lengths 0.400 mm and 0.600 mm. The sheet is immersed in a uniform electric field of magnitude 76.7 N/CN/C that is directed at 20 ∘∘ from the plane of the sheeta- Find the magnitude of the electric flux through the sheet?
Answer:
[tex]6.29591\times 10^{-6}\ N/C^2[/tex]
Explanation:
Flux is given by
[tex]\phi=EAcos\theta[/tex]
A = Area
[tex]A=0.4\times 10^{-3}\times 0.6\times 10^{-3}[/tex]
E = Electric field = 76.7 N/C
Angle is given by
[tex]\theta=90-20\\\Rightarrow \theta=70^{\circ}[/tex]
[tex]\phi=76.7\times 0.4\times 10^{-3}\times 0.6\times 10^{-3}\times cos70\\\Rightarrow \phi=6.29591\times 10^{-6}\ N/C^2[/tex]
The flux through the sheet is [tex]6.29591\times 10^{-6}\ N/C^2[/tex]
A circular rod with a gage length of 3.2 mm and a diameter of 2 cmcm is subjected to an axial load of 57 kNkN . If the modulus of elasticity is 200 GPaGPa , what is the change in length?
To solve this problem we will apply the concepts related to the change in length given by the following relation,
[tex]\delta_l = \frac{Pl}{AE}[/tex]
Here the variables mean the following,
P = Load
l = Length
A = Area
E = Modulus of elasticity
Our values are,
[tex]l = 3.2 m[/tex]
[tex]\phi = 2cm = 0.02m[/tex]
[tex]P = 57kN = 57*10^3N[/tex]
[tex]E = 200Gpa[/tex]
We can obtain the value of the Area through the geometrical relation:
[tex]A = \frac{\pi}{4} \phi^2[/tex]
Replacing,
[tex]A = \frac{\pi}{4} (0.02)^2[/tex]
[tex]A = 3.14*10^{-4}m^2[/tex]
Using our first equation,
[tex]\delta_l = \frac{Pl}{AE}[/tex]
[tex]\delta_l = \frac{(57*10^3)(3.2)}{(3.14*10^{-2})(200*10^9)}[/tex]
[tex]\delta_l = 0.000029044m[/tex]
[tex]\delta_l = 0.029044mm[/tex]
Therefore the change in length is 0.029mm
Oscilloscopes are found in most science laboratories. Inside, they contain deflecting plates consisting of more-or-less square parallel metal sheets, typically about 2.50 cm on each side and 2.00 mm apart. In many experiments, the maximum potential across these plates is about 30.0 V.
1.For this maximum potential, what is the strength of the electric field between the plates? [V/m]
2.For this maximum potential, what magnitude of acceleration would this field produce on an electron midway between the plates? [m/s^2]
Answer:
15000 V/m
[tex]2.634467618\times 10^{15}\ m/s^2[/tex]
Explanation:
V = Voltage = 30 V
d = Separation = 2 mm
q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]
m = Mass of electron = [tex]9.11\times 10^{-31}\ kg[/tex]
Electric field is given by
[tex]E=\dfrac{V}{d}\\\Rightarrow E=\dfrac{30}{2\times 10^{-3}}\\\Rightarrow E=15000\ V/m[/tex]
The electric field between the plates is 15000 V/m
Acceleration is given by
[tex]a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times 15000}{9.11\times 10^{-31}}\\\Rightarrow a=2.634467618\times 10^{15}\ m/s^2[/tex]
The acceleration is [tex]2.634467618\times 10^{15}\ m/s^2[/tex]
Your car's blinker has a period of 0.85 s and at the moment is in phase with a faster blinker on the car in front of you. They drift out of phase but then get back in phase after 16 s. What is the period of the other car's blinker in s?
Answer:
The time period of the other car's blinker is 0.807
Solution:
As per the question:
Time period of the car blinker, T = 0.85 s
Time taken by the blinkers to get back in phase, t = 16 s
Now,
To find the time period of the other car's blinker:
No. of oscillations, [tex]n = \frac{t}{T}[/tex]
Thus for the blinker:
[tex]n = \frac{16}{0.85}[/tex]
Now,
For the other car's blinker with time period, T':
[tex]n' = \frac{16}{T'}[/tex]
Time taken to get back in phase is t = 16 s:
n' - n = 1
[tex]\frac{16}{T'} - \frac{16}{0.85} = 1[/tex]
[tex]\frac{1}{T'} = \frac{1}{16} + \frac{1}{0.85}[/tex]
[tex]\frac{1}{T'} = 1.2389[/tex]
[tex]T' = \frac{1}{1.2389} = 0.807[/tex]
The period of the other car's blinker is approximately 1.06 s.
Explanation:To solve this problem, we need to understand the concept of phase and period. The period is the time it takes for a complete cycle of a periodic motion. In this case, the period of your car's blinker is given as 0.85 s. The phase refers to the position within a cycle at a given time. If your blinker is in phase with the other car's blinker initially, it means they are both starting their cycles at the same time.
However, they drift out of phase and then get back in phase after 16 s. This means that the other car's blinker completes a whole number of cycles in 16 s. Let's call the period of the other car's blinker T. So, in 16 s, the other car's blinker completes 16/T cycles. We know that the two cars get back in phase after 16 s, which means they complete the same number of cycles in that time.
Therefore, we can set up the following equation: 16/T = 16/0.85. Solving for T, we find that the period of the other car's blinker is approximately 1.06 s.