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?

The value of electric potential at point B which is 2 meters directly towards the north from point A is 600 V.

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.

Answer:

Explanation:

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.

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

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.

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.

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]

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.

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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