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3 years ago

A block of metal of density 3000 kg/m3 is 2m high and stands on a square base of side 0.5 m. Calculate:

Physics

1 answer:

labwork [276]3 years ago

8 0

Answer:

1) A = 0.25 m², 2) V = 0.5 m³, 3) m = 1500 kg, 4) W = 14700 N,

5) P = 58800 Pa

Explanation:

1) The area of the base is square

A = L²

A = 0.5²

A = 0.25 m²

2) The block is a parallelepiped

V = A h

V = 0.25 2

V = 0.5 m³

3) Density is defined

rho = m / V

m = rho V

m = 3000 0.5

m = 1500 kg

4) The weight of a body is

W = mg

W = 1500 9.8

W = 14700 N

5) The pressure is

P = F / A

in this case the force is equal to the weight of the body

P = 14700 / 0.25

P = 58800 Pa

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A hammer taps on the end of a 5.0-mm-long metal bar at room temperature. A microphone at the other end of the bar picks up two p

katrin2010 [14]

Answer:

810.37 m/s

Explanation:

s = Displacement = 5 m

$v_a$ = Speed of sound in air = 343 m/s

Time taken to travel the distance

$t'=\dfrac{s}{v_a}\\\Rightarrow t'=\dfrac{5}{343}\\\Rightarrow t'=0.01457\ s$

Time for the sound in metal

$t=0.01457-8.4\times 10^{-3}=0.00617\ s$

Speed of sound is given by

$v=\dfrac{s}{t}\\\Rightarrow v=\dfrac{5}{0.00617}\\\Rightarrow v=810.37277\ m/s$

The speed of sound in the metal is 810.37 m/s

6 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

2 years ago

How will a current change if the resistance of a circuit remains constant while the voltage across the circuit decreases to half

beks73 [17]

Answer:

1. The current will drop to half of its original value.

Explanation:

The problem can be solved by using Ohm's law:

$V=RI$

where

V is the voltage across the circuit

R is the resistance of the circuit

I is the current

We can rewrite it as

$I=\frac{V}{R}$

In this problem, we have:

- the resistance of the circuit remains the same: R' = R

- the voltage is decreased to half of its original value: $V'=\frac{V}{2}$

So, the new current will be

$I'=\frac{V'}{R'}=\frac{V/2}{R}=\frac{1}{2}\frac{V}{R}=\frac{I}{2}$

so, the current will drop to half of its original value.

4 0

3 years ago

A person pulls a bucket of water up from a well with a rope. Assume the initial and final speeds of the bucket are zero (Vi-Vf-0

n200080 [17]

Answer:

This a closed system because the mass of the system is conserved

The energy system that undergoes change is the Potential energy system

The energy system diagram is shown on the first uploaded image

Work done = Change in gravitational potential energy

So solving algebraically for work done would be

Work done = $m*g*h$

where m is mass

g is acceleration due to gravity

and h is the height

Work done in terms of force and distance is = mg

where m is mass of bucket and

g is acceleration due to gravity

Explanation:

a) At the start, potential and kinetic energy were zero. so, energy is zero.

As the person pulls the bucket up, the potential energy becomes mgh.

so,final energy will be consisting of only potential energy.

B) Here work done is equal to change in gravitational potential energy.

W = $\Delta P.E$

W = m*g*h

where g = 9.9 m/ $s^2$

C) Work = force * distance

mgh = force * h

force = mg

force = weight of bucket

6 0

3 years ago

How many million kilometers is one astronomical unit.

Andrews [41]

Answer:

150 million kilometres

Explanation:

The astronomical unit (symbol: au, or AU or AU) is a unit of length, roughly the distance from Earth to the Sun and equal to 150 million kilometres (93 million miles) or 8.3 light minutes.

6 0

2 years ago