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3 years ago

Which of these forces can increase the effect of electric force?

Physics

1 answer:

Sati [7]3 years ago

8 0

Answer:

Electrostatic force. can increase the effect of electric force .

hope it is helpful to you

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Light shined through a single slit will produce a diffraction pattern. Green light (565 nm) is shined on a slit with width 0.210

kondor19780726 [428]

Answer:(a)9.685 mm

(b)4.184 mm

Explanation:

Given

Wavelength of light $(\lambda )=565nm \approx 565\times 10^{-9}m$

Width of slit(b)=0.210

(a)Width of central maximum located 1.80m from slit

$=\frac{2\lambda L}{b}$

$=\frac{2\times 565\times 10^{-9}\times 1.8}{0.210\times 10^{-3}}$

=9.685 mm

(b)Width of the first order bright fringe

$Y_1=\frac{\lambda \times L}{b}$

$Y_1=\frac{565\times 10^{-9}\times 1.8}{0.210\times 10^{-3}}$

$Y_1=4.84mm$

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3 years ago

You forgot to put what IRITONCF is unscrambled

Lesechka [4]

I'm Pretty sure it's FRICTION

4 0

3 years ago

As an electromagnetic wave travels, what is the relationship between the

galina1969 [7]

The answer is a tell me if I am wrong.

7 0

3 years ago

Read 2 more answers

A solid sphere of mass 4.0 kg and radius 0.12 m starts from rest at the top of a ramp inclined 15°, and rolls to the bottom. The

cluponka [151]

Answer:

v = 4.1 m/s

Explanation:

As per mechanical energy conservation law we can say that initial total gravitational potential energy of the sphere is equal to final kinetic energy of rolling

so we will say

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

now for pure rolling condition we know that

$v = R\omega$

so we will have

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v^2}{R^2})$

$mgh = \frac{7}{10}mv^2$

now we will have

$v^2 = \sqrt{\frac{10}{7}gh}$

$v^2 = \sqrt{\frac{10}{7}(9.8)(1.2)}$

$v = 4.1 m/s$

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3 years ago

In January 2004, NASA landed exploration vehicles on Mars. Part of the descent consisted of the following stages:

fenix001 [56]

Acceleration is given by:

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time interval

Let's apply the formula to the different parts of the problem:

A) $-20.5 m/s^2$

Let's convert the quantities into SI units first:

$u = 19300 km/h \cdot \frac{1000 m/km}{3600 s/h} = 5361.1 m/s$

$v=1600 km/h \cdot \frac{1000 m/km}{3600 s/h} =444.4 m/s$

t = 4.0 min = 240 s

So the acceleration is

$a=\frac{444.4 m/s-5361.1 m/s}{240 s}=-20.5 m/s^2$

B) $-3.8 m/s^2$

As before, let's convert the quantities into SI units first:

$u = 444.4 m/s$

$v=321 km/h \cdot \frac{1000 m/km}{3600 s/h} =89.2 m/s$

t = 94 s

So the acceleration is

$a=\frac{89.2 m/s - 444.4 m/s}{94 s}=-3.8 m/s^2$

C) $-53.0 m/s^2$

For this part we have to use a different formula:

$v^2 - u^2 = 2ad$

where we have

v = 0 is the final velocity

u = 89.2 m/s is the initial velocity

a is the acceleration

d = 75 m is the distance covered

Solving for a, we find

$a=\frac{v^2-u^2}{2d}=\frac{0^2-(89.2 m/s)^2}{2(75 m)}=-53.0 m/s^2$

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3 years ago