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Elis [28]
1 year ago
13

An object, which is at the origin at time t=0

Physics
2 answers:
Paladinen [302]1 year ago
7 0

At time t=0, an object at the origin has a constant acceleration a=(6.0i + 3.0j)m/s2 and an initial velocity V0= (-14.0i - 7.0j)m/s (-16.3 I - 8.2 j) The position x at which the object comes to rest is denoted by m. (temporarily).

<h3>What exactly is Origin?</h3>

A body's origin of motion is defined as the point at which a body or object begins to move.

  • The act of being born, existing, or beginning; point of departure.
  • That causes something to happen.
  • That which serves as a source or is derived from something.
  • In the field of coordinate geometry, the origin is defined as the initial point or starting point from which we begin our calculations or measurements. We start at zero on a ruler.
  • Because the 0 on a ruler is where we start measuring, it is referred to as the scale's origin.
  • The center of a coordinate axis is defined as coordinate 0 in all axes.

Hence, at time t=0, an object at the origin has an initial velocity V0 of (-14.0i - 7.0j)m/s and a constant acceleration an of (6.0i + 3.0j)m/s2. r = (-16.3 I - 8.2 j) m is the position x at which the object comes to rest (temporarily).

  • To learn more about
  • origin
  • refer to:
  • brainly.com/question/28515326
  • #SPJ9
zheka24 [161]1 year ago
6 0

An object, which is at the origin at time t=0, has initial velocity V₀= (-14.0i - 7.0j)m/s and constant acceleration a=(6.0i + 3.0j)m/s². The position x where the object comes to rest (momentarily) is, r = ( -16.3 i - 8.2 j ) m

<h3>What is position?</h3>

Position is a term that means a object where placed. In other words a object is placed or stayed or comes to rest that place called the position of the object.

<h3 /><h3>How can we calculate the position?</h3>

To calculate the position of the object we are using two formulas, like

V= V₀+∫a dt

r=r₀+∫v dt

Here we are given,

V₀=initial velocity of the object=(-14.0i - 7.0j)m/s,

a=constant acceleration of the object= (6.0i + 3.0j)m/s²

Now we put the values in the first formula,

V= V₀+∫a dt

Or, V= (-14.0i - 7.0j)+ ∫(6.0i + 3.0j) dt

Or, V= (-14.0+6.0t)i + (- 7.0+ 3.0t)j

According to the question, the particle becomes rest then,

V(x)=0

Or, (-14.0+6.0t)=0

Or, t=7/3S

In other way, V(y)=0

Or, (- 7.0+ 3.0t)=0

Or, t=7/3S

Now we put the value of velocity in the second equation,

r=r₀+∫v dt

Or, r=0+∫(-14.0+6.0t)i + (- 7.0+ 3.0t)j dt

Or, r= (-14.0t+6.0t²)i + (- 7.0t+ 3.0t²)j

Now we put,  t=7/3S, then the r value becomes,

r= (-14.0(7/3)+6.0(7/3)²)i + (- 7.0(7/3)+ 3.0(7/3)²)j

r≈(-16.3 i - 8.2 j)m

From the above calculation we can conclude that, The position x where the object comes to rest (momentarily) is, r = ( -16.3 i - 8.2 j ) m

Learn more about  position:

brainly.com/question/2534565

#SPJ9

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Two light bulbs are 2.0 m apart. From what distance can these light bulbs be marginally resolved by a small telescope with a 4.5
andrezito [222]

Answer:

R = 1.2295 10⁵  m

Explanation:

After reading your problem they give us the diameter of the lens d = 4.50 cm = 0.0450 m, therefore if we use the Rayleigh criterion for the resolution in the diffraction phenomenon, we have that the minimum separation occurs in the first minimum of diffraction of one of the bodies m = 1 coincides with the central maximum of the other body

            θ = 1.22 λ / D

where the constant 1.22 leaves the resolution in polar coordinates and D is the lens aperture

             

how angles are measured in radians

          θ = y / R

where y is the separation of the two bodies (bulbs) y = 2 m and R the distance from the bulbs to the lens

            \frac{y}{R} = 1.22 \frac{ \lambda}{D}

            R = \frac{ y \ D}{1.22 \lambda}

let's calculate

            R = \frac{ 2 \ 0.045}{ 1.22 \ 600 \ 10^{-9}}

            R = 1.2295 10⁵  m

3 0
2 years ago
A bike travels 4 miles in half an hour, what is its speed?
omeli [17]

Answer:

8 mph

Explanation:

4 miles in half hour so you add 4 more for the second half

3 0
3 years ago
Two steel balls, of masses m1=1.00 kg and m2=2.00 kg, respectively, are hung from the ceiling with light strings next to each ot
Zolol [24]

Answer:

(a) The maximum height achieved by the first ball, m₁ is 0.11 m

(a) The maximum height achieved by the second ball, m₂ ball is 0.44 m

Explanation:

Given;

mass of the first ball, m₁ = 1 kg

mass of the second ball, m₂ = 2 kg

The velocity of the first when released from a height of 1 m before collision;

u₁² = u₀² + 2gh

u₀ = 0, since it was released from rest

u₁² =  2gh

u₁² = 2 x 9.8 x 1

u₁² = 19.6

u₁ = √19.6

u₁ = 4.427 m/s

The velocity of the second ball before collision, u₂ = 0

Apply the principle of conservation of linear momentum, to determine the velocity of the balls after an elastic collision.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

v₁ is the final velocity of the first ball after an elastic collision

v₂ is the final velocity of the second ball after an elastic collision

m₁u₁ + m₂(0) = m₁v₁ + m₂v₂

m₁u₁ =  m₁v₁ + m₂v₂

1 x 4.427 = v₁ + 2v₂

v₁ + 2v₂ = 4.427

v₁  = 4.427 - 2v₂  ----- equation (1)

one directional velocity;

u₁ + v₁ = u₂ + v₂

u₂ = 0

u₁ + v₁ = v₂

v₁ = v₂ - u₁

v₁ = v₂ - 4.427 ------ equation (2)

Substitute v₁ into equation (1)

v₂ - 4.427 = 4.427 - 2v₂

3v₂ = 4.427 + 4.427

3v₂  = 8.854

v₂ = 8.854 / 3

v₂  = 2.95 m/s (→ forward direction)

v₁ = v₂ - 4.427

v₁ = 2.95 - 4.427

v₁  = - 1.477 m/s

v₁  = 1.477 m/s ( ← backward direction)

Apply the law of conservation of mechanical energy

mgh_{max} = \frac{1}{2}mv_{max}^2

(a) The maximum height achieved by the first ball (v₁  = 1.477 m/s)

mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max}  =  \frac{1}{2g}v_{max}^2\\\\ h_{max}  = \frac{1}{2*9.8}(1.477^2)\\\\ h_{max}  = 0.11 \ m

(b) The maximum height achieved by the second ball (v₂  = 2.95 m/s)

mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max}  =  \frac{1}{2g}v_{max}^2\\\\ h_{max}  = \frac{1}{2*9.8}(2.95^2)\\\\ h_{max}  = 0.44 \ m

6 0
3 years ago
It would really help if anyone could answers please and thanks
mojhsa [17]

well it would be A because 55 degrees is going strait well 75 is going literally straight up

4 0
2 years ago
Read 2 more answers
A sand mover at a quarry lifts 2,000 kg of sand per minute a vertical distance of 12 m. The sand is initially at rest and is dis
Marianna [84]

Answer:

<h2>E. 3.95kW</h2>

Explanation:

Power is defined as the rate of workdone.

Power = Workdone/time taken

Given Workdone = Force * distance

Power = Force * distance/time taken

Power = mgd/t (F = mg)

m = mass of the sand in kg

g = acceleration due to gravity in m/s²

d = vertical distance covered in metres

t = time taken in seconds

Given m = 2000kg, d = 12m, t = 1min = 60secs, g = 9.8m/s²

Power = 2000*9.8*12/60

Power = 3920Watts

Minimum rate of power that must be supplied to this machine is 3920Watts or 3.92kW

5 0
3 years ago
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