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sweet [91]
3 years ago
6

A 70 kg stunt pilot begins pulling out of a dive into a vertical circle. If the plane's speed at the lowest point of the circle

is 80 m/s, what is the apparent
weight of the pilot at the lowest point of the pullout? The pilot experiences a force of 5g from the centripetal acceleration at the bottom of the dive.
O
3880 N
о
3430 N
0
4116 N
3986 N
Physics
1 answer:
jok3333 [9.3K]3 years ago
3 0

Answer:

Apparent weight of pilot due to centripetal acceleration:

m v^2 / R = 5 g m = 5 * 70 * 9.8 = 3430 N

Weight of pilot = 70 * 9.8 = 686 N

Total = 3430 + 686 = 4116 N

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Score
hammer [34]

Answer:

(a) J = 10560  kg-m/s (b) 251.42 N

Explanation:

Given that,

Mass, m = 4400 kg

Initial speed, u = 5.2 m/s

Final speed, v = 7.6 m/s

Time, t = 42 s

(a) Let J be the impulse exerted on the vehicle. Impuse is equal to the change in momentum such as :

J = m(v-u)

J = 4400 (7.6-5.2)

J = 10560  kg-m/s

(b) Impulse = Force × t

F=\dfrac{J}{t}\\\\F=\dfrac{10560}{42}\\F=251.42\ N

Hence, this is the required solution.

4 0
3 years ago
A toroidal solenoid has 600 turns, cross-sectional area 6.90 cm2, and mean radius 4.30 cm.
Dahasolnce [82]

(a) The coil's self-inductance is 7.26 mH.

(b) The self-induced emf in the coil is 7.26 V

(c) The direction of the induced emf is from b to a.

<h3>Coil's self-inductance</h3>

L = N²μA/I

L = (600² x 4π x 10⁻⁷ x 6.9 x 10⁻⁴)/(0.043)

L = 7.26 x 10⁻³ H

L = 7.26 mH

<h3>Self-induced emf in the coil</h3>

emf = N(ΔBA)/t

where;

  • B is magnetic field
  • A is area
  • N is number of turns
  • t is time

B = μNI/L

B1 = (4π x 10⁻⁷ x 600 x 5)/0.043

B1 = 0.0876 T

B2 =  (4π x 10⁻⁷ x 600 x 2)/0.043

B2 = 0.035 T

emf = NΔBA/t

emf = (600)(0.0876 - 0.035)(6.9 x 10⁻⁴) / (3 x 10⁻³)

emf = 7.26 V

The direction of the induced emf is always opposite to the direction of the applied current.

Thus, the direction of the induced emf is from b to a.

Learn more about induced emf here: brainly.com/question/13744192

#SPJ1

3 0
2 years ago
There is less difference between the speed of light in glass and the speed of light in water than there is between the speed of
CaHeK987 [17]

Answer: no, the magnifying glass has nothing to do with the speed of light.

Explanation: When light moves between 2 media (refraction), part of it properties that changes during this process is it speed.

The change in speed is dependent on the refractive index of the 2 media and as given by snell's law, the refractive index is inversely proportional to wave speed.

This implies that moving from dense to a more dense medium reduces wave speed and moving from dense to less dense medium increases wave speed.

For the first statement, light moved from glass to water, it implies that it moves from a dense to a less dense medium, it wave speed increases in water.

For the second statement, light moved from glass to air, it implies that it also moves from a dense to a less dense medium and it wave speed will also increase in air.

Looking at both second medium, for the first statement, the second medium is water, and for the second statement, the second medium is air.

Air is less dense that water, hence light travel faster in air than in water.

Thus we can see that the magnification property of a glass has nothing to do with the wave speed, just the refractive indices of the media.

8 0
3 years ago
Which resistors in the circuit are connected in parallel?
GenaCL600 [577]

Answer:

I think it's C - sorry if I'm wrong

8 0
3 years ago
MamaMia's Pizza purchases its pizza delivery boxes from a printing supplier. MamaMia's delivers on-average 200 pizzas each month
Ira Lisetskai [31]

Answer:

a) 138 units

b) 17 units

c) 17 units

d) Total Cost = $353.35

Explanation:

Given:

Average pizzas delivered = 200

Charge of inventory holding = 30% of cost

Lead time = 7 days

Now,

a) Economic Order Quantity =  \sqrt\frac{2\times\textup{Annual Demand}\times\textup{Cost per Order}}{\textup{Carrying cost}}

also,

Annual Demand = 200 × 12 = 2400

Cost per Order = Cost of Box + Processing Costs

= 30 cents + $10

= $10.30

and, Carrying Cost = \frac{\textup{Total Inventory Cost}}{\textup{total annual demand}}

=\frac{\textup{Total Cost per order}\times\textup{Annual demand}\times\frac{25}{100}}{\textup{Annual demand}}

= \frac{\$10.30\times2400}\times\frac{25}{100}}{2400}

= $2.575

Therefore,

Economic Order Quantity =  \sqrt\frac{2\times\textup{2400}\times\textup{10.30}}{\textup{2.575}}

= 138.56 ≈ 138 units

b) Reorder Point

= (average daily unit sales × the lead time in days) + safety stock

= (\frac{200}{30}\times7

= 46.67 ≈ 47 units

c) Number of orders per year = \frac{\textup{Annual Demand}}{\textup{Economic order quantity}}

= \frac{\textup{2400}}{\textup{138}}

= 17.39 ≈ 17 units

d) Total Annual Cost (Total Inventory Cost)

= Ordering Cost + Holding Cost

Now,

The ordering Cost = Cost per Order × Total Number of orders per year

= $10.30 × 17

= $175.1

and,

Holding Cost = Average Inventory Held × Carrying Cost per unit

Average Inventory Held = \frac{\textup{0+138}}{\textup{2}} =  69

Carrying Cost per unit = $2.575

Holding Cost = 69 × $2.575 =  

$177.675

Therefore,

Total Cost = Ordering Cost + carrying cost

= $175.1  + $177.675 = $353.35

5 0
3 years ago
Read 2 more answers
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