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.
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.
Learn more about Umbrella Positioning here:
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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.