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inna [77]
3 years ago
6

An ant on a picnic table travels 20 cm eastward then 25 cm northward 5 cm westward and finally 10 cm northward what is the ants

displacement relative to its original position
Physics
1 answer:
IrinaK [193]3 years ago
8 0
Turn it into a triangle pretty much. total displacement eastward is 15cm. total displacement northward is 35cm. now use the Pythagorean theorem which will be squareroot of 15squared+35squared. this gives you 38.08cm northeastward.<span />
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A 1.8-kg object is attached to a spring and placed on frictionless, horizontal surface. A force of 40 N stretches a spring 20 cm
Sergio [31]

Answer:

a) k = 200 N/m

b) E = 4 J

c) Δx = 6.3 cm

Explanation:

a)

  • In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:

       F = - k * \Delta x (1)

  • where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
  • Solving for k:

       k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)

b)

  • Assuming no friction present, total mechanical energy mus keep constant.
  • When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:

        U = \frac{1}{2}* k* (\Delta x)^{2} (3)

  • Replacing k and Δx by their values, we get:

       U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)

c)

  • When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
  • We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
  • So, we can write the following expression:

        \frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2}  = \frac{1}{2}*k*\Delta x^{2}   (5)

  • Replacing the right side of (5) with (4), k, m, and v by the givens, and simplifying, we can solve for Δx₁, as follows:

        \frac{1}{2}* 200N/m* \Delta x_{1} ^{2} + \frac{1}{2} * 1.8kg* (-2.0m/s)^{2}  = 4J   (6)

⇒      \frac{1}{2}* 200N/m* \Delta x_{1} ^{2}   = 4J  - 3.6 J = 0.4 J (7)

⇒     \Delta x_{1}   = \sqrt{\frac{0.8J}{200N/m} } = 6.3 cm (8)

6 0
3 years ago
A point charge of -3.0 x 10-5C is placed at the origin of coordinates. Find the electric field at the point 3. r= 50 m on the x-
Snezhnost [94]

Answer: -5×10-3

Explanation:

E=kq/r

4 0
2 years ago
these are the types of attractions found between molecules. their strength determines the state of the substance at room tempera
Elina [12.6K]
<h2>Answer</h2>

The physical state of the elements depends upon the <u>attraction forces </u>and their <u>kinetic energy</u>.

<h2>Explanation</h2>

The elements or substances are fixed with each other with the help of different chemical forces including ionic bonding, covalent bonding, H- bonding etc. The strength of these forces is also one of the factors that affect their physical natures. For example, covalent or ionic bonds are the strongest bonds than all other bonds and metals that contain these forces are mostly in solid form. The kinetic motion of electrons in the element also affects the physical state of the element and potential of bonding.

7 0
3 years ago
Read 2 more answers
A body oscillates with simple harmonic motion along the x axis. Its displacement varies with time according to the equation x =
frutty [35]

Answer:

θ = (7π / 3) rad

Explanation:

given,

displacement of simple harmonic motion along x-axis

equation is given as

                   x = 5 sin (π t + π/3 )

general equation of simple harmonic motion

                   x = A sin θ

           θ is the phase angle

      θ = π t + π/3

at   t = 2 s

      \theta =\pi \times 2 +\dfrac{\pi}{3}

      \theta =\dfrac{7\pi}{3}\ rad

Phase of the motion at t =2 s is θ = (7π / 3) rad

7 0
3 years ago
An 80-cm uniform 10-kg bar is resting on two scales, one at either end. A smaller 4-kg mass (m) is placed at a distance of d = 2
Varvara68 [4.7K]

Answer

given,

length of bar = 80 cm

mass of the bar = 10 kg

smaller mass = 4 kg

distance = 20 cm

s_1 + s_2 = 10 + 4

s_1 + s_2 = 14\ kg

taking moment about B

s_1 \times 0.8 - 10 \times 0.4 - 4 \times 0.6 = 0

s_1 \times 0.8 = 6.4

s_1 = 8\ N

s_2 = 14 - s_1

s_2 = 14 - 8

s_2 = 6 N

difference between two scale = 8 - 6

                                                  = 2 N

7 0
3 years ago
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