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Anarel [89]
3 years ago
11

Reasons for studying physics​

Physics
2 answers:
Aleksandr-060686 [28]3 years ago
5 0

Answer:

cause its for a grade UvU

Explanation:

andrey2020 [161]3 years ago
4 0
Well depending on ur major you could need it.
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A horizontal rectangular surface has dimensions 2.80 cm by 3.20 cm and is in a uniform magnetic field that is directed at an ang
koban [17]

Answer:

magnitude of the magnetic field 0.692 T

Explanation:

given data

rectangular dimensions = 2.80 cm by 3.20 cm

angle of 30.0°

produce a flux Ф = 3.10 × 10^{-4}  Wb

solution

we take here rectangular side a and b as a = 2.80 cm and b = 3.20 cm

and here angle between magnitude field and area will be ∅ = 90 - 30

∅ = 60°

and flux  is express as

flux Ф = \int \vec{B}.d\vec{A}   .................1

and Ф = BA cos∅    ............2

so B = \frac{\phi }{Acos\theta }    

and we know

A = ab

so

B = \frac{\phi }{abcos\theta }    ..............3

put here value

B =  \frac{3.10\times 10^{-4} }{2.80 \times 10^{-2}\times 3.20 \times 10^{-2}\times cos60}  

solve we get

B = 0.692 T

8 0
3 years ago
What is the difference between revolution and rotation
Sphinxa [80]
A rotation is when the earth turns in a circle on it's axis. A revolution is the oblong path the the earth is constantly moving in around the sun. A revolution takes 365.25 days where as a rotation happens in 24 hours. <span />
8 0
3 years ago
Read 2 more answers
Owen and Dina are at rest in frame S' , which is moving at 0.600 c with respect to frame S . They play a game of catch while Ed
Taya2010 [7]

The speed of the ball with respect to Dina is 0.800c.

A frame of reference is a set of reference points—geometric points whose positions are known mathematically and physically—that define the origin, orientation, and scale of an abstract coordinate system.

If a body does not continuously adjust its position in relation to its environment throughout the course of time, it is said to be at rest.

In frame S', Dina and Owen are at rest.

The speed of the ball with respect to Owen, u =0.800c

The speed of the fram S' with respect to frame S,  v = 0.600c

Distance between Dina and Owen, L(p) = 1.8 × 10¹² m

Speed of light, c = 3 × 10⁸ m/s

Therefore, the speed of the ball according to Dina is 0.800c. As Dina and Owen are in the same frame.

Learn more about the frame of reference here:

brainly.com/question/10962551

#SPJ4

6 0
1 year ago
child slides down a snow‑covered slope on a sled. At the top of the slope, her mother gives her a push to start her off with a s
Strike441 [17]

Answer:

θ = 13.7º

Explanation:

  • According to the work-energy theorem, the change in the kinetic energy of the combined mass of the child and the sled, is equal to the total work done on the object by external forces.
  • The external forces capable to do work on the combination of child +sled, are the friction force (opposing to the displacement), and the component of the weight parallel to the slide.
  • As this last work is just equal to the change in the gravitational potential energy (with opposite sign) , we can write the following equation:

       \Delta K + \Delta U = W_{nc} (1)

  • ΔK, is the change in kinetic energy, as follows:

       \Delta K = \frac{1}{2}* m* (v_{f} ^{2}  - v_{0} ^{2}) (2)

  • ΔU, is the change in the gravitational potential energy.
  • If we choose as our zero reference level, the bottom of the slope, the change in gravitational potential energy will be as follows:

        \Delta U = 0 - m*g*h = -m*g*d* sin\theta (3)

  • Finally, the work done for non-conservative forces, is the work done by the friction force, along the slope, as follows:

        W_{nc} = F_{f} * d * cos 180\º \\\\  = 0.2*m*g*d* cos 180\º = -0.2*m*g*d (4)

  • Replacing (2), (3), and (4) in (1), simplifying common terms, and rearranging, we have:

      \frac{1}{2}* (v_{f} ^{2}  - v_{0} ^{2}) = g*d* sin\theta -0.2*g*d

  • Replacing by the givens and the knowns, we can solve for sin θ, as follows:              \frac{1}{2}*( (4.30 m/s) ^{2}  - (0.75 m/s)^{2}) = 9.8 m/s2*25.5m* sin\theta -0.2*9.8m/s2*25.5m\\ \\ 8.56 (m/s)2 = 250(m/s)2* sin \theta -50 (m/s)2\\ \\ sin \theta = \frac{58.6 (m/s)2}{250 (m/s)2}  = 0.236⇒ θ = sin⁻¹ (0.236) = 13.7º
8 0
3 years ago
Two stars of masses M and 6M are separated by a distance D. Determine the distance (measured from M) to a point at which the net
kotykmax [81]

Answer:

0.29D

Explanation:

Given that

F = G M m / r2

F = GM(6m) / (D-r)2

G Mm/r2 = GM(6m) / (D-r)2

1/r2 = 6 / (D-r)2

r = D / (Ö6 + 1)

r = 0.29 D

See diagram in attached file

7 0
3 years ago
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