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3 years ago

7

the diameter and the length of a thin wire, approximately 1m in length, are measured as accurately as possible. what are the bes

t instruments to use?

1 answer:

slamgirl [31]3 years ago

5 0

Answer:

1.) Micrometres screw gauge

2.) Tape rule.

Explanation:

Given that the diameter and the length of a thin wire, approximately 1m in length, are measured as accurately as possible.

what are the best instruments to use ?

To measure the diameter of a thin wire, the best instrument to use is known as micrometres screw gauge.

And to measure the length of a thin wire up to 1 m, the measuring device can be tape rule or long metre rule.

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The field lines around one end of a bar magnet are shown

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Explanation:

Field lines are lines of forces around a bar magnet. The show the direction of force field in a magnet.

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A person hits a tennis ball with a mass of 0.058 kg against a wall.

horrorfan [7]

Explanation:

Mass of the ball, m = 0.058 kg

Initial speed of the ball, u = 11 m/s

Final speed of the ball, v = -11 m/s (negative as it rebounds)

Time, t = 2.1 s

(a) Let F is the average force exerted on the wall. It is given by :

$F=\dfrac{m(v-u)}{t}$

$F=\dfrac{0.058\times (11-(-11))}{2.1}$

F = 0.607 N

(b) Area of wall, $A=3\ m^2$

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3 years ago

Determine the angle of an incline that would yield a constant velocity, given the coefficient of kinetic friction is 0.10.

azamat

Answer:

$\theta=5.71^{o}$

Explanation:

In order to solve this problem, we mus start by drawing a free body diagram of the given situation (See attached picture).

From the free body diagram we can now do a sum of forces in the x and y direction. Let's start with the y-direction:

$\sum F_{y}=0$

$-W_{y}+N=0$

$N=W_{y}$

so:

$N=mgcos(\theta)$

now we can go ahead and do a sum of forces in the x-direction:

$\sum F_{x}=0$

the sum of forces in x is 0 because it's moving at a constant speed.

$-f+W_{x}=0$

$-\mu_{k}N+mg sen(\theta)=0$

$-\mu_{k}mg cos(\theta)+mg sen(\theta)=0$

so now we solve for theta. We can start by factoring mg so we get:

$mg(-\mu_{k} cos(\theta)+sen(\theta))=0$

we can divide both sides into mg so we get:

$-\mu_{k} cos(\theta)+sen(\theta)=0$

this tells us that the problem is independent of the mass of the object.

$\mu_{k} cos(\theta)=sen(\theta)$

we now divide both sides of the equation into $cos(\theta)$ so we get:

$\mu_{k}=\frac{sen(\theta)}{cos(\theta)}$

$\mu_{k}=tan(\theta)$

so we now take the inverse function of tan to get:

$\theta=tan^{-1}(\mu_{k})$

so now we can find our angle:

$\theta=tan^{-1}(0.10)$

so

$\theta=5.71^{o}$

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3 years ago