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3 years ago

What is the term for the positive charge in an atom?

Physics

2 answers:

kotykmax [81]3 years ago

7 0

The term for the positive charge in an atom is specifically called, "A proton"

dlinn [17]3 years ago

3 0

The answer is A. Proton. Protons have the opposite charge to electrons, which have negative charges.

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Cart 1 has an initial velocity and hits cart 2 which is stationary. after a perfectly inelastic collision, the combined carts ar

Tomtit [17]

Option(a) the mass of cart 2 is twice that of the mass of cart 1 is the right answer.

The mass of cart 2 is twice that of the mass of cart 1 is correct about the mass of cart 2.

Let's demonstrate the issue using variables:

Let,

m1=mass of cart 1

m2=mass of cart 2

v1 = velocity of cart 1 before collision

v2 = velocity of cart 2 before collision

v' = velocity of the carts after collision

Using the conservation of momentum for perfectly inelastic collisions:

m1v1 + m2v2 = (m1 + m2)v'

v2 = 0 because it is stationary

v' = 1/3*v1

m1v1 = (m1+m2)(1/3)(v1)

m1 = 1/3*m1 + 1/3*m2

1/3*m2 = m1 - 1/3*m1

1/3*m2 = 2/3*m1

m2 = 2m1

From this we can conclude that the mass of cart 2 is twice that of the mass of cart 1.

To learn more about inelastic collision visit:

brainly.com/question/14521843

#SPJ4

4 0

2 years ago

Sodium and ice a physical or chemical reaction

Alik [6]

Answer:

physical change uwu owo hope this helps you uwu silly goose

8 0

3 years ago

Read 2 more answers

Compare the kinetic energy of a 20,000-kg truck moving at 110 km/h with that of an 80.0-kg astronaut in orbit moving at 27,500 k

laiz [17]

Kinetic energy is the energy that is possessed by an object that is moving. It is calculated by one-half the product of the mass and the square of the velocity of the object. We calculate as follows:<span>

For the truck,
KE = mv^2 / 2
KE = 20000 kg (110 km/h (1 h/3600 s)(1000 m / 1 km))^2 / 2
KE = 611111.11 J

For the astronaut,
KE = 80.0 kg (27500 km/h</span>(1 h/3600 s)(1000 m / 1 km)<span>)^2/2
KE = 611111.11 J

The kinetic energy possessed by the two bodies are the same.

</span>

3 0

3 years ago

Vectors A and B lie in the xy ‑plane. Vector A has a magnitude of 17.1 and is at an angle of 150.5∘ counterclockwise from the +x

IRINA_888 [86]

Answer:

A = -14.87 i ^ + 8.42 j ^ + 0 k ^

B = -25.41 i ^ -12.0 j ^ + 0 k ^

Explanation:

For this exercise let's use trigonometry by decomposing to vectors

vector A

module 17.1 with an angle of 150.5 counterclockwise.

Sin 150.5 = $A_{y}$ / A

cos 150.5 = Ax / A

A_{y} = A sin 150.5 = 17.1 sin 150.5

Aₓ = A cos 1505 = 172 cos 150.5

A_{y} = 8,420

Aₓ = -14.870

the vector is

A = -14.87 i ^ + 8.42 j ^ + 0 k ^

Vector B

$B_{y}$ = 28.1 sin 205.3

Bₓ = 28.1 cos 205.3

B_{y} = -12.009

Bₓ = -25.405

the vector is

B = -25.41 i ^ -12.0 j ^ + 0 k ^

5 0

3 years ago

An insect 1.1 mm tall is placed 1.0 mm beyond the focal point of the objective lens of a compound microscope. The objective lens

Pavlova-9 [17]

Answer:

Explanation:

For image formation in objective lens

object distance u = 14 +1 = 15 mm

focal length f = 14 mm .

image distance v = ?

lens formula

$\frac{1}{v} -\frac{1}{u} =\frac{1}{f}$

Putting the values

$\frac{1}{v} +\frac{1}{15} =\frac{1}{14}$

v = 210 mm .

B )

magnification = v / u

= 210 / 15

= 14

size of image = 14 x 1.1 mm

= 15.4 mm

= 15 mm approx

C )

For final image to be at infinity , image produced by objective lens must fall at the focal point of eye piece . so objective lens's distance from the image formed by objective must be equal to focal length of eye piece that is 21 mm .

21 mm is the answer .

D )

overall magnification =

$\frac{210}{15} \times \frac{D}{f_e}$

D = 25 cm , f_e = focal length of eye piece

= 14 x 250 / 21

= 166.67

= 170 ( in two significant figures )

7 0

3 years ago