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3 years ago

The velocity of a 48.0 g shell leaving a 2.95 kg rifle is 391. m/s. What is the recoil velocity of the rifle?

Physics

1 answer:

Monica [59]3 years ago

8 0

Hi there!

$\large\boxed{ -6.36m/s}$

Use the equation:

$v_{2} = -\frac{m_{1}}{m_{2}} v_{1}$

Where m2 and v2 deal with the larger object, and m1 and v1 with the smaller object. Plug in the given values:

v2 = ?

m1 = 0.048 kg (converted)

m2 = 2.95

v1 = 391

$v_{2} = -\frac{0.048}{2.95} *391$

$v_{2} = -6.36m/s$

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3 years ago

A car travels at a steady 40.0 m/s around a horizontal curve of radius 200 m. What is the minimumcoefficient of static friction

zhenek [66]

Answer:

c. 0.816

Explanation:

Let the mass of car be 'm' and coefficient of static friction be 'μ'.

Given:

Speed of the car (v) = 40.0 m/s

Radius of the curve (R) = 200 m

As the car is making a circular turn, the force acting on it is centripetal force which is given as:

Centripetal force is, $F_c=\frac{mv^2}{R}$

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$f=\mu N$

As there is no vertical motion, therefore, $N=mg$ . So,

$f=\mu mg$

Now, the centripetal force is provided by the frictional force. Therefore,

Frictional force = Centripetal force

$f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu=\frac{v^2}{Rg}$

Plug in the given values and solve for 'μ'. This gives,

$\mu = \frac{(40\ m/s)^2}{200\ m\times 9.8\ m/s^2}\\\\\mu=\frac{1600\ m^2/s^2}{1960\ m^2/s^2}\\\\\mu=0.816$

Therefore, option (c) is correct.

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3 years ago

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5. Neutrons, Neutrons

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An atom has 3 subatomic particles:

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3 years ago