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3 years ago

A person speed walking down a street covers 20m in 5s. What is the speed of the individual? *

Physics

1 answer:

omeli [17]3 years ago

8 0

Answer:

4 m/s.

Explanation:

The following data were obtained from the question:

Distance travalled (d) = 20 m

Time (t) = 5 secs

Speed (S) =?

Speed is defined as the rate of change of distance moved with time. Mathematically, it is expressed as:

Speed (S) = Distance (d) /time (t)

S = d/t

With the above formula, we can easily calculate the speed of the individual as follow:

Distance travalled (d) = 20 m

Time (t) = 5 secs

Speed (S) =?

S = d/t

S = 20/5

S = 4 m/s

Therefore, the speed of the individual is 4 m/s

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A toy car accelerates at a constant rate from rest to a speed of 4 m/s in a time of 0.55 s. What was the magnitude of the accele

Korvikt [17]

initial velocity=u=0m/s
Final velocity=v=4m/s
Time=t=0.5s

$\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \sf\longmapsto Acceleration=\dfrac{4-0}{0.5}$

$\\ \sf\longmapsto Acceleration=\dfrac{4}{0.5}$

$\\ \sf\longmapsto Acceleration=8m/s^2$

3 0

3 years ago

The freezer compartment in a conventional refrigerator can be modeled as a rectangular cavity 0.3 m high and 0.25 m wide with a

Anon25 [30]

Answer:

Thickness of Styrofoam insulation is 0.02741 m.

Explanation:

Given that,

Height = 0.25 m

Depth = 0.5 m

Power = 400 W

Temperature = 33°C

We need to calculate the area of Styrofoam

Using formula of area

$A=2(lb+bh+hl)$

Put the value into the formula

$A=2(0.3\times0.5+0.25\times0.5+0.5\times0.3)$

$A=0.85\ m^2$

Inner surface temperature of freezer

$T_{i}=-10°C=263\ K$

Outer surface temperature of freezer

$T_{o}=33+273=306\ K$

We need to calculate the thickness of Styrofoam insulation

Using Fourier law,

$q=\dfrac{kA}{L}(T_{o}-T_{i})$

$L=\dfrac{kA}{q}(T_{o}-T_{i})$

Put the value into the formula

$L=\dfrac{0.30\times0.85}{400}(306-263)$

$L=0.02741\ m$

Hence, Thickness of Styrofoam insulation is 0.02741 m.

8 0

3 years ago

La velocidad de un tren se reduce uniformemente de 12m/s a 5m/s. Sabiendo que durante ese tiempo recorre una distancia de 100 m.

MA_775_DIABLO [31]

Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

s = 21,0 m

Por lo tanto, la distancia que recorre hasta detenerse asumiendo la misma aceleración es 21.0 m

6 0

3 years ago

If the equation of an average velocity of a sport car was given by the equation V(t) = 3t^2 -6t +24. Determine its displacement

alisha [4.7K]

Answer:

Explanation:

The displacement of an object can be found from the velocity of the object by integrating the expression for the velocity.

In this problem, the velocity of the sport car is given by the expression

$v(t)=3t^2-6t+24$

In order to find the expression for the position of the car, we integrate this expression. We find:

$x(t)=\int v(t) dt=t^3-3t^2+24t+C$

where C is an arbitrary constant.

Here we want to find the displacement after 3 seconds. The position at t = 0 is

$x(0)=0^3-0+0+C=C$

While the position after t = 3 s is

$x(3)=3^3-3(3)^2+24(3)+C=72+C$

Therefore, the displacement of the car in 3 seconds is

$d=x(3)-x(0)=72+C-C=72$

7 0

3 years ago

Una placa rectangular de metal tiene una longitud 8.430 m. y una anchura de 5.201 m. calcular el área de la placa con el número

bezimeni [28]

Responder:

43.84 m²

Explicación:

Dada la dimensión de una placa rectangular como:

Longitud (L) = 8.430 m

Ancho (W) = 5.201 m

El Área de una placa rectangular se puede obtener usando la fórmula:

Área = Largo × Ancho

Área = 8.430m * 5.201m

Superficie = 43.84443 m²

Área = 43.84 m²

3 0

3 years ago