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3 years ago

Describe the forces involved in Throwing a spear

Physics

1 answer:

Alexeev081 [22]3 years ago

8 0

Explanation:

The center of gravity is near the grip and does not change during throw. "Throwing through the tip," a popular term of how to throw a javelin, means throwing through the grip or center of gravity. The center of pressure is the aerodynamic force of drag and lift on the javelin.

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Cara is building a model of the solar system, which includes the Sun. She plans to include a written description to provide deta

Paladinen [302]

Answer:

She should explain that the Sun is made up of gaseous layers that surround an iron core.

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3 years ago

An example of a constant acceleration is

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An example is free fall ,

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3 years ago

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You want to examine the hairy details of your favorite pet caterpillar, using a lens of focal length 8.9 cm 8.9 cm that you just

Zepler [3.9K]

Answer:

The angular magnification is $M = 2.808$

Explanation:

From the question we are told

The focal length is $f = 8.9cm$

The near point is $n_p = 25.0cm$

The angular magnification is mathematically represented as

$M = \frac{n_p}{f}$

Substituting values

$M = \frac{25}{8.9}$

$= 2.808$

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3 years ago

3. A runner has constant acceleration for 6.4 seconds over a 20 m distance starting from rest.

garri49 [273]

Answer:

You need to use the equations of motion which I wrote down in the image. Since the runner starts from rest, the initial velocity(v0) is 0 and initial position(x0) is also 0. The solution is in the image attached below.

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3 years ago

Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point

Salsk061 [2.6K]

Answer:

The fluids speed at a) $0.105\,m^{2}$ and b) $0.047\,m^{2}$ are $2.33\,\frac{m}{s^{2}}$ and $5.21\,\frac{m}{s^{2}}$ respectively

c) Th volume of water the pipe discharges is: $882\,m^{3}$

Explanation:

To solve a) and b) we should use flow continuity for ideal fluids:

$\Delta Q=0$ (1)

With Q the flux of water, but Q is $Av$ using this on (1) we have:

$A_{2}v_{2}-A_{1}v_{1}=0$ (2)

With A the cross sectional areas and v the velocities of the fluid.

a) Here, we use that point 2 has a cross-sectional area equal to $A_{2}=0.105\,m^{2}$ , so now we can solve (2) for $v_{2}$ :

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.105}\approx2.33\,\frac{m}{s^{2}}$

b) Here we use point 2 as $A_{2}=0.047\,m^{2}$ :

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.047}\approx5.21\,\frac{m}{s^{2}}$

c) Here we need to know that in this case the flow is the volume of water that passes a cross-sectional area per unit time, this is $Q=\frac{V}{t}$ , so we can write:

$A_{1}v_{1}=\frac{V}{t}$ , solving for V:

$V=A_{1}v_{1}t=(0.070m^{2})(3.5\frac{m}{s})(3600s)=882\,m^{3}$

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4 years ago