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sveta [45]
3 years ago
15

Please need help on this

Physics
1 answer:
ale4655 [162]3 years ago
7 0

the answer is the third one down

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If all the magnetic domains in a sample of ferromagnetic material point in the same direction, then around the sample we definit
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Answer:

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Explanation:

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5 0
4 years ago
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The half life of uranium-235 is 4. 5 billion years. If 0. 5 half-lives have elapsed, how many years have gone by?
igor_vitrenko [27]

Answer:

2.25 billion years

Explanation:

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5 0
2 years ago
An object with a momentum of 1500 kg-m/s directed east is acted upon by an impulse of 100.0 kg-m/s in the same direction. What i
fiasKO [112]

Answer:

The final momentum of the body = 1600 kgm/s

Thus the impulse that acted on the body. = 500 N.s

Explanation:

Momentum: This can be defined as the product of mass and velocity. The S.I unit of momentum is kgm/s. Momentum is a vector quantity.

Mathematically, momentum can be expressed as

M =mv.

Where M = momentum of the object, m = mass of the object, v = velocity of the object.

Impulse acting on the object = Final momentum of the object - initial momentum of the object

I = M₂ - M₁

M₂ = I + M₁......................... Equation 1

Where I = impulse, M₁ and M₂ = Final and initial momentum.

Note:

(i) The momentum and impulse act in the same direction

(ii) impulse is also a vector quantity.

Given: M₁= 1500 kgm/s, I = 100 kgm/s.

Substituting these values into equation 1

M₂ = 1500 + 100

M₂ = 1600 kgm/s.

Thus the final momentum of the body = 1600 kgm/s

14.

I = m(v-u)............................................... Equation 2.

Where I = impulse on the object, m = mass of the object, v = final velocity, u = initial velocity.

Given: m = 100 kg, v = 15 m/s, u = 10 m/s.

I = 100(15-10)

I = 100(5)

I = 500 N.s or 500 kgm/s.

Thus the impulse that acted on the body. = 500 N.s

7 0
4 years ago
PLS HELP, WILL GIVE BRAINLIEST!!!
Yuki888 [10]

Answer:

b bro it's b bro

6 0
3 years ago
Two point charges each have a value of 3.0 c and are separated by a distance of 4.0 m. what is the electric field at a point mid
swat32
 <span>Place a test charge in the middle. It is 2cm away from each charge. 
The electric field E= F/Q where F is the force at the point and Q is the charge causing the force in this point. 
The test charge will have zero net force on it. The left 30uC charge will push it to the right and the right 30uC charge will push it to the left. The left and right force will equal each other and cancel each other out. 
THIS IS A TRICK QUESTION. 
THe electric field exactly midway between them = 0/Q = 0. 
But if the point moves even slightly you need the following formula 
F= (1/4Piε)(Q1Q2/D^2) 
Assume your test charge is positive and make sure you remember two positive charges repel, two unlike charges attract. Draw the forces on the test charge out as vectors and find the magnetude of the force, then divide by the total charge to to find the electric field strength:)</span>
4 0
3 years ago
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