Answer:
Sorry cant find the answer but i hope you got it right and if you didn't you'll still do great. :)
Explanation:
Answer:
Robert Hooke
Was the first to use the word "cell"
Observed cork cells
Anton van Leeuwenhoek
Observed "animalcules"
Used polished lens
.
Explanation:
Anton van Leeuwenhoek is known as father of microbiology. He is credited to improve the quality of lens in microscope. His first observation of organisms called animalcules.
He is credited to have build microscope that could get magnified by 200 times. He used word animalcules for small organisms from pond water when first observed in microscope. He discovered protozoa and named it animalcules".
Robert Hooke is famed for discovering cell from a cork of plant. He observed a compartment or honey comb like divisions when observed these cork cells under the microscope and named it cell. He was only able to see the cell wall as the cork cells are dead cells.
Answer:
Explanation:
The answer is electric field intensity. Electric field intensity is the force per unit positive charge which the charge exerts at any point.
Answer:
maximum amplitude = 0.13 m
Explanation:
Given that
Time period T= 0.74 s
acceleration of gravity g= 10 m/s²
We know that time period of simple harmonic motion given as
ω = 8.48 rad/s
ω=angular frequency
Lets take amplitude = A
The maximum acceleration given as
a= ω² A
The maximum acceleration should be equal to g ,then block does not separate
a= ω² A
10= 8.48² A
A=0.13 m
maximum amplitude = 0.13 m
R = 0.407Ω.
The resistance R of a particular conductor is related to the resistivity ρ of the material by the equation R = ρL/A, where ρ is the material resistivity, L is the length of the material and A is the cross-sectional area of the material.
To calculate the resistance R of a wire made of a material with resistivity of 3.2x10⁻⁸Ω.m, the length of the wire is 2.5m and its diameter is 0.50mm.
We have to use the equation R = ρL/A but first we have to calculate the cross-sectional area of the wire which is a circle. So, the area of a circle is given by A = πr², with r = d/2. The cross-sectional area of the wire is A = πd²/4. Then:
R =[(3.2x10⁻⁸Ω.m)(2.5m)]/[π(0.5x10⁻³m)²/4]
R = 8x10⁻⁸Ω.m²/1.96x10⁻⁷m²
R = 0.407Ω
