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3 years ago

Which of the following would decrease the resistance in a wire?

Physics

2 answers:

Anarel [89]3 years ago

8 0

The last one I believe

Alina [70]3 years ago

5 0

Answer:

Increase the thickness of the wire

Explanation:

From:

https://www.bbc.co.uk/bitesize/guides/z9sb2p3/revision/3

The resistance in a wire increases as:

the length of the wire increases

the thickness of the wire decreases

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A flat disk, a solid sphere, and a hollow sphere each have the same mass m and radius r. The three objects are arranged so that

Paladinen [302]

Answer:

$I_{disc} = \frac{1}{2}mR^2$

$I_{sphere} = \frac{2}{5}mR^2$

$I_{hollow} = \frac{2}{3}mR^2$

so largest moment of inertia is for Hollow Sphere

Explanation:

Moment of inertia flat disc is given as

$I_{disc} = \frac{1}{2}mR^2$

moment of inertia of solid sphere

$I_{sphere} = \frac{2}{5}mR^2$

moment of inertia of hollow sphere

$I_{hollow} = \frac{2}{3}mR^2$

all the three objects are given with same mass and same radius

so largest moment of inertia is for Hollow Sphere

8 0

4 years ago

Almost all cells are microscopic.

Degger [83]

Answer:

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4 0

3 years ago

A cannonball is fired across a flat field at an angle of 43 degrees with an initial speed 32 m/s and height of 12 m.

eimsori [14]

1) $x= v_{0x} t = 23.4 t\\y=y_0 + v_{0y}t-\frac{1}{2}gt^2 = 12+21.8t -4.9t^2$

The initial data of the projectile are:

$y_0 = 12 m$ is the initial height

$v_0 = 32 m/s$ is the initial speed of the projectile, so its components along the x- and y- directions are

$v_{0x} = v_0 cos \theta = (32 m/s)(cos 43^{\circ})=23.4 m/s\\v_{0y} = v_0 sin \theta = (32 m/s)(sin 43^{\circ})=21.8 m/s$

The motion of the cannonball along the x-direction is a uniform motion with constant speed, while on the y-direction it is an uniformly accelerated motion with constant acceleration g=9.8 m/s^2 downward. So, the two equations of motion of the projectile along the two directions are:

$x= v_{0x} t = 23.4 t\\y=y_0 + v_{0y}t-\frac{1}{2}gt^2 = 12+21.8t -4.9t^2$

2) 4.94 s

To determine how long the cannon ball was in the air, we need to find the time t at which the cannonball hits the ground, so the time t at which y(t)=0:

$0=12+21.8t-4.9 t^2$

Solving the equation with the formula, we have:

$t_{1,2}=\frac{-21.8\pm \sqrt{(21.8)^2-4(-4.9)(12)}}{2(-4.9)}$

which has two solutions:

t = -0.50 s

t = 4.94 s

Discarding the first solution which is a negative time so it has no physical meaning, the correct solution is

t = 4.94 s

3) 115.6 m

To determine how far the cannonball travelled, we need to find the value of the horizontal position x(t) when the ball hits the ground, at t=4.94 s. Substituting this value into the equation of motion along x, we find:

$x=v_{0x}t=(23.4 m/s)(4.94 s)=115.6 m$

4) 2.22 s

The cannonball reaches its maximum height when the vertical velocity becaomes zero.

The vertical velocity at time t is given by

$v_y(t)= v_{0y} -gt$

where

g = 9.8 m/s^2 is the acceleration due to gravity

Substutiting $v_y(t)=0$ and solving for t, we find

$t=\frac{v_{0y}}{g}=\frac{21.8 m/s}{9.8 m/s^2}=2.22 s$

5) 36.2 m

The maximum height reached by the cannon is equal to the vertical postion y(t) when the vertical velocity is zero, so when t=2.22 s. Substituting this value into the equation of the vertical motion, we find:

$y(t)=y_0 + v_{0y}t-\frac{1}{2}gt^2=12+(21.8)(2.22)-(4.9)(2.22)^2=36.2 m$