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3 years ago

A bullet is at rest. It travels a distance of 0.34m in a time of 0.0095 seconds. Calculate its acceleration.

Physics

1 answer:

tamaranim1 [39]3 years ago

7 0

Answer:

7.5 × 10^3 m/s^2

Explanation:

use the formula that does not have v in it to solve for acceleration.

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A two slit pattern is viewed on a screen 1.00m from the slits if the two third-order minima are 22.0 cm apart what is the width

Bingel [31]

Answer:

4.4 cm

Explanation:

Given:

Distance of the screen from the slit, D = 1 m

Distance between two third order interference minimas, x = 22 cm

Let's say, minima occurs at:

$x_n = (n + \frac{1}{2}) \frac{wL}{d}$

We have:

$2x_2 = 2(2 + \frac{1}{2}) * \frac{w*22}{d}$

Calculating further for the width of the central bright fringe, we have:

$\frac{w}{d} = \frac{22}{5}$

= 4.4 cm

Note: w in representswavelength

8 0

2 years ago

When a certain amount of current flows through a resistor, it uses 3.00 W of power. If the current doubles, how much power will

asambeis [7]

Answer: 12

Explanation: Acellus

7 0

2 years ago

Ions are atoms of the same element that have a different number of _____________.

Alex17521 [72]

C, electrons. Ion of an element has the same nucleus, the same number of protons and neutrons, with a different number of electrons.

6 0

2 years ago

Read 2 more answers

What is the economic system in the united states?

AnnyKZ [126]

Answer:

A. Capitalist economy

Explanation:

The United States is a capitalist economy. Hope this helps, thank you !!

3 0

2 years ago

The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles

belka [17]

Answer:

$F_a=5.67\times 10^{-5}\ N$

$F_b=3.49\times 10^{-5}\ N$

$F_c=9.16\times 10^{-5}\ N$

Explanation:

Given:

mass of particle A, $m_a=363\ kg$

mass of particle B, $m_b=517\ kg$

mass of particle C, $m_c=154\ kg$

All the three particles lie on a straight line.

Distance between particle A and B, $x_{ab}=0.5\ m$

Distance between particle B and C, $x_{bc}=0.25\ m$

Since the gravitational force is attractive in nature it will add up when enacted from the same direction.

Force on particle A due to particles B & C:

$F_a=G. \frac{m_a.m_b}{x_{ab}^2} +G. \frac{m_a.m_c}{(x_{ab}+x_{bc})^2}$

$F_a=6.67\times 10^{-11}\times (\frac{363\times 517}{0.5^2}+\frac{363\times 154}{(0.5+0.25)^2})$

$F_a=5.67\times 10^{-5}\ N$

Force on particle C due to particles B & A:

$F_c=G.\frac{m_c.m_b}{x_{bc}^2} +G.\frac{m_c.m_a}{(x_{ab}+x_{bc})^2}$

$F_c=6.67\times 10^{-11}\times (\frac{154\times 517}{0.25^2}+\frac{154\times 363}{(0.25+0.5)^2} )$

$F_c=9.16\times 10^{-5}\ N$

Force on particle B due to particles C & A:

$F_b=G.\frac{m_b.m_c}{x_{bc}^2} -G.\frac{m_b.m_a}{x_{ab}^2}$

$F_b=6.67\times 10^{-11}\times (\frac{517\times 154}{0.25^2}-\frac{517\times 363}{0.5^2} )$

$F_b=3.49\times 10^{-5}\ N$

3 0

3 years ago