Answer:
h = v₀ g / a
Explanation:
We can solve this problem using the kinematic equations. As they indicate that the air does not influence the vertical movement, we can find the time it takes for the body to reach the floor
y = [tex]v_{oy}[/tex] t - ½ g t²
The vertical start speed is zero
t² = 2t / g
The horizontal document has an acceleration, with direction opposite to the speed therefore it is negative, the expression is
x = v₀ₓ t - ½ a t²
Indicates that it reaches the same exit point x = 0
v₀ₓ t = ½ a t2
v₀ₓ = ½ a (2h / g)
v₀ₓ = v₀
h = v₀ g / a
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.
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.
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:
https://brainly.com/question/36931356
#SPJ3
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.
Learn more about electric field here;
https://brainly.com/question/14372859
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.
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]
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.
Learn more about Average Velocity here:https://brainly.com/question/1375440
#SPJ3
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