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3 years ago

A block of size 20m x 10 mx 5 m exerts a force of 30N. Calculate the

Physics

1 answer:

Orlov [11]3 years ago

3 0

Answer:

We know that force applied per unit area is called pressure.

Pressure = Force/ Area

When force is constant than pressure is inversely proportional to area.

1- Calculating the area of three face:

A1 = 20m x 10 m =200 Square meter

A2 = 10 mx 5 m = 50 Square meter

A3 = 20m x 5 m = 100 Square meter

Therefore A1 is maximum and A2 is minimum.

2- Calculate pressure:

P = F/ A1 = 30 / 200 = 0.15 Nm⁻² ( minimum pressure)

P = F / A2 = 30 / 50 = 0.6 Nm⁻² ( maximum pressure)

Hence greater the area less will be the pressure and vice versa.

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a force is applied to a box of 10.0 kg for 4.0 s. the box goes from rest to 25 m/s in that time. What is the magnitude of that f

Paul [167]

Given:

m(mass of the box)=10 Kg

t(time of impact)=4 sec

u(initial velocity)=0.(as the body is initially at rest).

v(final velocity)=25m/s

Now we know that

v=u+at

Where v is the final velocity

u is the initial velocity

a is the acceleration acting on the body

t is the time of impact

Substituting these values we get

25=0+a x 4

4a=25

a=6.25m/s^2

Now we also know that

F=mxa

F=10 x6.25

F=62.5N

8 0

3 years ago

Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

Marrrta [24]

Answer:

Solid Wire $I = 0.01237 \ A$

Stranded Wire $I_2 = 0.00978 \ A$

Solid Wire $R = 0.0149 \ \Omega$

Stranded Wire $R_1 = 0.0189 \ \Omega$

Explanation:

Considering the first question

From the question we are told that

The radius of the first wire is $r_1 = 1.53 mm = 0.0015 \ m$

The radius of each strand is $r_0 = 0.306 \ mm = 0.000306 \ m$

The current density in both wires is $J = 1750 \ A/m^2$

Considering the first wire

The cross-sectional area of the first wire is

$A = \pi r^2$

= > $A = 3.142 * (0.0015)^2$

= > $A = 7.0695 *10^{-6} \ m^2$

Generally the current in the first wire is

$I = J*A$

=> $I = 1750*7.0695 *10^{-6}$

=> $I = 0.01237 \ A$

Considering the second wire wire

The cross-sectional area of the second wire is

$A_1 = 19 * \pi r^2$

=> $A_1 = 19 *3.142 * (0.000306)^2$

=> $A_1 = 5.5899 *10^{-6} \ m^2$

Generally the current is

$I_2 = J * A_1$

=> $I_2 = 1750 * 5.5899 *10^{-6}$

=> $I_2 = 0.00978 \ A$

Considering question two

From the question we are told that

Resistivity is $\rho = 1.69* 10^{-8} \Omega \cdot m$

The length of each wire is $l = 6.25 \ m$

Generally the resistance of the first wire is mathematically represented as

$R = \frac{\rho * l }{A}$

=> $R = \frac{ 1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }$

=> $R = 0.0149 \ \Omega$

Generally the resistance of the first wire is mathematically represented as

$R_1 = \frac{\rho * l }{A_1}$

=> $R_1 = \frac{ 1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }$

=> $R_1 = 0.0189 \ \Omega$

3 0

3 years ago

b. Comparing and Contrasting Compare the change in atmospheric pressure with elevation to the change in water pressure with dept

olasank [31]

Answer:

fftgjjjuhgybcrgjjuhg ui

Explanation:

trggfyh

4 0

3 years ago

The membrane of a living cell can be approximated by a parallel-plate capacitor with plates of area 4.50×10−9 m2 , a plate separ

Marizza181 [45]

The energy stored in the membrane is $6.44\cdot 10^{-14} J$

Explanation:

The capacitance of a parallel-plate capacitor is given by

$C=\frac{k\epsilon_0 A}{d}$

where

k is the dielectric constant of the material

$\epsilon_0$ is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

For the membrane in this problem, we have

k = 4.6

$A=4.50\cdot 10^{-9} m^2$

$d=8.1\cdot 10^{-9} m$

Substituting, we find its capacitance:

$C=\frac{(4.6)(8.85\cdot 10^{-12})(4.50\cdot 10^{-9})}{8.1\cdot 10^{-9}}=2.26\cdot 10^{-11} F$

Now we can find the energy stored: for a capacitor, it is given by

$U=\frac{1}{2}CV^2$

where

$C=2.26\cdot 10^{-11} F$ is the capacitance

$V=7.55\cdot 10^{-2} V$ is the potential difference

Substituting,

$U=\frac{1}{2}(2.26\cdot 10^{-11} F)(7.55\cdot 10^{-2})^2=6.44\cdot 10^{-14} J$

Learn more about capacitors:

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brainly.com/question/9617400

#LearnwithBrainly

6 0

4 years ago

Write forty million in scientific notation.

Ivanshal [37]

Answer: 4 x 106

That’s how you write forty million in scientific notation.

7 0

3 years ago

Read 2 more answers