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3 years ago

1. calculate agni's average speed during the race.

Physics

1 answer:

MakcuM [25]3 years ago

7 0

The formula to find the speed/velocity is :
v = Δx/Δt = ( $\frac{ x_{final} - x_{initial} }{ t_{travel} }$

B.
1. From the graph given, we can get an information such as the distance travelled, and the time given.
2. If the graph are horizontal, it means that the Agni's speed is zero because Agni doesn't move.
3. If the question ask the velocity during all time, we can say that the velocity is :
. $v = \frac{x}{t} = \frac{50 km}{2.5 hour} = 20 \frac{km}{h}$
4. So, the final velocity can be found by looking the third part of the graph. So,..
$v = \frac{x}{t} = \frac{(50-35) km }{0.5 hour} = 30 \frac{km}{h}$

Hope this will help you :)

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2 years ago

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Hello there, and thank you for posting your question here on brainly.

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3 years ago

A 50 mm diameter steel shaft and a 100 mm long steel cylindrical bushing with an outer diameter of 70 mm have been incorrectly s

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Answer:

The axial force is $P = 15.93 k N$

Explanation:

From the question we are told that

The diameter of the shaft steel is $d = 50mm$

The length of the cylindrical bushing $L =100mm$

The outer diameter of the cylindrical bushing is $D = 70 \ mm$

The diametral interference is $\delta _d = 0.005 mm$

The coefficient of friction is $\mu = 0.2$

The Young modulus of steel is $207 *10^{3} MPa (N/mm^2)$

The diametral interference is mathematically represented as

$\delta_d = \frac{2 *d * P_B * D^2}{E (D^2 -d^2)}$

Where $P_B$ is the pressure (stress) on the two object held together

So making $P_B$ the subject

$P_B = \frac{\delta _d E (D^2 - d^2)}{2 * d* D^2}$

Substituting values

$P_B = \frac{(0.005) (207 *10^{3} ) * (70^2 - 50^2))}{2 * (50) (70) ^2 }$

$P_B = 5.069 MPa$

Now he axial force required is

$P = \mu * P_B * A$

Where A is the area which is mathematically evaluated as

$\pi d l$

So $P = \mu P_B \pi d l$

Substituting values

$P = 0.2 * 5.069 * 3.142 * 50 * 100$

$P = 15.93 k N$

8 0

3 years ago

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vekshin1

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3 years ago

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kotykmax [81]

Answer:

M₂ = M then L₂ = L

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Explanation:

This is a static equilibrium exercise, to solve it we must fix a reference system at the turning point, generally in the center of the rod. By convention counterclockwise turns are considered positive

∑ τ = 0

The mass of the rock is M and placed at a distance, L the mass of the rod M₁, is considered to be placed in its center of mass, which by uniform e is in its geometric center (x = 0) and the triangular mass M₂, with a distance L₂

The triangular shape of the second object determines that its mass can be considered concentrated in its geometric center (median) that tapers with a vertical line if the triangle is equilateral, the most used shape in measurements.

M L + M₁ 0 - m₂ L₂ = 0

M L - m₂ L₂ = 0

L₂ = $\frac{M}{M_{2}}$ L

From this answer we have several possibilities

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3 years ago