Hii please help i’ll give brainliest if you give a correct answer please please hurry

A6 kg mass moving at 10m/s collides with a 4 kg mass moving in the

Nat2105 [25]

Answer:

Explanation:

mdeemmkdkwdmwdmw

4 0

3 years ago

There are only two charged particles in a particular region. Particle 1 carries a charge of 3q and is located on the negative x-

andrezito [222]

Answer:

The net field will be the sum of the fields created by each charge.

where the charge Q in a position r' is given by:

E(r) = k*Q/(r - r')^2

Where k is a constant, and r is the point where we are calculating the electric field.

Then for the charge 3q, in the position r₁ = (-d, 0, 0) the electric field will be:

E₁(r) = k*3q/(r - r₁)^2

While for the other charge of -2q in the position r₂ = (d, 0, 0)

The electric field is:

E₂(r) = -k*2*q/(r - r₂)^2

Then the net field at the point r is:

E(r) = E₁(r) + E₂(r) = k*3q/(r - r₁)^2 + -k*2*q/(r - r₂)^2

E(r) = k*q*( 3/(r - r₁)^2 - 2/(r - r₂)^2)

Then if the we want to find the points r = (x, y, z) such that:

E(r) = 0 = k*q*( 3/(r - r₁)^2 - -k*2*q/(r - r₂)^2)

Then we must have:

0 = ( 3/(r - r₁)^2 - 2/(r - r₂)^2)

Also remember that the distance between two points:

(x, y, z) and (x', y', z') is given by:

D = √( (x - x')^2 + (y - y)^2 + (z -z')^2)

Then we can rewrite:

r - r₁ = √( (x - (-d))^2 + (y - 0 )^2 + (z -0)^2)

= √( (x + d))^2 + y^2 + z^2)

and

r - r₂ = √( (x - d)^2 + (y - 0 )^2 + (z -0)^2)

= √( (x - d))^2 + y^2 + z^2)

Replacing that in our equation we get:

0 = ( 3/(√( (x + d))^2 + y^2 + z^2))^2 - -k*2*q/(√( (x - d))^2 + y^2 + z^2))^2)

0 = (3/((x + d))^2 + y^2 + z^2) - 2/ (x - d))^2 + y^2 + z^2)

We want to find the values of x, y, z such that the above equation is true.

2/ (x - d))^2 + y^2 + z^2) = (3/((x + d))^2 + y^2 + z^2)

2*[((x + d))^2 + y^2 + z^2] = 3*[(x - d))^2 + y^2 + z^2]

2*(x + d)^2 + 2*y^2 + 2*z^2 = 3*(x - d)^2 + 3*y^2 + 3*z^2

2*(x + d)^2 - 3*(x - d)^2 = 3*y^2 + 3*z^2 - 2*y^2 - 2*z^2

2*(x + d)^2 - 3*(x - d)^2 = y^2 + z^2

2*x^2 + 2*2*x*d + 2*d^2 - 3*x^2 + 3*2*x*d - 3*d^2 = y^2 + z^2

-x^2 + 10*x*d - d^2 = y^2 + z^2

we can rewrite this as:

- ( x^2 - 10*x*d + d^2) = y^2 + z^2

now we can add and subtract 24*d^2 inside the parenthesis to get

- ( x^2 - 10*x*d + d^2 + 24*d^2 - 24*d^2) = y^2 + z^2

-( x^2 - 2*x*(5d) + 25d^2 - 24d^2) = y^2 + z^2

-(x^2 - 2*x*(5d) + (5*d)^2) + 24d^2 = y^2 + z^2

The thing inside the parenthesis is a perfect square:

-(x - 5d)^2 + 24d^2 = y^2 + z^2

we can rewrite this as:

24d^2 = y^2 + z^2 + (x - 5d)^2

This equation gives us the points (x, y, z) such that the electric field is zero.

Where we need to replace two of these values to find the other, for example, if y = z = 0

24d^2 = (x - 5d)^2

√(24d^2) = x - 5d

√24*d = x - 5d

√24*d + 5d = x

so in the point (√24*d + 5d, 0, 0) the net field is zero.

7 0

3 years ago

What might cause a shotgun to explode? <br> this is due in like 10 mins so help me please

Margarita [4]

The terminating pin strikes the preliminary, making it detonate. The flash from the groundwork touches off the black powder. Gas changed over from the consuming powder quickly extends in the cartridge. ... The shot's speed and getting away from gases produce a "blast."

5 0

3 years ago

Before Collision:

marissa [1.9K]

Answer:

4/3 m/s

Explanation:

Assuming momentum is conserved, the sum of products of mass and speed before the collision is the same as after:

(2000 kg)(4 m/s) +(4000 kg)(0 m/s) = (2000 +4000 kg)(Vt)

Vt = (8000 kg·m/s)/(6000 kg) = 4/3 m/s

The speed of the combined objects after the collision is 4/3 m/s.

6 0

2 years ago

Determine the centroid of the shaded area shown in figure 2. Determine the moment of inertia about y-axis of the shaded area sho

Nady [450]

Answer:

centroid: (x, y) = (81.25 mm, 137.5 mm)
I = 8719.31 mm^2 for unit mass

Explanation:

Finding the desired measures requires we know a differential of area. That, in turn, requires we have a way to describe a differential of area. Here, we choose to use a vertical slice, which requires we know the area boundaries as a function of x.

The upper boundary is a line with a slope of 125/156.25 = 0.8, and a y-intercept of 125. That is, ...

y1 = 0.8x +125

The lower boundary is given in terms of y, but we can solve for y to find ...

100x = y^2

y2 = 10√x

Then our differential of area is ...

dA = (y1 -y2)dx

__

The centroid is found by computing the first moment about the x- and y-axes, and dividing those values by the area of the figure.

The area will be ...

$\displaystyle A=\int_0^{156.25}{dA}=\int_0^{156.25}{(y_1-y_2)}\,dx$

The y-coordinate of the centroid is ...

$\displaystyle \overline{y}=\dfrac{S_x}{A}=\dfrac{1}{A}\int_0^{156.25}{\dfrac{y_1+y_2}{2}}\,dA=\dfrac{1}{A}\int_0^{156.25}{\dfrac{y_1+y_2}{2}(y_1-y_2)}\,dx=137.5$

Similarly, the x-coordinate is ...

$\displaystyle \overline{x}=\dfrac{S_y}{A}=\dfrac{1}{A}\int_0^{156.25}{x}\,dA=\dfrac{1}{A}\int_0^{156.25}{x(y_1-y_2)}\,dx=81.25$

That is, centroid coordinates are (x, y) = (81.25, 137.5) mm.

__

The moment of inertia is the second moment of the area. If we normalize by the "mass" (area), then the integral looks a lot like the one for $\overline{x}$ , but multiplies dA by x^2 instead of x.

The attachment shows that value to be ...

I ≈ 8719.31 mm^2 (normalized by area)

The area is 16276.0416667 mm^2, if you want to "un-normalize" the moment of inertia.

7 0

3 years ago