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lawyer [7]
3 years ago
15

A factory line moves parts for an automobile at a constant speed of 0.22 m/s. How long does it take the parts to move 8.5 m alon

g the line? A. 1.9 s B. 12 s C. 39 s D. 56 s
Physics
1 answer:
OleMash [197]3 years ago
5 0

Answer: C. 39 s

Explanation:

We know the constant speed is 0.22 m/s. We have to get to 8.5 m. We divide <u>8.5 m by 0.22</u> = 38.6. After we estimate, 6 is greater than 5, so 39 s.

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Explain relative velocity briefly​
fomenos

Answer:

Explanation:

Relative velocity is defined as the velocity of an object B in the rest frame of another object A.

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3 years ago
London is making shelves from their room using a large piece of American red oak which has a density of .74 g/cm^3. They saw the
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London is making in their room by the way I’m just using this as

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3 years ago
Sam heaves a 16-lb shot straight up, giving it a constant upward acceleration from rest of 35.0 m/s2 for 64.0 cm. He releases it
stich3 [128]

Answer:

a. 6.69m/s

b. y=4.48m

c. t=1.43secs

Explanation:

Data given, acceleration,a=35m/s^2

distance covered,d=64cm=0.64m,

a. to determine the speed, we use the equation of motion

initial velocity,u=0m/s

if we substitute values we arrive at

v^{2}=u^{2}+2as\\v^{2}=0+2*35*0.64\\v^{2}=44.8m/s\\v=\sqrt{44.8}\\ v=6.69m/s\\

b. After taking the shot,the acceleration value is due to gravity i.e a=9.81m/s^2

and the distance becomes (y-2.2) above the ground. When it reaches the maximum height, the final velocity becomes zero and the initial velocity becomes 6.69m/s.

Hence we can write the equation above again

v^{2}=u^{2}-2a(y-2.2)\\

if we substitute values we have

v^{2}=u^{2}-2a(y-2.2)\\0=6.69^{2}-2*9.81(y-2.2)\\y-2.2=\frac{44.76}{19.62} \\y=2.28+2.2\\y=4.48m

c. the time it takes to arrive at 1.83m is obtain by using the equation below

1.83-2.2=6.69t-\frac{1}{2} *9.81t^{2}\\4.9t^{2}-6.69t-0.37\\using \\t= \frac{-b±\sqrt{b^{2}-4ac} }{2a}\\ where \\a=4.9, b=-6.69, c=-0.37

if we insert the values, we solve for t , hence t=1.43secs

6 0
3 years ago
9) A bug of mass 0.020 kg is at rest on the edge of a solid cylindrical disk (M = 0.10 kg, R = 0.10 m) rotating in a horizontal
natima [27]

Answer:

a. ω₂ = 14rad/s

b. ∇K.E = 0.014J

c. The bug does not conserve force while moving on the disk (non-conservative force).

Explanation:

Mass of the bug (m) = 0.02kg

Mass of the cylindrical disk (M) = 0.10kg

Radius of the disk (r) = 0.10m

Initial angular velocity ω₁ = 10rad/s

final angular velocity ω₂ = ?

a.

To calculate the new angular velocity, we relate it to the conservation of angular momentum of the system I.e when the bug was at the edge of the disk and when it is located at the centre of the disk.

I = Mr² / 2

I₁ = Mr₂ / 2 + mr²

I₁ = moment of inertia when the bug was at the edge

I₁ = [(0.10 * 0.10²) / 2 ] + (0.02 * 0.1²)

I₁ = 0.0005 + 0.0002

I₁ = 7.0*10⁻⁴kgm²

I₂ = moment of inertia when yhe bug was at the center of the disk.

I₂ = Mr² / 2

I₂ = (0.01 * 0.01²) 2

I₂ = 0.0005kgm²

for conservation of angular momentum,

I₁ω₁ = I₂ω₂

solve for ω₂

ω₂ = (I₁ * ω₁) / I₂

ω₂ = (7.0*10⁻⁴ * 10) / 5.0*10⁻⁴

ω₂ = 14 rad/s

b. the change in kinetic energy of the system is

∇K = K₂ - K₁

∇K = ½I₂*ω₂² - ½I₁*ω₁²

∇K = ½(I₂*ω₂² - I₁ω₁²)

∇K = ½[(5.0*10⁻⁴ * 14²) - (7.0*10⁻⁴ * 10²)]

∇k = ½(0.098 - 0.07)

∇K = ½ * 0.028

∇K = 0.014J

c.

The cause of the decrease and increase in kinetic energy is because the bug uses a non-conservative force. To conserve the mechanical energy of a system, all the forces acting in it must be conservative.

The work W produced by this force brings the difference in kinetic energy of the system

W = K₂ - K₁

6 0
3 years ago
Which statement correctly describes the differences between positive and negative acceleration?
damaskus [11]
In my opinion, the third answer makes the most sense
7 0
3 years ago
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