All categories

3 years ago

HEY GUYS:)WHATS MICROSCOPE??

2 answers:

8 0

A microscope is an instrument from which you can see the cells of an object.
It helps you to observe the cells of substance.
Powerful microscopes helps better in observing minute particles or bacterias.

Hope it helps...

White raven [17]3 years ago

4 0

A microscope is an instrument used to see objects that are too small for the naked eye. The science of investigating small objects using such an instrument is called microscopy. Microscopic means invisible to the eye unless aided by a microscope

You might be interested in

When i stand halfway between two speakers, with one on my left and one on my right, a musical note from the speakers gives me co

Nitella [24]

Destructive interference occurs when the difference is π, 3π, 5π ..., whereas constructive interference occurs when the phase difference between the waves is a multiple of 2π. 
If the difference between the phases is intermediate between these two extremes, then the magnitude of the displacement of the summed waves lies between the minimum and maximum values. 

One wavelength is equal to 2π (360°) and 1/2 wavelength is equal to π (180°). 

so Half a Wavelength.

3 0

3 years ago

Read 2 more answers

How fast must a 1000 kg car be moving to have a kinetic energy of 2.0×10^5j

ankoles [38]

Answer:

E=0.5 m v^2

2*10^-5 = 0.5 1000*v^2

4*10^-5=1000v^2

4*10^-8=v^2

2*10^-4 =v

7 0

3 years ago

At what position or positions on the x-axis is the electric field zero?

ElenaW [278]

Answer:

The electric field will be zero at x = ± ∞.

Explanation:

Suppose, A -2.0 nC charge and a +2.0 nC charge are located on the x-axis at x = -1.0 cm and x = +1.0 cm respectively.

We know that,

The electric field is

$E=\dfrac{kq}{r^2}$

The electric field vector due to charge one

$\vec{E_{1}}=\dfrac{kq_{1}}{r_{1}^2}(\hat{x})$

The electric field vector due to charge second

$\vec{E_{2}}=\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})$

We need to calculate the electric field

Using formula of net electric field

$\vec{E}=\vec{E_{1}}+\vec{E_{2}}$

$\vec{E_{1}}+\vec{E_{2}}=0$

Put the value into the formula

$\dfrac{kq_{1}}{r_{1}^2}(\hat{x})+\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})=0$

$\dfrac{kq_{1}}{r_{1}^2}(\hat{x})=\dfrac{kq_{2}}{r_{2}^2}(\hat{x})$

$(\dfrac{r_{2}}{r_{1}})^2=\dfrac{q_{2}}{q_{1}}$

$\dfrac{r_{2}}{r_{1}}=\sqrt{\dfrac{q_{2}}{q_{1}}}$

Put the value into the formula

$\dfrac{2.0+x}{x}=\pm\sqrt{\dfrac{2.0}{2.0}}$

$2.0+x=x$

If x = ∞, then the equation is be satisfied.

Hence, The electric field will be zero at x = ± ∞.

4 0

4 years ago

2. A hydraulic lift is used to lift heavy machine pushing down on a 5 square meters piston with a force of 1000 N. What force ne

IRINA_888 [86]

Answer:

The force needs to be applied on the 1 square meter piston to lift the machine is 200 N.

Explanation:

Given that,

Force, $F_1=1000\ N$

Area, $A_1=5\ m^2$

We need to find the force needs to be applied on the 1 square meter piston to lift the machine. We know that the pressure at every point is same due to Pascal's law such that :

$P_1=P_2\\\\\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2}\\\\\dfrac{1000}{5}=\dfrac{F_2}{1}\\\\F_2=200\ N$

So, the force needs to be applied on the 1 square meter piston to lift the machine is 200 N. Hence, this is the required solution.

3 0

4 years ago

Read 2 more answers

When a force of 20.0 N is applied to a spring, it elongates 0.20 m. Determine the period of oscillation of a 4.0-kg object suspe

mash [69]

Answer:

1.26 secs.

Explanation:

The following data were obtained from the question:

Force (F) = 20 N

Extention (e) = 0.2 m

Mass (m) = 4 Kg

Period (T) =.?

Next, we shall determine the spring constant, K for spring.

The spring constant, K can be obtained as follow:

Force (F) = 20 N

Extention (e) = 0.2 m

Spring constant (K) =..?

F = Ke

20 = K x 0.2

Divide both side by 0.2

K = 20/0.2

K = 100 N/m

Finally, we shall determine the period of oscillation of the 4 kg object suspended on the spring. This can be achieved as follow:

Mass (m) = 4 Kg

Spring constant (K) = 100 N/m

Period (T) =..?

T = 2π√(m/K)

T = 2π√(4/100)

T = 2π x √(0.04)

T = 2π x 0.2

T = 1.26 secs.

Therefore, the period of oscillation of the 4 kg object suspended on the spring is 1.26 secs.

6 0

3 years ago