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3 years ago

If the truck has a mass of 2,000 kilograms , what's its momentum?(v=35 m/s)

Physics

1 answer:

katen-ka-za [31]3 years ago

8 0

Answer:

$\boxed {\boxed {\sf 70,000 \ kg*m/s}}$

Explanation:

Momentum is the product of mass and velocity.

$p=m*v$

The mass of the truck is 2,000 kilograms and the velocity is 35 meters per second.

$m= 2000 \ kg \\v= 35 \ m/s$

Substitute the values into the formula and multiply.

$p= 2000 \ kg * 35 \ m/s \\p= 70,000 \ kg*m/s$

The truck's momentum is <u>70,000 kilograms meters per second.</u>

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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

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

$P = 15.93 k N$

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brainly.com/question/12791045

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