All categories

3 years ago

The momentum of a piston is +9 kg-m/s. If it is moving at +60 m/s, what is its mass?

Physics

1 answer:

vesna_86 [32]3 years ago

5 0

Momentum = mass times velocity
Therefore mass= momentum/velocity
M= 9/60=0,15kg

You might be interested in

. An object whose mass is 375 lb falls freely under the influence of gravity from an initial elevation of 253 ft above the surfa

lozanna [386]

Answer:

(a) Vf = 128 ft/s

(b) K.E = 122.8 Btu

Explanation:

(a)

In order to find the velocity of the object just before striking the surface of earth or the final velocity, we use 3rd equation of motion:

2gh = Vf² - Vi²

where,

g = 32.2 ft/s²

h = height = 253 ft

Vf = Final Velocity = ?

Vi = Initial Velocity = 10 ft/s

Therefore,

(2)(32.2 ft/s²)(253 ft) = Vf² - (10 ft/s)²

16293.2 ft²/s² + 100 ft²/s² = Vf²

Vf = √(16393.2 ft²/s²)

<u></u>

(b)

The kinetic energy of the object before it hits the surface of earth is given by:

K.E = (0.5)(m)(Vf)²

where,

m = mass of object = 375 lb

K.E = Kinetic energy of object before it strikes the surface of earth = ?

Therefore,

K.E = (0.5)(375 lb)(128 ft/s)²

K.E = 3073725 lb.ft²/s²

Now, converting this to Btu:

K.E = (3073725 lb.ft²/s²)(1 Btu/25037 lb.ft²/s²)

3 0

3 years ago

A 600 N force acts on an object with a mass of 50 kg. What is the resulting acceleration of the object?

DochEvi [55]

Answer:

<h3>The answer is 12 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

f is the force

m is the mass

From the question we have

$a = \frac{600}{50} = \frac{60}{5} \\$

We have the final answer as

Hope this helps you

4 0

3 years ago

A pendulum is suspended from the cusp of a cycloid cut in rigid support. The path described by the pendulum bob is cycloidal and

oksano4ka [1.4K]

Answer:

Verified that he oscillations are exactly isosynchronous with frequency ω0 = p g/l, independent of the amplitude.

Explanation:

Starting from the first principle for the derivation and to prove that the oscillations are exactly isosynchronous with frequency ω0 = p g/l, independent of the amplitude. The mathematical manipulations was applied, trigonometric identities was also applied.The steps and explanation are shown in the attachment.