The 5.00-cm-long second hand on a watch rotates smoothly. What is its angular velocity?

When jumping, a flea reaches a takeoff speed of 1.0 m/s over a distance of 0.47 mm .What is the flea's acceleration during the j

garri49 [273]

We can use kinematics here if we assume a constant acceleration (not realistic, but they want a single value answer, so it's implied). We know final velocity, vf, is 1.0 m/s, and we cover a distance, d, of 0.47mm or 0.00047 m (1m = 1000mm for conversion). We also can assume that the flea's initial velocity, vi, is 0 at the beginning of its jump. Using the equation vf^2 = vi^2 + 2ad, we can solve for our acceleration, a. Like so: a = (vf^2 - vi^2)/2d = (1.0^2 - 0^2)/(2*0.00047) = 1,064 m/s^2, not bad for a flea!

8 0

3 years ago

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Why is the weight of a body more at the pole than at the equator of the earth ??

Yuki888 [10]

At pole the gravitation is more than that of equator .

And according to newtons second law

Force=Mass×Acceleration

And

$\\ \sf\longmapsto F\propto a$

Hence force varies directly with acceleration hence weight will be more .

3 0

3 years ago

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A solid sphere of radius 40.0cm has a total positive charge of 26.0μC uniformly distributed throughout its volume. Calculate the

Rudiy27

The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C

R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

$Q\;(\text{total charge of the solid sphere})=(26\;\mathrm{\mu C})\left(\dfrac{1\;\mathrm{C}}{10^6\;\mathrm{\mu C}} \right)={26\times 10^{-6}\;\mathrm{C}}$

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

$E=\dfrac{Q}{4\pi\epsilon_0 r^2}$

Substitute numerical values:

$E&=\dfrac{24\times 10^{-6}}{4\pi (8.8542\times 10^{-12})(0.6)}\\ &={6.49\times 10^5\;\mathrm{N/C}\;\text{directed radially outward}}}$

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.

As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).

Learn more about Gaussian sphere here:

brainly.com/question/2004529

#SPJ4

6 0

2 years ago

A diffraction grating with 230 lines per mm is used in an experiment to study the visible spectrum of a gas discharge tube. At w

mr_godi [17]

Answer:

θ₁ = 5.4°

θ₂ = 10.86°

Explanation:

The angle ca be found by using grating equation:

mλ = d Sinθ

where,

m = order of diffraction

λ = wavelength = 405.3 nm = 4.053 x 10⁻⁷ m

d = grating element = 1/230 lines/mm = 0.0043 mm/line = 4.3 x 10⁻⁶ m/line

θ = angle = ?

FOR m = 1:

(1)(4.053 x 10⁻⁷ m) = (4.3 x 10⁻⁶ m/line) Sin θ₁

Sin θ₁ = 0.09425

θ₁ = Sin⁻¹(0.09425)

θ₁ = 5.4°

FOR m = 2:

(2)(4.053 x 10⁻⁷ m) = (4.3 x 10⁻⁶ m/line) Sin θ₁

Sin θ₂ = 0.1885

θ₂ = Sin⁻¹(0.1885)

θ₂ = 10.86°

6 0

3 years ago

What would happen if rats was eliminated from the environment

I am Lyosha [343]

Then the workers that made poison for rats would have to throw all of the poison in the garbage, sad, I know. Oh, if rats was eliminated from the environment, then cats that LOVE rats wont be poisoned by the rats that EAT the poison :). Yes, one of my mama cats died );

4 0

3 years ago