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skelet666 [1.2K]

3 years ago

5

A girl pulls on a 10-kg wagon with a constant horizontal force of 30 N. If there are no other horizontal forces, what is the wag

on's acceleration in meters per second per second?

1 answer:

strojnjashka [21]3 years ago

7 0

Force equals mass*distance
F = ma

Given m = 10 kg, F = 30 N

30 = 10a
30/10 = a
3 = a

The wagon's acceleration is 3 m/s^2

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A barbell consists of two small balls, each with mass m at the ends of a very low mass rod of length d. The barbell is mounted o

sveta [45]

The total angular momentum of the system about point B is $L=m_1r_1\omega_1+m_2r_2\omega_2$

Angular momentum, also known as moment of momentum or rotational momentum, is the rotating counterpart of linear momentum.

A rigid object's angular momentum is defined as the product of its moment of inertia and its angular velocity. If there is no external torque on the object, it is analogous to linear momentum and is subject to the fundamental constraints of the conservation of angular momentum principle. The vector quantity angular momentum It is derived from the expression for a particle's angular momentum.

Given,

mass of ball 1 = m1

m₂ mass of ball 2=m2

v₁ is the velocity of ball=r₁ω₁

v₂ is the velocity of ball 2=r₂ω₂

The total angular momentum is given as;

$V_{total}=r_1\omega_1+r_2\omega_2\\\\L=m_1r_1\omega_1+m_2r_2\omega_2$

Hence the total angular momentum will be $L=m_1r_1\omega_1+m_2r_2\omega_2$

To learn more about angular momentum refer here

brainly.com/question/29512279

#SPJ4

6 0

1 year ago

Giving brainiest to correct answer.

mixas84 [53]

Answer:

$5.33\ m/s$

Explanation:

$We\ know\ that,\\Momentum=Mass*Velocity\\p=mv\\Hence,\\Lets\ first\ consider\ the\ case\ of\ the\ two\ balls\ 'Before\ Collision':\\\\Mass\ of\ the\ green\ ball=0.2\ kg\\Initial\ Velocity\ of\ the\ green\ ball=5\ m/s\\Initial\ Momentum\ of\ the\ green\ ball=5*0.2=1\ kg\ m/s\\\\Mass\ of\ the\ pink\ ball=0.3\ kg\\Initial\ Velocity\ of\ the\ pink\ ball=2\ m/s\\Initial\ Momentum\ of\ the\ pink\ ball=0.3*2=0.6\ kg\ m/s\\\\Total\ momentum\ of\ both\ the\ balls\ 'Before\ Collision'=1+0.6=1.6\ kg\ m/s$

$Hence,\\Lets\ now\ consider\ the\ case\ of\ the\ two\ balls\ 'After\ Collision':\\\\Mass\ of\ the\ green\ ball=0.2\ kg\\Final\ Velocity\ of\ the\ green\ ball=0\ m/s\\Final\ Momentum\ of\ the\ green\ ball=0\ kg\ m/s\\\\Mass\ of\ the\ pink\ ball=0.3\ kg\\Final\ Velocity\ of\ the\ pink\ ball=v\ m/s\\Final\ Momentum\ of\ the\ pink\ ball=0.3*v=0.3v\ kg\ m/s\\\\Total\ momentum\ of\ both\ the\ balls\ 'After\ Collision'=0+0.3v=0.3v\ kg\ m/s$

$As\ we\ know\ that,\\Through\ the\ law\ of\ conservation\ of\ momentum,\\In\ an\ isolated\ system:\\Total\ Momentum\ Before\ Collision=Total\ Momentum\ After\ Collision\\Hence,\\1.6=0.3v\\v=\frac{1.6}{0.3}=5.33\ m/s$

5 0

3 years ago

How do you do this?Or what are the formulus

LuckyWell [14K]

You got the formulas on the sheet on the top :) So just use those, exchanging v (as in velocity, expressed in m/s) and the d (in meters) and t (in seconds). Hope you will manage it.

4 0

3 years ago

The tension in a string from which a 4.0 kg object is suspended in an elevator is equal to 44 N. What is the acceleration of the

Ronch [10]

Answer: 1

Explanation:

Given

Tension is the string $T=44\ N$

mass of object $m=4\ kg$

Tension is greater than the weight of the object i.e. elevator is moving upward

we can write

$\Rightarrow T-mg=ma\\\Rightarrow T=m(g+a)\\\Rightarrow 44=4(10+a)\\\Rightarrow 11=10+a\\\Rightarrow a=1\ m/s^2$

8 0

3 years ago

how much time would it take for the sound of a thunder to travel 1,500 meters if sound travelers at a speed of 330 m/sec?

inysia [295]

(1,500 meters) x (1 sec/330 meters) =

(1,500 / 330) (meters-sec/meters) =

4.55 seconds

8 0

3 years ago