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Scilla [17]
3 years ago
15

A man started at his house and went to the post office, then to the pet store, and finally home again. What is the average veloc

ity if this trip took three hours?
Explain how you arrived at your answer.
Physics
1 answer:
docker41 [41]3 years ago
6 0

Answer:

0

Explanation:

Defining velocity :

Velocity is a vector which is the ratio of a person's total displacement with time.

Displacement, in simple terms refers to the distance between an individual's initial position to his final position.

Man's initial position = House

After all his navigation and points covered irrespective of the distance ;

Final position = House

Hence, we can conclude that the man's Displacement is ;

Final position - Initial position = 0

Hence,

Velocity = Displacement / time taken

Velocity = 0 / 3

Velocity = 0

You might be interested in
A 99.5 N grocery cart is pushed 12.9 m along an aisle by a shopper who exerts a constant horizontal force of 34.6 N. The acceler
Romashka [77]

1) 9.4 m/s

First of all, we can calculate the work done by the horizontal force, given by

W = Fd

where

F = 34.6 N is the magnitude of the force

d = 12.9 m is the displacement of the cart

Solving ,

W = (34.6 N)(12.9 m) = 446.3 J

According to the work-energy theorem, this is also equal to the kinetic energy gained by the cart:

W=K_f - K_i

Since the cart was initially at rest, K_i = 0, so

W=K_f = \frac{1}{2}mv^2 (1)

where

m is the of the cart

v is the final speed

The mass of the cart can be found starting from its weight, F_g = 99.5 N:

m=\frac{F_g}{g}=\frac{99.5 N}{9.8 m/s^2}=10.2 kg

So solving eq.(1) for v, we find the final speed of the cart:

v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(446.3 J)}{10.2 kg}}=9.4 m/s

2) 2.51\cdot 10^7 J

The work done on the train is given by

W = Fd

where

F is the magnitude of the force

d is the displacement of the train

In this problem,

F=4.28 \cdot 10^5 N

d=586 m

So the work done is

W=(4.28\cdot 10^5 N)(586 m)=2.51\cdot 10^7 J

3)  2.51\cdot 10^7 J

According to the work-energy theorem, the change in kinetic energy of the train is equal to the work done on it:

W=\Delta K = K_f - K_i

where

W is the work done

\Delta K is the change in kinetic energy

Therefore, the change in kinetic energy is

\Delta K = W = 2.51\cdot 10^7 J

4) 37.2 m/s

According to the work-energy theorem,

W=\Delta K = K_f - K_i

where

K_f is the final kinetic energy of the train

K_i = 0 is the initial kinetic energy of the train, which is zero since the train started from rest

Re-writing the equation,

W=K_f = \frac{1}{2}mv^2

where

m = 36300 kg is the mass of the train

v is the final speed of the train

Solving for v, we find

v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(2.51\cdot 10^7 J)}{36300 kg}}=37.2 m/s

7 0
4 years ago
A stone is dropped from the upper observation deck of a tower, 950 m above the ground. (assume g = 9.8 m/s2.) (a) find the dista
sergey [27]
<span>d = 950 m - 4.9t^2 m The distance an object moves under constant acceleration is d = 0.5at^2 where d = distance a = acceleration t = time. Since we're falling and since we're starting at 950 m above ground, the formula becomes: d = 950 m - 0.5at^2 Substituting known values, and simplifying gives us d = 950 m - 0.5*9.8 m/s^2 * t^2 d = 950 m - 4.9 m/s^2 * t^2 Since time is in seconds, we can cancel out the seconds in the units, getting d = 950 m - 4.9t^2 m</span>
8 0
3 years ago
Read 2 more answers
AYUDA!!!!!
ZanzabumX [31]

Answer:

a) El valor de la densidad es 0.79 \frac{g}{cm^{3} } o 790 \frac{g}{cm^{3} }

b) El peso especifico es 7749.9\frac{N}{m^{3} }

Explanation:

a) La densidad se define como la propiedad que tiene la materia, ya sean sólidos, líquidos o gases, para comprimirse en un espacio determinado. En otras palabras, la densidad es una magnitud que permite medir la cantidad de masa que hay en determinado volumen de una sustancia. Entonces, la expresión para el cálculo de la densidad es el cociente entre la masa de un cuerpo y el volumen que ocupa:

d=\frac{m}{V}

En este caso:

  • masa= 237 g= 0,237 kg (siendo 1000 g= 1 kg)
  • volumen= 300 cm³= 0,0003 m³ (siendo 1 cm³= 0,000001 m³)

Reemplazando:

d=\frac{237 g}{300cm^{3} } →  d=0.79 \frac{g}{cm^{3} }

d=\frac{0,237 kg}{0,0003m^{3} }→  d=790 \frac{g}{cm^{3} }

<u><em>El valor de la densidad es 0.79 </em></u>\frac{g}{cm^{3} }<u><em> o 790 </em></u>\frac{g}{cm^{3} }<u><em></em></u>

b) El peso específico es la relación existente entre el peso y el volumen que ocupa una sustancia en el espacio.

Entonces, en este caso, siendo el peso:

P= m*g= 0,237 kg* 9,81 \frac{m}{s^{2} }= 2,32497 N

el peso especifico es calculado como:

Pe=\frac{Peso}{Volumen}= \frac{2,32497N}{0,0003 m^{3} }

Pe= 7749.9\frac{N}{m^{3} }

<u><em>El peso especifico es 7749.9</em></u>\frac{N}{m^{3} }<u><em></em></u>

8 0
3 years ago
Help Please! Been stuck for some time
Naya [18.7K]

Answer:

pretty sure A is correct, as they are pushing in opposite directions, hence canceling each other

Explanation:

6 0
3 years ago
What constant horizontal force is required to drag a 5 kg block along a
BARSIC [14]

Answer:

25N

Explanation:

formula=> Frictional force= coefficient of friction x normal reaction

normal reaction is mass x gravity(10N/kg)

so Frictional force= 0.5x5x10

=25N

but that is the Frictional force not the constant force

to get the constant force, subtract the Frictional force from the weight of the block

weight= mass x gravity

= 5x10

= 50N

50N-25N

=25N

5 0
3 years ago
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