All categories

3 years ago

George walks to a friend's house. He walks 750 meters North, then realizes he walked

Physics

1 answer:

Sauron [17]3 years ago

4 0

Answer:

I think you are looking for the velocity, so the answer is 38.5 m/s

Explanation:

Velocity is giving by displacement/time. North will be positive and South will be negative, George travels 750 meters then travels -250 meters his displacement is then 750 - 250 or 500 meters. Divide that by 13 seconds to get 38.5 m/s.

You might be interested in

3 Draw energy transfer diagrams

timofeeve [1]

Answer:

The outline of the energy transfer are;

a) Kinetic energy → Clockwork spring → Potential energy

b) Potential energy in clockwork car → Clockwork spring coil unwound → Clockwork car run

c) Chemical potential energy → Batteries in the car → Electric motors → Kinetic energy

Please find attached the drawings of the energy transfer created with MS Visio

Explanation:

The energy transfer diagrams are diagrams that can be used to indicate the part of a system where energy is stored and the form and location to which the energy is transferred

a) The energy transfer diagram for the winding up a clockwork car is given as follows;

Mechanical kinetic energy is used to wind up (turn) the clockwork car such that the kinetic energy is transformed into potential energy and stored in the wound up clockwork as follows;

Kinetic energy → Clockwork spring → Potential energy

b) Letting a wound up clockwork car run results in the conversion of mechanical potential energy into kinetic (energy due tom motion) energy as follows;

Potential energy in clockwork car → Clockwork spring coil unwound → Clockwork car run

c) The energy stored in the battery of a battery powered car is chemical potential energy. When the battery powered car runs, the chemical potential energy produces an electromotive force which is converted into kinetic energy as electric current flows from the batteries

Therefore, we have;

Chemical potential energy → Batteries in the car → Electric motors → Kinetic energy

6 0

3 years ago

A very long conducting cylinder (length L) of radius R(R<R and (b)r

mina [271]

Answer:

Explanation:

We have to find electric potential V at a distance r.

a) For r>R,

The electric field in the cylinder is given by

E.A equating it to the other electric field given by

б.A/ε₀

Here the area of cylinder is given by= 2*3.14*r*L

While for the outside, the area= 2*3.14*R*L

Equating both, we get

E= бR/rε₀

Now,

The potential difference is given as:

ΔV= -бR/rε₀ and integrating right side with respect to dr under limits r and R.

Where ΔV= V₀-V

So solving we get

V₀=V-бR/ε₀ln (r/R)

b) For r<R i.e. inside the cylinder

There will be no electric field produced as E=0

So ultimately Vin= V

c) V=0 at r= infinity.

4 0

3 years ago

The equation P^xV^yT^z= constant is Boyle law for what is the values of x,y,z

Setler79 [48]

Answer:

x = 1, y = 1 and z = 0

Explanation:

Given equation;

$P^x V^y T^z = constant$

Boyle's law states that at constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure.

Mathematically the law is written as;

$PV = constant$

From the given equation, the values of x, y and z that will match this law is calculated as follows;

$P^1 V^1 T^0 = constant\\\\P^1 = P\\\\V^1 = V\\\\T^0 = 1\\\\P \times V \times 1 = PV = constant\\\\Thus, x = 1, \ y = 1 \ \ and \ z = 0$

6 0

3 years ago

Two taut strings of identical mass and length are stretched with their ends fixed, but the tension in one string is 1.10 times g

ollegr [7]

Answer:

The beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

Explanation:

Given;

velocity of wave on the string with lower tension, v₁ = 35.2 m/s

the fundamental frequency of the string, F₁ = 258 Hz

velocity of wave on the string with greater tension;

$v_1 = \sqrt{\frac{T_1}{\mu }$

where;

v₁ is the velocity of wave on the string with lower tension

T₁ is tension on the string

μ is mass per unit length

$v_1 = \sqrt{\frac{T_1}{\mu} } \\\\v_1^2 = \frac{T_1}{\mu} \\\\\mu = \frac{T_1}{v_1^2} \\\\ \frac{T_1}{v_1^2} = \frac{T_2}{v_2^2}\\\\v_2^2 = \frac{T_2v_1^2}{T_1}$

Where;

T₁ lower tension

T₂ greater tension

v₁ velocity of wave in string with lower tension

v₂ velocity of wave in string with greater tension

From the given question;

T₂ = 1.1 T₁

$v_2^2 = \frac{T_2v_1^2}{T_1} \\\\v_2 = \sqrt{\frac{T_2v_1^2}{T_1}} \\\\v_2 = \sqrt{\frac{1.1T_1*(35.2)^2}{T_1}}\\\\v_2 = \sqrt{1.1(35.2)^2} = 36.92 \ m/s$

Fundamental frequency of wave on the string with greater tension;

$f = \frac{v}{2l} \\\\2l = \frac{v}{f} \\\\thus, \frac{v_1}{f_1} =\frac{v_2}{f_2} \\\\f_2 = \frac{f_1v_2}{v_1} \\\\f_2 =\frac{258*36.92}{35.2} \\\\f_2 = 270.6 \ Hz$