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Tresset [83]
3 years ago
12

A piece of wood 400N in weight and 50cm X 30cm X 20cm in size lies on 50cm X 20cm face. Calculate the pressure exerted by it.

Physics
1 answer:
Serga [27]3 years ago
6 0

Answer:

P = 4000 [Pa]

Explanation:

Pressure is defined as the relationship between Force and the area where the body rests.

The support area is equal to:

A=50*20=1000[cm^{2} ]

But we must convert from square centimeters to square meters.

1000[cm^{2}]*\frac{1^{2}m^{2}  }{100^{2}m^{2}  }=0.1[m^{2} ]

And the pressure is:

P=\frac{F}{A} \\P=400/0.1\\P=4000[N/m^{2} ]or 4000[Pa]

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The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me
Sav [38]

Answer:

(a):  \rm meter/ second^2.

(b):  \rm meter/ second^3.

(c):  \rm 2ct-3bt^2.

(d):  \rm 2c-6bt.

(e):  \rm t=\dfrac{2c}{3b}.

Explanation:

Given, the position of the particle along the x axis is

\rm x=ct^2-bt^3.

The units of terms \rm ct^2 and \rm bt^3 should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of \rm ct^2=meter

Therefore, unit of \rm c= meter/ second^2.

(b):

Unit of \rm bt^3=meter

Therefore, unit of \rm b= meter/ second^3.

(c):

The velocity v and the position x of a particle are related as

\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.

(d):

The acceleration a and the velocity v of the particle is related as

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.

(e):

The particle attains maximum x at, let's say, \rm t_o, when the following two conditions are fulfilled:

  1. \rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.
  2. \rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Applying both these conditions,

\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.

For \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c

Since, c is a positive constant therefore, for \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0

Thus, particle does not reach its maximum value at \rm t = 0\ s.

For \rm t_o = \dfrac{2c}{3b},

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.

Here,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Thus, the particle reach its maximum x value at time \rm t_o = \dfrac{2c}{3b}.

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3 years ago
A point charge of 5.0 x 10^-7 C moves to the right at 2.6 x 10^5 m/s in a magnetic field that is directed into the screen and ha
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The magnitude of the magnetic force acting on the charge is 2.34×10⁻³ N.

<h3>What is magnetic force?</h3>

A magnetic force is the force that act in a magnetic field.

To calculate the magnetic force, we use the formula below.

Formula:

  • F = qvB.........Equation 1

Where:

  • F = magnetic force
  • q = point charge
  • v = Velocity of the the charge
  • B = Field strength

From the question,

Given:

  • q = 5.0×10⁻⁷ C
  • v = 2.6×10⁵ m/s
  • B = 1.8×10⁻² T

Substitute these values into equation 2

  • F = (5.0×10⁻⁷)(2.6×10⁵)(1.8×10⁻²)
  • F = 23.4×10⁻⁴
  • F = 2.34×10⁻³ N

Hence, the magnitude of the magnetic force acting on the charge is 2.34×10⁻³ N.

Learn more about magnetic force here: brainly.com/question/2279150

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2 years ago
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3 years ago
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Double the force on the object
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Calculate the wavelength in centimeters of radar energy at a frequency of 10 GHz. What is the frequency in gigahertz of radar en
konstantin123 [22]

Explanation:

(a) Frequency of radar energy, f = 10\ GHz =10^{10}\ Hz

The relation between frequency and wavelength is given by :

c=f\lambda

\lambda=\dfrac{c}{f}

\lambda=\dfrac{3\times 10^8}{10^{10}}

\lambda=0.03\ m

or

\lambda=3\ cm

(b) If wavelength, \lambda=25\ cm=0.25\ m

c=f\lambda

f=\dfrac{c}{\lambda}

f=\dfrac{3\times 10^8}{0.25}

f=1.2\times 10^9\ Hz

or

f = 1.2 GHz

Hence, this is the required solution.

7 0
3 years ago
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