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3 years ago

What is a distraction that a pedestrian may engage in while crossing the street?

Physics

1 answer:

goldenfox [79]3 years ago

6 0

Answer:

a dog walking or their phone rings or heard a neighbor talking to them

You might be interested in

A particle with charge − 3.74 × 10 − 6 C is released at rest in a region of constant, uniform electric field. Assume that gravit

dalvyx [7]

Answer:

57.5022228905 s

Explanation:

$V_i$ = Initial voltage

$V_f$ = Final voltage = 0.401 V

$t_1$ = Initial time = 8.36 s

$t_2$ = Final time

K = Kinetic Energy = $3.18\times 10^{-8}\ J$

C = Charge = $3.74\times 10^{-6}\ C$

Voltage is given by

$V=\dfrac{K}{C}\\\Rightarrow V=\dfrac{3.17\times 10^{-8}}{3.74\times 10^{-6}}\\\Rightarrow V=0.00847593582888\ V$

We have the relation

$\dfrac{V_i}{V_f}=(\dfrac{t_i}{t_f})^2\\\Rightarrow t_f=\sqrt{\dfrac{t_i^2V_f}{V_i}}\\\Rightarrow t_f=\sqrt{\dfrac{8.36^2\times 0.401}{0.00847593582888}}\\\Rightarrow t_f=57.5022228905\ s$

The time taken is 57.5022228905 s

4 0

2 years ago

Many hot-water heating systems have a reservoir tank connected directly to the pipeline, so as to allow for expansion when the w

Andre45 [30]

Answer:

$108.306\times 10^{-6}m^3$

Explanation:

According to volume thermal expansion the expansion in volume due to temperature is given by $\Delta V=V_0\beta \Delta T$ here $\beta$ is coefficient of volume expansion

The volume of copper pipe before expansion is $V_0=\pi r^2L=3.14\times (9.20\times 10^{-3})^2\times 59.7=0.0158m^3$

Now the increase of copper pipe due to increase in temperature = $\Delta V_c=V_0\beta _c\Delta T=0.0158\times 51\times 10^{-6}\times (64.5-20.5)=35.6\times 10^{-6}m^3$

As $\beta$ for copper is $51\times 10^{-6}$

Now the increase of water due to increase in temperature = $\Delta V_w=V_0\beta _c\Delta T=0.0158\times 207\times 10^{-6}\times (64.5-20.5)=143.906\times 10^{-6}m^3$

As $\beta$ for water is $207\times 10^{-6}$

So the minimum volume of reservoir tank to hold the overflow of water = $\Delta V=\Delta V_w-\Delta V_c=143.906\times 10^{-6}-35.6\times 106{-6}=108.306\times 10^{-6}m^3$

3 0

3 years ago

In which of the following units is acceleration expressed

Fittoniya [83]

There are no correct choices on the list you provided. Any unit of acceleration is (a unit of length) divided by (a unit of time, squared).

4 0

3 years ago

An external resistor with resistance R is connected to a battery that has emf ε and internal resistance r. Let P be the electric

ELEN [110]

Answer:

a. 0 W b. ε²/R c. at R = r maximum power = ε²/4r d. For R = 2.00 Ω, P = 227.56 W. For R = 4.00 Ω, P = 256 W. For R = 6.00 Ω, P = 245.76 W

Explanation:

Here is the complete question

An external resistor with resistance R is connected to a battery that has emf ε and internal resistance r. Let P be the electrical power output of the source. By conservation of energy, P is equal to the power consumed by R. What is the value of P in the limit that R is (a) very small; (b) very large? (c) Show that the power output of the battery is a maximum when R = r . What is this maximum P in terms of ε and r? (d) A battery has ε= 64.0 V and r=4.00Ω. What is the power output of this battery when it is connected to a resistor R, for R=2.00Ω, R=4.00Ω, and R=6.00Ω? Are your results consistent with the general result that you derived in part (b)?

Solution

The power P consumed by external resistor R is P = I²R since current, I = ε/(R + r), and ε = e.m.f and r = internal resistance

P = ε²R/(R + r)²

a. when R is very small , R = 0 and P = ε²R/(R + r)² = ε² × 0/(0 + r)² = 0/r² = 0

b. When R is large, R >> r and R + r ⇒ R.

So, P = ε²R/(R + r)² = ε²R/R² = ε²/R

c. For maximum output, we differentiate P with respect to R

So dP/dR = d[ε²R/(R + r)²]/dr = -2ε²R/(R + r)³ + ε²/(R + r)². We then equate the expression to zero

dP/dR = 0

-2ε²R/(R + r)³ + ε²/(R + r)² = 0

-2ε²R/(R + r)³ = -ε²/(R + r)²

cancelling out the common variables

2R = R + r

2R - R = R = r

So for maximum power, R = r

So when R = r, P = ε²R/(R + r)² = ε²r/(r + r)² = ε²r/(2r)² = ε²/4r

d. ε = 64.0 V, r = 4.00 Ω

when R = 2.00 Ω, P = ε²R/(R + r)² = 64² × 2/(2 + 4)² = 227.56 W

when R = 4.00 Ω, P = ε²R/(R + r)² = 64² × 4/(4 + 4)² = 256 W

when R = 6.00 Ω, P = ε²R/(R + r)² = 64² × 6/(6 + 4)² = 245.76 W

The results are consistent with the results in part b

8 0

3 years ago

Problem 9: Suppose you wanted to charge an initially uncharged 85 pF capacitor through a 75 MΩ resistor to 90.0% of its final vo

Bingel [31]

Answer:

$t=14.678\times 10^{-3}s$

Explanation:

Given:

Capacitance, C = 85 pF = 85 × 10⁻¹² F

Resistance, R = 75 MΩ = 75×10⁶Ω

Charge in capacitor at any time 't' is given as:

$Q=Q_o(1-e^{-\frac{t}{RC}})$

where,

Q₀ = Maximum charge = CE

E = Initial voltage

t = time

also, Q = CV

V= Final voltage = 90% of E = 0.9E

thus, we have

$C\times 0.9E=CE(1-e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}})$

$0.9=1-e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}}$

$e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}}=1-0.9=0.1$

taking log both sides, we get

$ln(e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}})=ln(0.1)=ln(\frac{1}{10})$

$-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}=-ln(10)$

$t=75\times 10^6\ \times\ 85\times 10^{-12}\times ln{10}$

$t=14.678\times 10^{-3}s$

3 0

3 years ago