Find the solid angle made by the part of surface area 2 cm square of square and radius 4 centimetre

Why can light microscopes produce images in their natural color

Fiesta28 [93]

Answer:

The electron microscopes are uses the electron to produce the image of the any object.The area where the electron passes then that area appeaser white but on the other hand where the electron does not pass that area appears black.But this does not give any information about the color of light it shows only white and black areas.We also know that color is the property of photons ( light).So the microscopes does not produce color of the image.

6 0

3 years ago

How much force must be applied on a blade of length 4cm and thickness of 0.1mm to exert a pressure of 4000000pa?

Viktor [21]

Answer:

F= 403429 kpa

Explanation:

Pressure is the product of force and area

Mathematically,

P=F*A -------where F is force and A is area.

A= 40 *0.1 = 4mm² -----convert to m²

A= 4e⁻⁶ m²

P= 4000000 pa

F= P/A = 4000000/4e⁻⁶

F= 403428793.493 pa

F= 403429 kpa

7 0

3 years ago

A mass m = 3.9 kg hangs from a massless string wrapped around a uniform cylinder with mass Mp = 10.53 and radius R= 0.96 m. The

luda_lava [24]

Answer:

the rotational inertia of the cylinder = 4.85 kgm²

the mass moved 7.942 m/s

Explanation:

Formula for calculating Inertia can be expressed as:

$I =\frac{1}{2}mR^2$

For calculating the rotational inertia of the cylinder ; we have;

$I = \frac{1}{2}m_pR^2$

$I = \frac{1}{2}*10.53*(0.96)^2$

$I=5.265*(0.96)^2$

$I=4.852224$

I ≅ 4.85 kgm²

mg - T ma and RT = I ∝

T = $\frac{Ia}{R^2}$

$a = \frac{g}{1+\frac{I}{mR^2}}$

$a = \frac{9.8}{1+\frac{4.85}{3.9*(0.96)^2}}$

a = 4.1713 m/s²

Using the equation of motion

$v^2 = u^2+2as \\ \\ v^2 = 2as \\ \\ v = \sqrt{2*a*s} \\ \\ v= \sqrt{2*4.1713*7.56} \\ \\ v = 7.942 \ m/s$

3 0

3 years ago

Read 2 more answers

What are the horizontal and vertical components of a lizard’s displacement if it has climbed 7m directly up a tree?

madam [21]

Answer:

The horizontal component is zero.

The vertical component is $7\sin\theta$

Explanation:

Given that,

The lizard climb 7m directly up on a tree.

We know that,

The horizontal component is

$x=\cos\theta$

The vertical component is

$y=\sin\theta$

If the lizard climb 7m directly up on a tree then,

We need to find the components

Using given data

The horizontal component of lizard is

$x=0$

The vertical component is

$y=7\sin\theta$

Hence, The horizontal component is zero.

The vertical component is $7\sin\theta$

7 0

3 years ago

A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 35.5 m/s. a) What is the coeff

postnew [5]

Answer:

The coefficient of friction present between the roadway and the wheels of the truck is <u>0.833</u>.

Explanation:

Given:

Radius of the curve (R) = 150 m

Maximum speed of truck (v) = 35.5 m/s

Let the coefficient of friction between the roadway and the wheels of the truck be "μ".

As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.

The centripetal force acting on the truck is given as:

$F_c=\frac{mv^2}{R}$

Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.

Frictional force is given as:

$f=\mu N$

Where, 'N' is the normal force. Since there is no vertical motion, the normal force is equal to weight of truck. So,

$N=mg$

Therefore, frictional force, $f=\mu mg$

Now, frictional force = centripetal force

$f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu = \frac{v^2}{Rg}$

Plug in the given values and solve for 'μ'. This gives,

$\mu=\frac{(35\ m/s)^2}{(150\ m)(9.8\ m/s^2)}\\\\\mu=\frac{1225\ m^2/s^2}{1470\ m^2/s^2}\\\\\mu=0.833$

Therefore, the coefficient of friction present between the roadway and the wheels of the truck is 0.833

7 0

3 years ago

Find the solid angle made by the part of surface area 2 cm square of square and radius 4 centimetre​

Find the solid angle made by the part of surface area 2 cm square of square and radius 4 centimetre