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luda_lava [24]
3 years ago
7

Who was the first to hypothesize that electron orbit a positively charged nucleus

Physics
1 answer:
Eva8 [605]3 years ago
6 0
I think the answer is ruthorford
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A car drives to the right. There is a large amount or air resistance and the cars engine provides the cars forward motion. Which
scZoUnD [109]

Answer:

the answer to some may be Force A

Explanation:

7 0
3 years ago
Read 2 more answers
Velocity vector and acceleration vector in a uniform circular motion are related as.
mr_godi [17]

They are related as \bold{\underline {v}\,.\,\underline a }= \bold{0}

  • In a uniform circular motion, the magnitude of the speed does not change during the travel and only the instantaneous direction changes.
  • This speed is always directed along the tangent to the circle at a given point. (refer to the figure attached)
  • For any circular motion, the must-have acceleration is the centripetal acceleration that is directed towards the centre of the circular locus (if the motion has a tangential acceleration, it has a tangential acceleration additionally).
  • Therefore, both the directions of the tangential speed and the centripetal acceleration are orthogonal to each other (perpendicular: one is 90 degrees apart from the other).
  • In mathematics, 2 vectors (\underline p , \underline q) that are perpendicular to each other have a quality that their dot product (\underline p\,.\, \underline q) equal to zero vector (\bold 0) which is written as \undeline p\,.\, \underline q = \bold 0.
  • This quality can be considered when dealing with the velocity vector and the acceleration vector in a manner \underline v\,.\, \underline a =\bold 0.

#SPJ4

8 0
9 months ago
A rotating object starts from rest at t = 0 s and has a constant angular acceleration. At a time of t = 2.5 s the object has an
Evgesh-ka [11]

Answer:

52 rad

Explanation:

Using

Ф = ω't +1/2αt²................... Equation 1

Where Ф = angular displacement of the object, t = time, ω' = initial angular velocity, α = angular acceleration.

Since the object states from rest, ω' = 0 rad/s.

Therefore,

Ф = 1/2αt²................ Equation 2

make α the subject of the equation

α = 2Ф/t².................. Equation 3

Given: Ф = 13 rad, t = 2.5 s

Substitute into equation 3

α = 2(13)/2.5²

α = 26/2.5

α = 4.16 rad/s².

using equation 2,

Ф = 1/2αt²

Given: t = 5 s, α = 4.16 rad/s²

Substitute into equation 2

Ф = 1/2(4.16)(5²)

Ф = 52 rad.

6 0
3 years ago
Read 2 more answers
The tires on your truck have 0.35 m radius. In a straight line, you drive 2600 m. What is the angular displacement of the tire,
Travka [436]
1) In a circular motion, the angular displacement \theta is given by
\theta =  \frac{S}{r}
where S is the arc length and r is the radius. The problem says that the truck drove for 2600 m, so this corresponds to the total arc length covered by the tire: S=2600 m. Using the information about the radius, r=0.35 m, we find the total angular displacement:
\theta =  \frac{2600 m}{0.35 m} =7428 rad

2) If we put larger tires, with radius r=0.60 m, the angular displacement will be smaller. We can see this by using the same formula. In fact, this time we have:
\theta =  \frac{2600 m}{0.60 m}=4333 rad
8 0
3 years ago
A robot is on the surface of Mars. The angle of depression from a camera in the robot to a rock on the surface of Mars is 13.69
ra1l [238]

Answer:

The distance between the camera and the rock is 836.6 cm

Explanation:

A right triangle is formed where the hypotenuse (h) is the distance between the rock and the camera. One of the leg (l) is the distance between the camera and the surface. The angle between the hypotenuse and this leg is α = 90° - 13.69° = 76.31°. By definition:

cos α = adjacent/hypotenuse

cos(76.31) = 198.0/h

h = 198.0/cos(76.31)

h = 836.6 cm

3 0
3 years ago
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