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cestrela7 [59]
2 years ago
5

Look at the graph above. It shows how three runners ran a 100 meter race

Physics
2 answers:
Rzqust [24]2 years ago
8 0

Answer:

1. Albert

2. Charlie

3. 05 Seconds

4. 14 seconds

5. 8.33 m/s

Explanation:

The problems in the given scenarios can be solved just by analyzing the given the graph. Here the x-axis of the graph shows the time in seconds and y-axis shows the distance covered.

1. We can see that Albert finished the race in 12 seconds. This is lesser in comparison with Bob (14 seconds) and Charlie (17 seconds). Thus Albert is the winner of the race.

2. We can see that when Charlie reached at 50 meters, the time is increasing while the distance is same. It implies that he stopped there for some rest.

3. This rest was of 05 seconds.

4. Bob completed the race in 14 seconds.

5. Albert covered 100 m in 12 seconds, so his speed (v) will be,

v = \frac{100}{12}  = 8.33 m/s

anastassius [24]2 years ago
7 0

Answer:

1. Albert

2. Charlie

3. 5 seconds

4. 14 seconds

5. 8.33... meters per second

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A solid object has a mass of 104 kg and a volume of 1,278 m3. What is its density?
MrMuchimi
The density is 81.4 g/m3. Before you start plugging numbers into the density formula (D=M/V), you should convert 104 kg to grams, which ends up being 104,000 grams. Then you can plug in the 104,000 grams and 1,278 m3 into the formula. When you divide the mass by the volume, you get a really long decimal, which you can round to 81.4 g/m3, or whatever place your teacher wants you to round to.
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3 years ago
Write a sentence on how these words are used in real life situations
ipn [44]

Answer:

Explanation:

You can approach an expression for the instantaneous velocity at any point on the path by taking the limit as the time interval gets smaller and smaller. Such a limiting process is called a derivative and the instantaneous velocity can be defined as.#3

For the special case of straight line motion in the x direction, the average velocity takes the form: If the beginning and ending velocities for this motion are known, and the acceleration is constant, the average velocity can also be expressed as For this special case, these expressions give the same result. Example for non-constant acceleration#1

6 0
2 years ago
A puck of mass 0.110 kg slides across ice in the positive x-direction with a kinetic friction coefficient between the ice and pu
lara [203]

Answer:

a) Ffr = -0.18 N

b) a= -1.64 m/s2

c) t = 9.2 s

d) x = 68.7 m.

e) W= -12.4 J

f) Pavg = -1.35 W

g) Pinst = -0.72 W

Explanation:

a)

  • While the puck slides across ice, the only force acting in the horizontal direction, is the force of kinetic friction.
  • This force is the horizontal component of the contact force, and opposes to the relative movement between the puck and the ice surface, causing it to slow down until it finally comes to a complete stop.
  • So, this force can be written as follows, indicating with the (-) that opposes to the movement of the object.

       F_{frk} = -\mu_{k} * F_{n} (1)

       where μk is the kinetic friction coefficient, and Fn is the normal force.

  • Since the puck is not accelerated in the vertical direction, and there are only two forces acting on it vertically (the normal force Fn, upward, and  the weight Fg, downward), we conclude that both must be equal and opposite each other:

      F_{n} = F_{g} = m*g (2)

  • We can replace (2) in (1), and substituting μk by its value, to find the value of the kinetic friction force, as follows:

       F_{frk} = -\mu_{k} * F_{n} = -0.167*9.8m/s2*0.11kg = -0.18 N (3)

b)

  • According Newton's 2nd Law, the net force acting on the object is equal to its mass times the acceleration.
  • In this case, this net force is the friction force which we have already found in a).
  • Since mass is an scalar, the acceleration must have the same direction as the force, i.e., points to the left.
  • We can write the expression for a as follows:

        a= \frac{F_{frk}}{m} = \frac{-0.18N}{0.11kg} = -1.64 m/s2  (4)

c)

  • Applying the definition of acceleration, choosing t₀ =0, and that the puck comes to rest, so vf=0, we can write the following equation:

        a = \frac{-v_{o} }{t} (5)

  • Replacing by the values of v₀ = 15 m/s, and a = -1.64 m/s2, we can solve for t, as follows:

       t =\frac{-15m/s}{-1.64m/s2} = 9.2 s (6)

d)

  • From (1), (2), and (3) we can conclude that the friction force is constant, which it means that the acceleration is constant too.
  • So, we can use the following kinematic equation in order to find the displacement before coming to rest:

        v_{f} ^{2} - v_{o} ^{2} = 2*a*\Delta x  (7)

  • Since the puck comes to a stop, vf =0.
  • Replacing in (7) the values of v₀ = 15 m/s, and a= -1.64 m/s2, we can solve for the displacement Δx, as follows:

       \Delta x  = \frac{-v_{o}^{2}}{2*a} =\frac{-(15.0m/s)^{2}}{2*(-1.64m/s2} = 68.7 m  (8)

e)

  • The total work done by the friction force on the object , can be obtained in several ways.
  • One of them is just applying the work-energy theorem, that says that the net work done on the object is equal to the change in the kinetic energy of the same object.
  • Since the final kinetic energy is zero (the object stops), the total work done by friction (which is the only force that does work, because the weight and the normal force are perpendicular to the displacement) can be written as follows:

W_{frk} = \Delta K = K_{f} -K_{o} = 0 -\frac{1}{2}*m*v_{o}^{2} =-0.5*0.11*(15.0m/s)^{2}   = -12.4 J  (9)

f)

  • By definition, the average power is the rate of change of the energy delivered to an object (in J) with respect to time.
  • P_{Avg} = \frac{\Delta E}{\Delta t}  (10)
  • If we choose t₀=0, replacing (9) as ΔE, and (6) as Δt, and we can write the following equation:

       P_{Avg} = \frac{\Delta E}{\Delta t} = \frac{-12.4J}{9.2s} = -1.35 W (11)

g)

  • The instantaneous power can be deducted from (10) as W= F*Δx, so we can write P= F*(Δx/Δt) = F*v (dot product)
  • Since F is constant, the instantaneous power when v=4.0 m/s, can be written as follows:

       P_{inst} =- 0.18 N * 4.0m/s = -0.72 W (12)

7 0
3 years ago
1) A force of 20 Newton acts on a bar having a cross sectional area of 0.8m^2 and length 10cm.calculate the stress developed in
Elanso [62]

Answer:25N/M^2

Explanation:

Force=20N Area=0.8M^2

Stress=force/area

Stress=20/0.8

Stress=25N/M^2

4 0
3 years ago
Please help!!
IgorLugansk [536]

Answer:

Si un objeto se mueve en relación a un marco de referencia (por ejemplo, si una profesora se mueve a la derecha con respecto al pizarrón, o un pasajero se mueve hacia la parte trasera de un avión), entonces la posición del objeto cambia. A este cambio en la posición se le conoce como desplazamiento. La palabra desplazamiento implica que un objeto se movió, o se desplazó.

Explanation:

El desplazamiento se define como el cambio en la posición de un objeto. Se puede definir de manera matemática con la siguiente ecuación:

\text{desplazamiento}=\Delta x=x_f-x_0desplazamiento=Δx=x  

f

​  

−x  

0

​  

start text, d, e, s, p, l, a, z, a, m, i, e, n, t, o, end text, equals, delta, x, equals, x, start subscript, f, end subscript, minus, x, start subscript, 0, end subscript

x_fx  

f

​  

x, start subscript, f, end subscript se refiere al valor de la posición final.

x_0x  

0

​  

x, start subscript, 0, end subscript se refiere al valor de la posición inicial.

\Delta xΔxdelta, x es el símbolo que se usa para representar el desplazamiento.

Debemos ser cuidados al usar la palabra distancia, ya que hay dos maneras de usar el término en física. Podemos hablar acerca de la distancia entre dos puntos, o podemos hablar de la distancia recorrida por un objeto.

La distancia se define como la magnitud o el tamaño del desplazamiento entre dos posiciones. Observa que la distancia entre dos posiciones no es la misma que la distancia recorrida entre ellas.

Es importante darse cuenta que la distancia recorrida no tiene que ser igual a la magnitud del desplazamiento (es decir, la distancia entre dos puntos). De manera específica, si un objeto cambia de dirección en su trayecto, la distancia total recorrida será mayor que la magnitud del desplazamiento entre esos dos puntos. Ve los ejemplos resueltos a continuación.

8 0
2 years ago
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