3 years ago

A chain link fence should be cut quickly with a

Physics

1 answer:

kumpel [21]3 years ago

7 0

Answer: it should be cut with a chainsaw

Explanation:

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"5 N, up" is an example of a ___.<br> OA) force<br> OB) mass<br> OC) weight<br> OD) magnitude

Sunny_sXe [5.5K]

Answer:

A) Force

Explanation:

It is an example of force since force is a vector quantity so it has magnitude and direction. In this case the magnitude is equal to 5 [N] and the direction is upward.

The weight can not be, as it always acts downward.

Mass is not a force, its unit is given usually in kilogram [kg]

5 0

3 years ago

Consider the minute hand on a clock. (a) Compute the frequency of its motion in cycles per second. State your answer to three si

kicyunya [14]

Answer:

(a)0.0002778

(b) $1.16\times10^{-5}$

Explanation:

(a) The minute hand has a period of 60 minutes ( or 60 * 60 = 3600 seconds) for 1 circle. Its frequency per second would be

1 / 3600 = 0.0002778

(b) The hour hand has a period of 24 hours ( or 24*60 * 60 = 86400

seconds) for 1 circle. Its frequency per second would be

$1 / 86400 = 1.16\times10^{-5}$

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3 years ago

What is the term that describes the amount of water in a volume of air at a specific temperature compared to the amount of water

Natasha_Volkova [10]

I believe it is relative humidity.

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3 years ago

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Which structure found in echinoderms can sometimes perform a similar function as a radula found in some mollusks? This Is A Scie

hammer [34]

I think it’s D which is the stomach if not I can get you someone that can help you

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A 1.0 meter long rod is held horizontally in the east-west direction on a distant planet and is dropped from a height of 1.23 m.

enyata [817]

Answer:

The induced emf between two end is $34.02 \times 10^{-5}$ V

Explanation:

Given:

Length of rod $l = 1$ m

Height $h = 1.23$ m

Magnetic field $B = 6.93 \times 10^{-5}$ T

For finding induced emf,

$\epsilon = Blv$

Where $v =$ velocity of rod,

For finding the velocity of rod.

From kinematics equation,

$v^{2} = v_{o} ^{2} + 2gh$

Where $v_{o} =$ initial velocity, $g = 9.8 \frac{m}{s^{2} }$

$v = \sqrt{2gh}$

$v = \sqrt{2 \times 9.8 \times 1.23}$

$v = 4.91$ $\frac{m}{s}$

Put the velocity in above equation,

$\epsilon = 6.93 \times 10^{-5} \times 1 \times 4.91$

$\epsilon = 34.02 \times 10^{-5}$ V

Therefore, the induced emf between two end is $34.02 \times 10^{-5}$ V

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3 years ago