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Maksim231197 [3]
3 years ago
15

Can someone help me??

Physics
1 answer:
Dmitry_Shevchenko [17]3 years ago
4 0

Answer:

Bones

Explanation:

There is no explanation

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Nady [450]
It would be 3.15 in meters
8 0
3 years ago
Pure sodium metal placed in water will spark and ignite, as well as form bubbles and gas. What are the signs that this is a chem
ad-work [718]
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  • Heat is produced
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7 0
3 years ago
Read 2 more answers
Two objects are moving in the xy-plane. Object A has a mass of 3.2 kg and has a velocity = (2.3 m/s)i+ (4.2 m/s)j and object B h
prisoha [69]

The total momentum of the system is 2.14i + 21.27j.

A vector quantity with both direction and magnitude is momentum. Kg m/s (kilogram meter per second) or N s serve as its units (newton second).

The total starting momentum of a system must match the entire final momentum of the system since momentum is a conserved quantity. The overall momentum does not change.

The total momentum of the system is defined as follows:

As momentum is vector quantity and vectors can be added, so, the momentum of a system is given by

P = Pₓ + P'

where Pₓ is the x-component of momentum

P' is the y-component of the momentum

Also, we know that

P=mv

where m is mass

v is velocity

Thus,

P = Pₓ + P'

P = m₁vₓ + m₂v'

vₓ is the x-component of the velocity

v' is the y- component of the velocity

Given, m₁= 3.2kg

m₂ = 2.9kg

Now,

P = 3.2 (2.3i + 4.2j) + 2.9 (-1.8i +2.7j)

P = (7.36i + 13.44j) + (-5.22i + 7.83j)

P = 2.14i + 21.27j

Thus, the total momentum of the given system is 2.14i + 21.27j.

Learn more about the momentum here:

brainly.com/question/4956182

#SPJ1

8 0
1 year ago
In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co
labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m

a)

The final velocity of cart 1 after collision is given as:

v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\  Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s

The final velocity of cart 2 after collision is given as:

v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\  Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s

b) Using the law of conservation of energy:

\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm

7 0
4 years ago
A projectile is launched from the ground at an angle of 60o above the horizontal. At what point in its trajectory does it have t
Elodia [21]

Answer:

It's constant everywhere in its trajectory.

Explanation:

the projectile was launched with an initial velocity, the only acceleration that is affecting the projectile's velocity is gravity.

The acceleration of gravity is practically equal everywhere on earth, so during its trajectory, we have to take into consideration only the acceleration because of gravity.

This is only correct because the projectile was launched with an initial velocity and it's not accelerating from rest and then falls.

5 0
3 years ago
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