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3 years ago

5.0 kg, with a bullet of mass 0.1 kg. The target was mounted on

Physics

1 answer:

kari74 [83]3 years ago

8 0

Answer:

306 m/s

Explanation:

Law of conservation of momentum

m1v1 + m2v2 = (m1+m2)vf

m1 is the bullet's mass so it is 0.1 kg

v1 is what we're trying to solve

m2 is the target's mass so it is 5.0 kg

v2 is the targets velocity, and since it was stationary, its velocity is zero

vf is the velocity after the target is struck by the bullet, so it is 6.0 m/s

plugging in, we get

(0.1 kg)(v1) + (5.0 kg)(0 m/s) = (0.1 kg + 5.0 kg)(6.0 m/s)

(0.1)(v1) + 0 = 30.6

(0.1)(v1) = 30.6

v1 = 306 m/s

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The radius of the aorta is «10 mm and the blood flowing through it has a speed of about 300 mm/s. A capillary has a radius of ab

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Answer:

(I). The effective cross sectional area of the capillaries is 0.188 m².

(II). The approximate number of capillaries is $3.74\times10^{9}$

Explanation:

Given that,

Radius of aorta = 10 mm

Speed = 300 mm/s

Radius of capillary $r=4\times10^{-3}\ mm$

Speed of blood $v=5\times10^{-4}\ m/s$

(I). We need to calculate the effective cross sectional area of the capillaries

Using continuity equation

$A_{1}v_{1}=A_{2}v_{2}$

Where. v₁ = speed of blood in capillarity

A₂ = area of cross section of aorta

v₂ =speed of blood in aorta

Put the value into the formula

$A_{1}=A_{2}\times\dfrac{v_{2}}{v_{1}}$

$A_{1}=\pi\times(10\times10^{-3})^2\times\dfrac{300\times10^{-3}}{5\times10^{-4}}$

$A_{1}=0.188\ m^2$

(II). We need to calculate the approximate number of capillaries

Using formula of area of cross section

$A_{1}=N\pi r_{c}^2$

$N=\dfrac{A_{1}}{\pi\times r_{c}^2}$

Put the value into the formula

$N=\dfrac{0.188}{\pi\times(4\times10^{-6})^2}$

$N=3.74\times10^{9}$

Hence, (I). The effective cross sectional area of the capillaries is 0.188 m².

(II). The approximate number of capillaries is $3.74\times10^{9}$

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3 years ago

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Answer:

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Explanation:

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3 years ago

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While camping in Denali National Park in Alaska, a wise camper hangs his pack of food from a rope tied between two trees, to kee

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The tension in each of the ropes is 625 N.

Draw a free body diagram for the bag of food as shown in the attached diagram. Since the bag hangs from the midpoint of the rope, the rope makes equal angles θ with the horizontal. The tensions <em>T</em> in both the ropes are also equal.

Resolve the tension T in the ropes into horizontal and vertical components T cosθ and T sinθ respectively, as shown in the figure. At equilibrium,

$2T sin\theta = 50 N$ ......(1)

Calculate the value of sinθ using the right angled triangles from the diagram.

$sin\theta =\frac{0.06 m}{1.5 m} =0.04$

Substitute the value of sinθ in equation (1) and simplify to obtain T.

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4 years ago

A bridge that was 5.0 m long has been washed out by the rain several days ago. How fast must a car be going to successfully jump

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Answer:

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It is given that,

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v = 7.83 m/s

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4 years ago