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Mekhanik [1.2K]
3 years ago
9

The table represents the speed of a car in a northern direction over several seconds. Column 1 would be on the x-axis, and Colum

n 2 would be on the y-axis. Which best lists the titles of each column? Column 1 0 2. 4 6 8 10 Column 2 5 10 15 20 25 30 O Column 1 should be titled "Time," and Column 2 should be titled "Velocity." O Column 1 should be titled "Velocity," and Column 2 should be titled "Time." O Column 1 should be titled "Time," and Column 2 should be titled "Acceleration." O Column 1 should be titled "Acceleration," and Column 2 should be titled "Time."​
Physics
1 answer:
DerKrebs [107]3 years ago
5 0
Answer is A hope this helps luv
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A 5 kg ball rolling at 1.0 m/s hits a 15 kg ball at rest. The balls stick together after the collision What is the
Reika [66]

Answer:

0.25m/s

Explanation:

Given parameters

m₁  = 5kg

v₁ = 1.0m/s

m₂ = 15kg

v₂ = 0m/s

Unknown:

velocity after collision = ?

Solution:

Momentum before collision and after collision will be the same. For inelastic collision;

    m₁v₁ + m₂v₂  = v(m₁ + m₂)

Insert parameters and solve for v;

   5 x 1  + 15 x 0 = v (5 + 15 )

          5  = 20v

          v = \frac{5}{20}   = 0.25m/s

5 0
3 years ago
How can you determine the amount of work done on an object? (joules, work, power...etc.)
Novay_Z [31]
Work done is equal to force by distance; so you take the force exerted, in newtons, and multiply that by the direction it's moved (from the starting point in a line, not along the path it's taken.)
7 0
3 years ago
What is the formula for the moment of inertia of the person/single particle rotating in a circle? (Give these values with a subs
Ann [662]

Moment of inertia of single particle rotating in circle is I1 = 1/2 (m*r^2)

The value of the moment of inertia when the person is on the edge of the merry-go-round is I2=1/3 (m*L^2)

Moment of Inertia refers to:

  • the quantity expressed by the body resisting angular acceleration.
  • It the sum of the product of the mass of every particle with its square of a distance from the axis of rotation.

The moment of inertia of single particle rotating in a circle I1 = 1/2 (m*r^2)

here We note that the,

In the formula, r being the distance from the point particle to the axis of rotation and m being the mass of disk.

The value of the moment of inertia when the person is on the edge of the merry-go-round is determined with parallel-axis theorem:

I(edge) = I (center of mass) + md^2

d be the distance from an axis through the object’s center of mass to a new axis.

I2(edge) = 1/3 (m*L^2)

learn more about moment of Inertia here:

<u>brainly.com/question/14226368</u>

#SPJ4

7 0
2 years ago
A bungee jumper launches herself off a bridge, how would you describe her motion?
Blizzard [7]
It would depend on how she jumped off but based on it sounds it would be a curving motion
4 0
3 years ago
Read 2 more answers
1. A DC-10 jumbo jet maintains an airspeed of 550 mph in a southwesterly direction. The velocity of the jet stream is a constant
Vladimir79 [104]

Answer:

The magnitude of actual velocity is <u>496.67 mph</u> and its direction is <u>51.54° with the x axis in the third quadrant</u>.

Explanation:

Given:

Speed of jumbo jet in southwesterly direction (v_j) = 550 mph

Velocity of jet stream from west to east direction (v_s)=80\ mph

First let us draw a vectorial representation of the above velocity vectors.

Consider the south direction as negative y axis and west direction as negative x axis.

From the diagram,

The velocity of the jet can be represented as:

\vec{v_j}=-550\cos(45)\vec{i}+(-550\sin(45)\vec{j} )\\\\\vec{v_j}=-388.91\vec{i}-388.91\vec{j}\ mph

Similarly, the velocity of the stream is, \vec{v_s}=80\vec{i}

Now, the vector sum of the above two vectors gives the actual velocity of the aircraft. So, the resultant velocity is given as:

\vec{v}=\vec{v_j}+\vec{v_s}\\\\\vec{v}=-388.91\vec{i}-388.91\vec{j}+80\vec{i}\\\\\vec{v}=(-388.91+80)\vec{i}-388.91\vec{j}\\\\\vec{v}=(-308.91)\vec{i}-388.91\vec{j}

Now, magnitude is given as the square root of sum of the squares of the 'i' and 'j' components. So,

|\vec{v}|=\sqrt{(-308.91)^2+(-388.91)^2}\\\\|\vec{v}|=496.67\ mph

As the horizontal and vertical components of actual velocity negative, the resultant vector makes an angle \theta with the x axis in the third quadrant.

The direction is given as:

\theta=\tan^{-1}(\frac{v_y}{v_x})\\\\\theta=\tan^{-1}(\frac{-388.91}{-308.91})\\\\\theta=51.54\°(Third\ quadrant)

Therefore, the magnitude of actual velocity is 496.67 mph and its direction is 51.54° with the x axis in the third quadrant.

5 0
3 years ago
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