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2 years ago

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To withstand "g-forces" of up to 10 g's, caused by suddenly pulling out of a steep dive, fighter jet pilots train on a "human ce

ntrifuge." 10 g's is an acceleration of 98m/s2. Part A If the length of the centrifuge arm is 10.0 m , at what speed is the rider moving when she experiences 10 g's?

1 answer:

makvit [3.9K]2 years ago

4 0

Answer:

31.3 m/s

Explanation:

Centripetal acceleration = v² / r where r is the length of the arm of the chair

98 m/s² = v² / r

v² = √(98 m/s² × 10) = 31.3 m/s

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A plane flying at 70.0 m/s suddenly stalls. If the acceleration during the stall is 9.8 m/s2 directly downward, the stall lasts

tino4ka555 [31]

Answer:

v = 66.4 m/s

Explanation:

As we know that plane is moving initially at speed of

$v = 70 m/s$

now we have

$v_x = 70 cos25$

$v_x = 63.44 m/s$

$v_y = 70 sin25$

$v_y = 29.6 m/s$

now in Y direction we can use kinematics

$v_y = v_i + at$

$v_y = 29.6 - (9.81 \times 5)$

$v_y = -19.5 m/s$

since there is no acceleration in x direction so here in x direction velocity remains the same

so we will have

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{63.44^2 + 19.5^2}$

$v = 66.4 m/s$

4 0

2 years ago

A man is dragging a trunk up the loading ramp of a mover’s truck. The ramp has a slope angle of 20.0°, and the man pulls upward

AleksandrR [38]

Answer:

(a) 104 N

(b) 52 N

Explanation:

Given Data

Angle of inclination of the ramp: 20°

F makes an angle of 30° with the ramp

The component of F parallel to the ramp is Fx = 90 N.

The component of F perpendicular to the ramp is Fy.

(a)

Let the +x-direction be up the incline and the +y-direction by the perpendicular to the surface of the incline.

Resolve F into its x-component from Pythagorean theorem:

Fx=Fcos30°

Solve for F:

F= Fx/cos30°

Substitute for Fx from given data:

Fx=90 N/cos30°

=104 N

(b) Resolve r into its y-component from Pythagorean theorem:

Fy = Fsin 30°

Substitute for F from part (a):

Fy = (104 N) (sin 30°)

= 52 N

5 0

2 years ago

Honeybees accumulate charge as they fly, and they transfer charge to the flowers they visit. Honeybees are able to sense electri

Vilka [71]

Answer:

ΔE> E_minimo

We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference

Explanation:

For this exercise let's use the electric field expression

E = k q / r²

where k is the Coulomb constant that is equal to 9 109 N m² /C², q the charge and r the distance to the point of interest positive test charge, in this case the distance to the bee

let's calculate the field for each charge

Q = 24 pC = 24 10⁻¹² C

E₁ = 9 10⁹ 24 10⁻¹² / 0.20²

E₁ = 5.4 N / C

Q = 32 pC = 32 10⁻¹² C

E₂ = 9 10⁹ 32 10⁻¹² / 0.2²

E₂ = 7.2 N / C

let's find the difference between these two fields

ΔE = E₂ -E₁

ΔE = 7.2 - 5.4

ΔE = 1.8 N / C

the minimum detection field is

E_minimum = 0.77 N / C

ΔE> E_minimo

We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference

8 0

2 years ago

The human circulation system has approximately 1 × 109 capillary vessels. Each vessel has a diameter of about 8 μm .Assuming the

DerKrebs [107]

To solve this exercise we must apply the concept of Flow as the measure given to determine the volume of a liquid flowing per unit of time, and that can be calculated through velocity and Area, mathematically this can be determined as

$\bar{v}=\frac{Q}{A}$

Q = Discharge of Flow

A = Cross sectional Area

$\bar{v} =$ Velocity

The area of the cross section of the capillary tube is

$A_c = \pi r^2$

$A_c = \pi (\frac{d}{2})^2$

$A_c = \pi (\frac{8*10^{-6}}{2})^2$

$A_c = 5.02685*10^{-11}m^2$

The total Area by this formula:

$A_1 = nA_c$

Where,

$A_c =$ Stands for area of capillary

n = Stands for number of blood vessels

$A_1 = (1*10^9)(5.0265*10^{-11})$

$A_1 = 5.0265*10^{-2}m^2$

Finally replacing at our first equation,

$\bar{v} = (\frac{5L/min}{5.0265*10^{-2}m^2})(\frac{1000cm^3}{1L})(\frac{1min}{60s})$

$\bar{v} = 1.66cm^3/s$

Therefore the average speed, in centimeters per second, of blood flow through each capillary vessel is 1.66cm^3/s

7 0

2 years ago

A CARPENTER WORKING AT A CONSTRUCTION SITE MISTAKENLY DROPPED HIS HAMMER FROM A HEIGHT OF 125m. How long does it take to reach t

viktelen [127]

Answer:

It all depends how heavy the hammer is

Explanation:

6 0

2 years ago

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