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3 years ago

8

A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to

the victim with an initial velocity of 2.12 m/s and observes that it takes 1.5 s to reach the water.
(a) List the knowns in this problem.
(b) How high above the water was the preserver released? Note that the downdraft of the helicopter reduces the effects of air resistance on the falling life preserver, so that an acceleration equal to that of gravity is reasonable.

1 answer:

Y_Kistochka [10]3 years ago

8 0

2.12m/s is water and if it goes 1.5 seconds fast

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A pressure cylinder has a diameter of 150-mm and has a 6-mm wall thickness. What pressure can this vessel carry if the maximum s

myrzilka [38]

Answer:

p = 8N/mm2

Explanation:

given data ;

diameter of cylinder = 150 mm

thickness of cylinder = 6 mm

maximum shear stress = 25 MPa

we know that

hoop stress is given as = $\frac{pd}{2t}$

axial stress is given as = $\frac{pd}{4t}$

maximum shear stress = (hoop stress - axial stress)/2

putting both stress value to get required pressure

$25 = \frac{ \frac{pd}{2t} -\frac{pd}{4t}}{2}$

$25 = \frac{pd}{8t}$

t = 6 mm

d = 150 mm

therefore we have pressure

p = 8N/mm2

7 0

3 years ago

A loop of wire is carrying current of 2 A . The radius of the loop is 0.4 m. What is the magnetic field at a distance 0.09 m alo

HACTEHA [7]

Answer:

$B=2.91\ \mu T$

Explanation:

Given that,

The current in the loop, I = 2 A

The radius of the loop, r = 0.4 m

We need to find the magnetic field at a distance 0.09 m along the axis and above the center of the loop. The formula for the magnetic field at some distance is given as follows :

$B=\dfrac{\mu_o}{4\pi }\dfrac{2\pi r^2 I}{(r^2+d^2)^{3/2}}$

Put all the values,

$B=10^{-7}\times \dfrac{2\pi \times 0.4^2 \times 2}{(0.4^2+0.09^2)^{3/2}}\\\\=2.91\times 10^{-6}\ T\\\\or\\\\B=2.91\ \mu T$

So, the required magnetic field is equal to $2.91\ \mu T$ .

3 0

3 years ago

The function s(t) represents the position of an object at time t moving along a line. Suppose s (2 )equals 146and s (6 )equals

sineoko [7]

Answer:

v(t) = 27 units

Explanation:

The function s(t) represents the position of an object at time t moving along a line such that,

$s(2)=146$

and

$s(6)=254$

We need to find the average velocity of the object over the interval of time [2,6]. The velocity of the object is equal to the total distance divided by time. It is given by :

$v(t)=\dfrac{s(6)-s(2)}{6-2}$

$v(t)=\dfrac{254-146}{6-2}$

v(t) = 27 units

So, the average velocity of the object is 27 units. Hence, this is the required solution.

5 0

3 years ago

What times what equals 400

Radda [10]

If you're willing to consider fractions or decimals,
then there are an infinite number of answers.
Like (2.5 x 160), and (15 x 26-2/3).

If you want to stick to only whole numbers,
then these 8 combinations do:

1, 400
2, 200
4, 100
5, 80
8, 50
10, 40
16, 25
20, 20

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4 years ago

Milk with a density of 1020 kg/m3 is transported on a level road in a 9-m-long, 3-m-diameter cylindrical tanker. The tanker is c

jeka57 [31]

Solution :

Given data is :

Density of the milk in the tank, $\rho = 1020 \ kg/m^3$

Length of the tank, x = 9 m

Height of the tank, z = 3 m

Acceleration of the tank, $a_x = 2.5 \ m/s^2$

Therefore, the pressure difference between the two points is given by :

$P_2-P_1 = -\rho a_x x - \rho(g+a)z$

Since the tank is completely filled with milk, the vertical acceleration is $a_z = 0$

$P_2-P_1 = -\rho a_x x- \rho g z$

Therefore substituting, we get

$P_2-P_1=-(1020 \times 2.5 \times 7) - (1020 \times 9.81 \times 3)$

$=-17850 - 30018.6$

$=-47868.6 \ Pa$

$=-47.868 \ kPa$

Therefore the maximum pressure difference in the tank is Δp = 47.87 kPa and is located at the bottom of the tank.

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3 years ago