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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.

You might be interested in

The Displacement is 5m. We found that using the

Llana [10]

Answer:

A vector can be written as:

(R, θ)

Where R is the magnitude, in this case, we know that the magnitude of the displacement is 5m

Then:

R = 5m

and θ defines the direction, it's an angle measured from the positive x-axis.

(In the image, θ would be the angle located at the point A)

Now, if you look at the image, you can see a triangle rectangle.

Where the adjacent cathetus has a length of 4,

the opposite cathetus has a length of 3 units

the hypotenuse has a length of 5 units.

So we can use any trigonometric rule to find the value of θ, like:

sin(θ) = (opposite cathetus)/hypotenuse

Then:

sin(θ) = 3m/5m

Now we can use the inverse sin function, Asin(x), in both sides

Asin( sin(θ)) = θ = Asin( 3/5) = 36.87°

then the vector is:

(5m, 36.87°)

Now, if we define the positive y-axis as the North, and the positive x-axis as the East.

This vector would point at 36.87° North of East.

(or almost Northeast)

7 0

2 years ago

A wire carrying a current of 26.9 A is bent into a circular arc with a radius of 0.6 cm that sweeps out 0.900 radians. What is t

melomori [17]

The magnetic field at the center of the arc is 4 × 10^(-4) T.

To find the answer, we need to know about the magnetic field due to a circular arc.

<h3>What's the mathematical expression of magnetic field at the center of a circular arc?</h3>

According to Biot savert's law, magnetic field at the center of a circular arc is

B=(μ₀ I/4π)× (arc/radius²)

As arc is given as angle × radius, so

B=( μ₀I/4π)×(angle/radius)

<h3>What will be the magnetic field at the center of a circular arc, if the arc has current 26.9 A, radius 0.6 cm and angle 0.9 radian?</h3>

B=(μ₀ I/4π)× (0.9/0.006)

= (10^(-7)× 26.9)× (0.9/0.006)

= 4 × 10^(-4) T

Thus, we can conclude that the magnitude of magnetic field at the center of the circular arc is 4 × 10^(-4) T.

Learn more about the magnetic field of a circular arc here:

brainly.com/question/15259752

#SPJ4

5 0

2 years ago

a carbon atom with 6 proton and 6 neutron in its nucleus is called carbon-12 the carbon atom with 8 neutrons is called what

daser333 [38]

Answer:

carbon-14 must be the answer

4 0

3 years ago

An object of mass m = 4.0 kg, starting from rest, slides down an inclined plane of length l = 3.0 m. The plane is inclined by an

kirill [66]

Answer:

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf= -20.4 J is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane:

Wg= 58.8 J is positive

Explanation:

Nomenclature

vf: final velocity

v₀ :initial velocity

a: acceleleration

d: distance

Ff: Friction force

W: weight

m:mass

g: acceleration due to gravity

Graphic attached

The attached graph describes the variables related to the kinetics of the object (forces and accelerations)

Calculation de of the components of W in the inclined plane

W=m*g

Wx₁ = m*g*sin30°

Wy₁= m*g*cos30°

Object kinematics on the inclined plane

vf₁²=v₀₁²+2*a₁*d₁

v₀₁=0

vf₁²=2*a₁*d₁

$v_{f1} = \sqrt{2*a_{1}*d_{1} }$ Equation (1)

Object kinetics on the inclined plane (μ= 0.2)

∑Fx₁=ma₁ :Newton's second law

-Ff₁+Wx₁ = ma₁ , Ff₁=μN₁

-μ₁N₁+Wx₁ = ma₁ Equation (2)

∑Fy₁=0 : Newton's first law

N₁-Wy₁= 0

N₁- m*g*cos30°=0

N₁ = m*g*cos30°

We replace N₁ = m*g*cos30 and Wx₁ = m*g*sin30° in the equation (2)

-μ₁m*g*cos30₁+m*g*sin30° = ma₁ : We divide by m

-μ₁*g*cos30°+g*sin30° = a₁

g*(-μ₁*cos30°+sin30°) = a₁

a₁ =9.8(-0.2*cos30°+sin30°)=3.2 m/s²

We replace a₁ =3.2 m/s² and d₁= 3m in the equation (1)

$v_{f1} = \sqrt{2*3.2*3} }$

$v_{f1} =\sqrt{2*3.2*3}$

$v_{f1} = 4.38 m/s$

Rough surface kinematics

vf₂²=v₀₂²+2*a₂*d₂ v₀₂=vf₁=4.38 m/s

0 =4.38²+2*a₂*d₂ Equation (3)

Rough surface kinetics (μ= 0.3)

∑Fx₂=ma₂ :Newton's second law

-Ff₂=ma₂

--μ₂*N₂ = ma₂ Equation (4)

∑Fy₂= 0 :Newton's first law

N₂-W=0

N₂=W=m*g

We replace N₂=m*g inthe equation (4)

--μ₂*m*g = ma₂ We divide by m

--μ₂*g = a₂

a₂ =-0.2*9.8= -1.96m/s²

We replace a₂ = -1.96m/s² in the equation (3)

0 =4.38²+2*-1.96*d₂

3.92*d₂ = 4.38²

d₂=4.38²/3.92

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf = - Ff₁*d₁

Ff₁= μ₁N₁= μ₁*m*g*cos30°= -0.2*4*9.8*cos30° = 6,79 N

Wf= - 6.79*3 = 20.4 N*m

Wf= -20.4 J is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane

Wg=W₁x*d= m*g*sin30*3=4*9.8*0.5*3= 58.8 N*m

Wg= 58.8 J is positive

6 0

3 years ago

You are riding on a school bus and suddenly get thrown forward. What did the bus just do?

sergij07 [2.7K]

The bus is going forward and suddenly stops

5 0

3 years ago