1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
masha68 [24]
4 years ago
4

A firecracker breaks up into several pieces, one of which has a mass of 200 g and flies off along

Physics
1 answer:
labwork [276]4 years ago
5 0

Explanation:

It is given that,

Mass of first piece, m_1=200\ g=0.2\ kg

Speed of first piece, v_x=82\ m/s

Mass of second piece, m_2=300\ g=0.3\ kg

Speed of first piece, v_x=45\ m/s

The momentum along x axis is given by :

p_x=m_1\times v_x

p_x=0.2\times 82=16.4\ kg-m/s

The momentum along y axis is given by :

p_y=m_2\times v_y

p_y=0.3\times 45=13.5\ kg-m/s

Let p is the total momentum of these two pieces. Its magnitude is given by :

p=\sqrt{p_x^2+p_y^2}

p=\sqrt{16.4^2+13.5^2}

p = 21.24 kg-m/s

The direction of total momentum is given by :

tan\theta=\dfrac{p_y}{p_x}

tan\theta=\dfrac{13.5}{16.4}

\theta=39.4^{\circ}

So, the magnitude and direction of the total momentum  of these two pieces are 21.2 kg m/s at 39.5 from the x-axis. Hence, this is the required solution.

You might be interested in
A uniform 1.0-N meter stick is suspended horizontally by vertical strings attached at each end. A 2.0-N weight is suspended from
fgiga [73]

Answer:

3.5 N

Explanation:

Let the 0-cm end be the moment point. We know that for the system to be balanced, the total moment about this point must be 0. Let's calculate the moment at each point, in order from 0 to 100cm

- Tension of the string attached at the 0cm end is 0 as moment arm is 0

- 2 N weight suspended from the 10 cm position: 2*10 = 20 Ncm clockwise

- 2 N weight suspended from the 50 cm position: 2*50 = 100 Ncm clockwise

- 1 N stick weight at its center of mass, which is 50 cm position, since the stick is uniform: 1*50 = 50 Ncm clockwise

- 3 N weight suspended from the 60 cm position: 3*60 = 180 Ncm clockwise

- Tension T (N) of the string attached at the 100-cm end: T*100 = 100T Ncm counter-clockwise.

Total Clockwise moment = 20 + 100 + 50 + 180 = 350Ncm

Total counter-clockwise moment = 100T

For this to balance, 100 T = 350

so T = 350 / 100 = 3.5 N

4 0
3 years ago
ASAP pls need help will give 25 points per person
faltersainse [42]

Explanation:

1. inertia

2.unbalanced?

3.inertia

4.more

5.inertia

6.inertia

7.unbalanced

8 0
3 years ago
An object is sitting on the floor. A 22.4 N force is pulling the object to the right and an 11 N force is pulling the object to
Nimfa-mama [501]

answer = 33.4 net force.

5 0
3 years ago
1 Table Exercise the object released from atop of building house of heigh 10m . Calculaie a final velocity if it time is 4s​
Alekssandra [29.7K]

Answer:

What is velocity?  

v  = d/t = distance / time     you need to know this formula !

= 10 m/ 4 s = 2.5 m/s    

Explanation:

7 0
3 years ago
PLEASE HELP <3
aliya0001 [1]

1) The landing spot of the projectile is given by d=v_x \sqrt{\frac{2h}{g}}

2) The common variable is the time

Explanation:

1)

The motion of a projectile consists of two independent motions:  

- A uniform motion (constant velocity) along the horizontal direction  

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction  

The landing spot can be determined in the following way:

- First of all, we analyze the vertical motion to find the time of flight of the projectile. This can be done by using the suvat equation

h=ut+\frac{1}{2}at^2

where

h is the vertical displacement of the projectile, which corresponds to the height from which the projectile has been fired, above the ground

u = 0 is the initial vertical velocity

a=g=9.8 m/s^2 is the acceleration of gravity

t is the time of flight

Solving for t,

t=\sqrt{\frac{2h}{g}}

- After finding the time of flight, we analyze the horizontal motion, which is a uniform motion with constant horizontal velocity v_x. Therefore, the horizontal distance covered is given by

d=v_x t

And substituting the time of flight,

d=v_x \sqrt{\frac{2h}{g}}

2)

Since the horizontal motion is uniform, the horizontal component of the displacement of the projectile is given by

x=v_x t

where v_x is the horizontal velocity and t is the time.

The vertical motion is accelerated, so the vertical component of the displacement is given by

y=\frac{1}{2}gt^2

where g is the acceleration of gravity and t is the time.

Therefore, from the two equations we see that the common variable is t, the time.

Learn more about projectile:

brainly.com/question/8751410

#LearnwithBrainly

6 0
3 years ago
Other questions:
  • Two capacitors A and B are connected in
    13·1 answer
  • What is the Scientific definition of friction?
    13·2 answers
  • Consider 1 mol an ideal gas at 28∘ C and 1.06 atm pressure. To get some idea how close these molecules are to each other, on the
    9·1 answer
  • Steam at 700 bar and 600 oC is withdrawn from a steam line and adiabatically expanded to 10 bar at a rate of 2 kg/min. What is t
    14·1 answer
  • Three balls are thrown off a tall building with the same speed but in different directions. Ball A is thrown in the horizontal d
    9·1 answer
  • From Smallest to Largest <br> Moon<br> asteroid <br> meteorite <br> meteoroid
    9·2 answers
  • a fire woman dropped a person onto the safety net. right before the person hit the net he had a velocity of 11.2m/s and 1800j of
    12·2 answers
  • The tachometer in the author's car has a maximum reading of 16,000 rpm. (A tachometer is a gauge that shows the angular speed of
    13·1 answer
  • For an object on a flat surface, the force of gravity is 15 newtons downward and the normal force is 15 newtons upward. The appl
    11·1 answer
  • a generator uses loops of area 0.239 m^2, rotating 373 rad/s in a 0.0639 t magnetic field. how many loops must the coil have to
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!