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bezimeni [28]
3 years ago
11

On the way to school, Jed traveled 100m north, 300m east, 100 north, 100m east, and 100m north. A.) Find the total distance trav

elled by Jed. B.) Determine the displacement by jed.​
Physics
1 answer:
Oduvanchick [21]3 years ago
5 0

Answer:

Total distance = 700 m

Displacement = 500 m

Explanation:

Notice that Jed travelled a total of 3 x 100 m = 300 m in the North direction, and 300 m + 100 m = 400 m in the East direction. Therefore the total distance he travelled is:  300 + 400 = 700 m.

But the actual displacement is given by the Pythagorean theorem as the hypotenuse of a right angle triangle of legs 300 m and 400 m:

displacement = \sqrt{300^2+400^2} =\sqrt{250000} =500\,m

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The magnetic strips on credit cards can become demagnetized. What type of magnet are they?
Damm [24]
I think you're fishing for "temporary magnet" or something like that,
but I don't agree with it. 

Credit card strips, refrigerator magnets, recording tape, bar magnets,
and big heavy horseshoe magnets are permanent magnets ... you don't
have to keep an electric current circulating around them to make them
magnetic. 

But that doesn't mean that they stay magnetic no matter WHAT you do
to them.  They can be DEmagnetized by being heated, dropped on the
floor, hit with a hammer, or in the presence of another, stronger magnet. 
4 0
3 years ago
Read 2 more answers
The volume of a cylindrical tin can with a top and a bottom is to be 16π cubic inches. If a minimum amount of tin is to be used
sergiy2304 [10]

Answer:

h = 4 in

Explanation:

GIVEN DATA:

volume of tin= 16 \pi

we know that

volume of cylinder is v = \pi r^2  h

so,

16 \pi = \pi r^2 h

16 = r^2 h

r = \sqrt{\frac{16}{h}}

construct formula for surface area

S = 2\pi r^2 +  2\pi rh

S = \frac{2v}{h} + 2 \sqrt{v \pi h}

minimize the function wrt  h

S' = \frac{2v}{h^2} + \sqrt{\frac{v \pi}{h} =  0

solving for h we have

h = [\frac{4 v}{\pi}]^{1/3}

we kow v = 16 \pi   so

h = 4 in

8 0
3 years ago
A rope passes over a fixed sheave with both ends hanging straight down. The coefficient of friction between the rope and sheave
Oliga [24]

Answer:3.51

Explanation:

Given

Coefficient of Friction \mu =0.4

Consider a small element at an angle \theta having an angle of d\theta

Normal Force=T\times \frac{d\theta }{2}+(T+dT)\cdot \frac{d\theta }{2}

N=T\cdot d\theta

Friction f=\mu \times Normal\ Reaction

f=\mu \cdot N

and T+dT-T=f=\mu Td\theta

dT=\mu Td\theta

\frac{dT}{T}=\mu d\theta

\int_{T_2}^{T_1}\frac{dT}{T}=\int_{0}^{\pi }\mu d\theta

\frac{T_2}{T_1}=e^{\mu \pi}

\frac{T_2}{T_1}=e^{0.4\times \pi }

\frac{T_2}{T_1}==e^{1.256}

\frac{T_2}{T_1}=3.51

7 0
3 years ago
Un puente de acero de 100 m de largo a 8° C aumenta su temperatura a 24°C ¿Cuánto medirá su longitud? Valor del coeficiente de d
BartSMP [9]

La longitud <em>final</em> del puente de acero es 100.018 metros.

Asumamos que la dilatación <em>térmica</em> experimentada por el puente de acero es <em>pequeña</em>, de modo que podemos emplear la siguiente aproximación <em>lineal</em> para determinar la longitud <em>final</em> del puente de acero (L), en metros:

L = L_{o}\cdot [1+\alpha\cdot (T_{f}-T_{o})] (1)

Donde:

  • L_{o} - Longitud inicial del puente, en metros.
  • \alpha - Coeficiente de dilatación, sin unidad.
  • T_{o} - Temperatura inicial, en grados Celsius.
  • T_{f} - Temperatura final, en grados Celsius.

Si tenemos que L_{o} = 100\,m, \alpha = 11.5\times 10^{-6}, T_{o} = 8\,^{\circ}C y T_{f} = 24\,^{\circ}C, entonces la longitud final del puente de acero es:

L = (100\,m)\cdot [1+(11.5\times 10^{-6})\cdot (24\,^{\circ}C - 8\,^{\circ}C)]

L = 100.018\,m

La longitud <em>final</em> del puente de acero es 100.018 metros.

Para aprender más sobre dilatación térmica, invitamos cordialmente a ver esta pregunta verificada: brainly.com/question/24953416

5 0
2 years ago
What is the number of electrons that move past a point in a wire carrying 500 A of current in 4.0 minutes
mr Goodwill [35]
The current is defined as the amount of charge Q that passes through a given point of a wire in a time \Delta t:
I= \frac{Q}{\Delta t}
Since I=500 A and the time interval is
\Delta t=4.0 min=240 s
the charge is
Q=I \Delta t=(500 A)(240 s)=1.2 \cdot 10^5 C

One electron has a charge of q=1.6 \cdot 10^{-19}C, therefore the number of electrons that pass a point in the wire during 4 minutes is
N= \frac{Q}{q}= \frac{1.2 \cdot 10^5 C}{1.6 \cdot 10^{-19}C}=7.5 \cdot 10^{23} electrons
3 0
3 years ago
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