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Flauer [41]
2 years ago
7

Inside most ball-point pens is a small spring that compresses as the pen is pressed against the paper. If a force of 0.1 N compr

esses the pen's spring a distance of 0.005 m, what is the spring constant of the tiny spring?
Physics
1 answer:
AnnZ [28]2 years ago
8 0

Answer:

20 N/m

Explanation:

From the question,

The ball-point pen obays hook's law.

From hook's law,

F = ke............................ Equation 1

Where F = Force, k = spring constant, e = compression.

Make k the subject of the equation

k = F/e........................ Equation 2

Given: F = 0.1 N, e = 0.005 m.

Substitute these values into equation 2

k = 0.1/0.005

k = 20 N/m.

Hence the spring constant of the tiny spring is 20 N/m

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When you hold a cold glass of water in your warm hand, which way does the heat flow ?
Tanya [424]

Heat flows naturally from a warmer object (higher energy) to a cooler one (lower energy).

6 0
3 years ago
A charge of 7.00 mC is placed at opposite corners corner of a square 0.100 m on a side and a charge of -7.00 mC is placed at oth
andrew-mc [135]

Answer:

4.03\times10^{7}N[/tex], 135°

Explanation:

charge, q = 7 mC = 0.007 C

charge, - q = - 7 mC = - 0.007 C

d = 0.1 m

Let the force on charge placed at C due to charge placed at D is FD.

F_{D}=\frac{kq^{2}}{DC^{2}}

F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N

The direction of FD is along C to D.

Let the force on charge placed at C due to charge placed at B is FB.

F_{B}=\frac{kq^{2}}{BC^{2}}

F_{B}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N

The direction of FB is along C to B.

Let the force on charge placed at C due to charge placed at A is FA.

F_{A}=\frac{kq^{2}}{AC^{2}}

F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1 \times\sqrt{2} \times 0.1 \times\sqrt{2}}=2.205 \times 10^{7}N

The direction of FA is along A to C.

The net force along +X axis

F_{x}=F_{A}Cos45-F_{D}

F_{x}=2.205\times10^{7}Cos45-4.41\times10^{7}=-2.85\times10^{7}N

The net force along +Y axis

F_{y}=F_{B}-F_{A}Sin45

F_{y}=4.41\times10^{7}-2.205\times10^{7}Sin45=2.85\times10^{7}N

The resultant force is given by

F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{(-2.85\times10^{7})^{2}+(2.85\times10^{7})^{2}}

F = 4.03\times10^{7}N

The angle from x axis is Ф

tan Ф = - 1

Ф = -45°

Angle from + X axis is 180° - 45° = 135°

5 0
3 years ago
Resonance structures have _______ connectivity of atoms and _______ distribution of electrons.
Arte-miy333 [17]

Answer:

Resonance structures have <u> </u><u>same</u><u>  </u>  connectivity of atoms and <u> differ only in</u> distribution of electrons.

Explanation:

Atoms supply the electrons from their outer electron shells. Electrons are found free in nature and are grouped around the nucleus into shells. Electrons can be further explained as negatively charged subatomic particle. Electrons have properties of both particles and waves and they can be moved around.

Resonance structures are imaginary structures and not all of them are created equally. Resonance structures have two or more possible electron structures, and, the resonance structures for a particular substance sometimes have different energy and stability. When resonance structures are identical, they are important descriptions  of the molecule. The position of the atoms is the same in the various resonance structures of a compound, but the electrons are distributed differently around the structure.

3 0
3 years ago
Suppose we have a 600 kilogram great "yellow" shark swimming to the right at a speed of 3 meters traveled each second as it trie
sammy [17]

Answer:

2.64 m/s

Explanation:

Given that a 600 kilogram great "yellow" shark swimming to the right at a speed of 3 meters traveled each second as it tries to get lunch. An unsuspecting 100 kilogram blue fin tuna is minding its own business swimming to the left at a speed of 0.5 meters traveled each second. GULP! After the great "yellow" shark "collides" with the blue fin tuna

Momentum = MV

Momentum of the yellow shark before collision = 600 × 3 = 1800 kgm/s

Momentum of the tun final before collision = 100 × 0.5 = 50 kgm/s

Total momentum before collision = 1800 + 50 = 1850 kgm/s

Let's assume that they move together after collision. Then,

1850 = ( 600 + 100 ) V

1850 = 700V

V = 1850 / 700

V = 2.64285 m/s

Therefore, the momentum of the shark after collision is 2.64 m/ s approximately

6 0
3 years ago
A jet accelerates from rest down a runway at 1.75m/s² for a distance of 1500 m before takeoff.
babunello [35]
A. Using the third equation of motion:
v2 = u2 + 2as
from the question;
the jet was initially at rest
hence u = 0
a = 1.75m/s2
s = 1500m
v2 = 02 + 2(1.75)(1500)
v2 = 5250
v = √5250
v = 72.46m/s
hence it moves with a velocity of 72.46m/s.
b. s = ut + 1/2at2
1500 = 0(t) + 1/2(1.75)t2
1500 × 2 = 2× 1/2(1.75)t2
3000 = 1.75t2
1714.29 = t2
41.4 = t
hence the time taken for the plane to down the runway is 41.4s.


Read more on Brainly.com - brainly.com/question/18743384#readmore
8 0
3 years ago
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