A 35 grams bullet travels with a velocity of magnitude 126 km/h. What is the bullet's linear momentum?

A 0.500-kg block, starting at rest, slides down a 30.0° incline with static and kinetic friction coefficients of 0.350 and 0.250

Leviafan [203]

Answer:x=23.4 cm

Explanation:

Given

mass of block $m=0.5 kg$

inclination $\theta =30$

coefficient of static friction $\mu =0.35$

coefficient of kinetic friction $\mu _k=0.25$

distance traveled $d=77.3 cm$

spring constant $k=35 N/m$

work done by gravity+work done by friction=Energy stored in Spring

$mg\sin \theta d-\mu _kmg\cos \theta d=\frac{kx^2}{2}$

$mgd\left ( \sin \theta -\mu _k\cos \theta \right )=\frac{kx^2}{2}$

$0.5\times 9.8\times 0.773\left ( \sin 30-0.25\cos 30\right )=\frac{35\times x^2}{2}$

$x=\sqrt{\frac{2\times 0.5\times 9.8\times 0.773(\sin 30-0.25\times \cos 30)}{35}}$

$x=0.234 m$

$x=23.4 cm$

6 0

3 years ago

What inertia is present in streched rybber

MrRissso [65]

<h2>You input potential (stored) energy into the rubber band system when you stretched the rubber band back. Because it is an elastic system, this kind of potential energy is specifically called elastic potential energy.</h2><h2>Hope it helps..</h2>

4 0

3 years ago

The linear momentum of a car of mass 1000 kg moving with a speed of 10 m/s is-----kg.m/s. HELP ME

Assoli18 [71]

Answer:

4

Explanation:

The kilogram-meter per second (kg · m/s or kg · m · s -1 ) is the standard unit of momentum . Reduced to base units in the International System of Units ( SI ), a kilogram-meter per second is the equivalent of a newton-second (N · s), which is the SI unit of impulse .

5 0

3 years ago

I NEED HELP ASAP!!!!!

Ratling [72]

Answer:

D) momentum of cannon + momentum of projectile= 0

Explanation:

The law of conservation of momentum states that the total momentum of an isolated system is constant.

In this case, the system cannon+projectile can be considered as isolated, because no external forces act on it (in fact, the surface is frictionless, so there is no friction acting on the cannon). Therefore, the total momentum of the two objects (cannon+projectile) must be equal before and after the firing:

$p_i = p_f$