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Aleks04 [339]
3 years ago
15

6) A ball free falls from the top of the roof for 5 seconds. How far did it fall? What is its final

Physics
1 answer:
gayaneshka [121]3 years ago
5 0

Answer:

distance fallen = 122.5 m

speed = 49 m/s

Explanation:

Δd = vi*t - 0.5at^2

Δd = 0 - 0.5(9.8)(5^2)

Δd = -122.5 (i.e. 122.5m down)

a = (vf - vi) / t

9.8 = (vf - 0) / 5

vf = 49

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A manufacturer provides a warranty against failure of a carbon steel product within the first 30 days after sale. Out of 1000 so
hodyreva [135]

Answer:A) Risk(R)= $1000

B) There is justification for spending an additional cost of $100 to prevent a corrosion whose consequence in monetary terms is $1000

Explanation:R= Risk,

P=Probability of failure

C= Consequence of failure

Mathematically, R=P ×C

10 out of 1000 carbon-steal products failed

Probability of failure= 10/1000 =0.01

The consequence of failure by corrosion given in monetary term =$100,000

Risk of failure = 0.01 × $100,000

R=$1000

4 0
3 years ago
What minimum speed must the block have at the base of the 70 m hill to pass over the pit at the far (right-hand) side of that hi
Drupady [299]

Answer:

initial velocity is v = 4.95 m / s

Explanation:

To solve this exercise we use the projectile launch ratios, when the block leaves the hill its speed is horizontal, let's find the time it takes to fall to the other point.

Initial vertical velocity is zero

          y = y₀ + v_{oy} t - ½ g t²

          y-y₀ = 0 -1/2 g t²

          t = \sqrt{ \frac{ 2(y_o -y)}{g} }

calculate

          t = \sqrt{ \frac{2 ( 70-50)}{9.8} }

          t = 2.02 s

with this time we can substitute in the horizontal displacement equation

          x = v₀ₓ t

          v₀ₓ = x / t

suppose that the distance between the two points is x = 10 m

          v₀ₓ = 10 / 2.02

          v₀ₓ = 4.95 m / s

initial velocity is v = 4.95 m / s

4 0
3 years ago
How can a small human retina detect objects larger than itself?
aliya0001 [1]

Exactly the same way that you can photograph a mountain or a skyscraper
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7 0
3 years ago
Read 2 more answers
Two long, parallel transmission lines, 40.0cm apart, carry 25.0-A and 73.0-A currents.A). Find all locations where the net magne
In-s [12.5K]

Answer:

a) If the currents are in the same direction, the magnetic field is zero at x = 0.298 m = 29.8 cm

That is, in between the wires, 29.8 cm from the 73.0 A wire and 10.2 cm from the 25.0 A wire.

b) If the currents are in opposite directions, the magnetic field is zero at x = 0.608 m = 60.8 cm

That is, along the positive x-axis, 60.8 cm from the 73.0 A wire and 20.8 cm from the 25.0 A wire.

Explanation:

The origin is at the 73.0 A wire and the 25.0 A wire is at x = 0.40 m

The magnetic field in a current carrying wire at a distance r from the wire is given by

B = (μ₀I/2πr)

μ₀ = magnetic constant = (4π × 10⁻⁷) H/m

a) If the currents are in the same direction, at what positions is the magnetic field equal to 0.

According to laws describing the direction.of magnetic fields, this position will be at some point between the two wires.

The magnetic field due to the 73.0 A wire points out of the book, at points along the positive x-axis while the magnetic field due to the 25.0 A wire points into the plane of the book, moving in the negative x-direction.

Hence,

For the 73.0 A wire, I₁ = 73.0 A, r₁ = x

For the 25.0 A wire, I₂ = 25.0 A, r₂ = (0.4 - x)

B = B₁ - B₂ = 0

(μ₀/2π) [(I₁/r₁) - (I₂/r₂)] = 0

(I₁/r₁) = (I₂/r₂)

(I₁/x) = [I₂/(0.4-x)]

(73/x) = [25/(0.4-x)]

73(0.4-x) = 25x

29.2 - 73x = 25x

73x + 25x = 29.2

98x = 29.2

x = (29.2/98) = 0.298 m

b) If the currents are in the opposite directions, at what positions is the magnetic field equal to 0?

According to laws describing the direction.of magnetic fields, this position will be at some point beyond the second wire (since we're initially concerned about the positive x-direction).

The magnetic field due to the 73.0 A wire points out of the book, at points along the positive x-axis while the magnetic field due to the 25.0 A wire (whose direction is now in the opposite direction to the current in the first wire) is also along the positive x-direction.

Hence,

For the 73.0 A wire, I₁ = 73.0 A, r₁ = x

For the 25.0 A wire, I₂ = 25.0 A, r₂ = (x - 0.4)

B = B₁ - B₂ = 0

(μ₀/2π) [(I₁/r₁) - (I₂/r₂)] = 0

(I₁/r₁) = (I₂/r₂)

(I₁/x) = [I₂/(x-0.4)]

(73/x) = [25/(x-0.4)]

73(x-0.4) = 25x

73x - 29.2 = 25x

73x - 25x = 29.2

48x = 29.2

x = (29.2/48) = 0.608 m

Hope this Helps!!!

5 0
3 years ago
A school bus transporting students to school is an example of
Andru [333]
The last one, Kinetic energy bc it is taking them to school
4 0
3 years ago
Read 2 more answers
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