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3 years ago

14

A learner sets up a circuit to determine the conductivity of certain solids. What piece of apparatus is not required for the nex

t experiment ?
A. battery
B. light bulb
C. conducting wire
D. stopwatch

1 answer:

Lilit [14]3 years ago

4 0

It is D as u dont need a stop watch aft that

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The magnitude of a velocity vector is called speed. Suppose that a wind is blowing from the direction N45°W at a speed of 50 km/

velikii [3]

To solve this problem, you must figure out (in vector form) both the wind vector and plane vector

w⃗ = wind vector

P⃗ = plane vector

To get the true course of the plane, you need to add the plane and wind vectors, the formula would be

w⃗ +P⃗ ,

which will result to the ground speed.

ground speed=||w⃗ +P⃗ ||

Using the planar representation of your situation, this will help you understand the equation, use this to make the equation more understandable.

w⃗ =AB¯¯¯¯¯¯¯¯, P⃗ =AC¯¯¯¯¯¯¯¯

the smaller circle is of radius 50 (similar to the wind speed) and the larger circle is of radius 200 (similar to the plane vector. To get the coordinates of these two vectors, use polar coordinates.

Let East be 0 degrees, so since the wind vector is on the circle of radius 50, we have:

w⃗ =⟨50cos(135),50sin(135)⟩=⟨−252√, 252√⟩.

P⃗ =⟨200cos(60),200sin(60)⟩=⟨100,\1003√⟩.

w⃗ +P⃗ =⟨100−252√ , 1003√+252√⟩

||w⃗ +P⃗ ||=(100−252√)2+(1003√+252√)2

√≈218.349218.

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8 0

3 years ago

To test the performance of its tires, a car

velikii [3]

<u>Answer</u>:

The coefficient of static friction between the tires and the road is 1.987

<u>Explanation</u>:

<u>Given</u>:

Radius of the track, r = 516 m

Tangential Acceleration $a_r$ = 3.89 m/s^2

Speed,v = 32.8 m/s

<u>To Find:</u>

The coefficient of static friction between the tires and the road = ?

<u>Solution</u>:

The radial Acceleration is given by,

$a_{R = \frac{v^2}{r}$

$a_{R = \frac{(32.8)^2}{516}$

$a_{R = \frac{(1075.84)}{516}$

$a_{R = 2.085 m/s^2$

Now the total acceleration is

$\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2$

=> $= \sqrt{ (a_r)^2+(a_R)^2}$

=> $\sqrt{ (3.89 )^2+( 2.085)^2}$

=> $\sqrt{ (15.1321)+(4.347)^2}$

=> $19.4791 m/s^2$

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of static friction is

$\mu =\frac{f}{W}$

From (1) and (2)

$\mu =\frac{ma}{mg}$

$\mu =\frac{a}{g}$

Substituting the values, we get

$\mu =\frac{19.4791}{9.8}$

$\mu =1.987$

8 0

3 years ago

three letters (JET) are placed in front of a plane mirror the image formed is in what arrangement???

mario62 [17]

Answer:

TEJ as this is a thing you wont get

3 0

3 years ago

Pls heplp 70 points!!!!!

Rasek [7]

Answer

the answer is d for sure

5 0

3 years ago

A ball is held at rest at the top of a hill. The ball is then released and starts rolling down the hill. At the bottom it reache

Alex17521 [72]

Answer:

Gravitational Potential Energy

Explanation:

a ball is held rest at the top of hill

gravitational potential energy will store due to its height

it. and body will start move downward and its potential energy will convert into kinetic energy due to motion of body

at the ground level it will stop and potential energy will became zero and kinetic energy get convert into internal energy due to collisions

3 0

3 years ago