Calculate the pOH of this solution. pH = 1.90 pOH = right answer get brainliest

A racecar experiences a centripetal acceleration of 36.0 m/s2 as it travels at a constant speed of 27.0 m/s along a circular arc

alexdok [17]

Answer:

20.25 m

Explanation:

Centripetal acceleration is given by; the square of the velocity, divided by the radius of the circular path.

That is;

ac = v²/r

Where; ac = acceleration, centripetal, m/s², v is the velocity, m/s and r is the radius, m

Therefore;

r = v²/ac

= 27²/36

= 20.25 m

Hence the radius is 20.25 meters

5 0

3 years ago

Read 2 more answers

HELP PLEASE!!! 30+ points!!

GarryVolchara [31]

1) 9.26 cm

Explanation:

The focal length of a plane mirror is virtually infinite. Considering the lens equation,

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where f is the focal length, p is the object distance, q the image distance. If we replace f with infinity, we get

$q=-p$

The magnification equation states that

$y' = -\frac{q}{p}y$

where y is the size of the object and y' the size of the image. Substituting q=-p, we get

$y'=y$

this means that the image produced by a plane mirror is always:

- Upright (y' is positive)

- The same size as the object

In this case, we have a book of height 9.26 cm (y=9.26 cm). This means that the magnitude of the size of the image (y') will be 9.26 cm as well.

2) 22.7 cm

As we said before, due to the infinite focal length of a plane mirror,

$q=-p$

this means that the image produced by a plane mirror is always:

- Virtual (because q is negative)

- At the same distance from the mirror as the object

In this case, we have a book placed at 22.7 cm from the mirror (p=22.7 cm). This means that the magnitude of the distance of the image from the mirror (q) will be 22.7 cm as well.

3) 1.60 m/s

We said previously that the image produced by a plane mirror is always at the same distance from the mirror as the real object. This implies that whenever we move the object toward/away from the mirror, the distance p will alway remain equal to the distance q. But this also means that the object and the distance are moving toward/away from the mirror at the same speed.

Therefore, since in this case the person is moving away from the mirror at 1.60 m/s, the image will also move away at a speed of 1.60 m/s.

4 0

3 years ago

Suppose A=BnCm, where A has dimensions LT, B has dimensions L2T-1, and C has dimensions LT2. Then the exponents n and m have the

julsineya [31]

Explanation:

The expression is :

$A=B^nC^m$

A =[LT], B=[L²T⁻¹], C=[LT²]

Using dimensional of A, B and C in above formula. So,

$A=B^nC^m\\\\\ [LT]=[L^2T^{-1}]^n[LT^2}]^m\\\\\ [LT]=L^{2n}T^{-n}L^mT^{2m}\\\\\ [LT]=L^{2n+m}T^{2m-n}$

Comparing the powers both sides,

2n+m=1 ...(1)

2m-n=1 ...(2)

Now, solving equation (1) and (2) we get :

$n=\dfrac{1}{5}\\\\m=\dfrac{3}{5}$

Hence, the correct option is (E).

5 0

3 years ago

A thin spherical shell with radius R1 = 2.00 cm is concentric with a larger thin spherical shell with radius R2 = 6.00 cm. Both

Dafna11 [192]

Answer:

a. i. 1350 V ii 0 V iii -450 V b. 6.75 kV. The inner shell is at a higher potential.

Explanation:

The formula for electric potential is given by V = Σkq/r, where k = 9 × 10⁹ Nm²/C², q = charge and r = distance.

q₁ = charge on smaller shell = +6.00 nC = +6.00 × 10⁻⁹ C, r₁ = radius of smaller shell = 2.00 cm = 2.00 × 10⁻² m.

q₂ = charge on larger shell = -9.00 nC = -9.00 × 10⁻⁹ C, r₂ = radius of larger shell = 6.00 cm = 6.00 × 10⁻² m.

a. At r = 0, inside both spheres V = kq₁/r₁ + kq₂/r₂. = k(q₁/r₁ + q₂/r₂) = 9 × 10⁹ [+6.00 × 10⁻⁹/2.00 × 10⁻² + (-9.00 × 10⁻⁹/6.00 × 10⁻²)] = 1350 V

ii. At r = 4.00 cm, the point outside of smaller shell but inside larger shell. r₁ = 4.00 cm = 4.00 × 10⁻² m and r₂ = 6.00 cm = 6.00 × 10⁻². So, V = kq₁/r₁ + kq₂/r₂. = k(q₁/r₁ + q₂/r₂) = 9 × 10⁹ [+6.00 × 10⁻⁹/4.00 × 10⁻² + (-9.00 × 10⁻⁹/6.00 × 10⁻²)] = 0 V.

iii. At r = 6.00 cm, the point outside both shells. r₁ = r₂ = r = 6.00 cm = 6.00 × 10⁻². So, V = kq₁/r₁ + kq₂/r₂. = k(q₁ + q₂)/r = 9 × 10⁹ [+6.00 × 10⁻⁹+ (-9.00 × 10⁻⁹)]/6.00 × 10⁻² = -450 V.

b. The potential of the surface of the smaller shell is V₁ = 9 × 10⁹ [+6.00 × 10⁻⁹/2.00 × 10⁻²] = 2700 V = 2.7 kV.

The potential of the surface of the larger shell is V₂ = 9 × 10⁹ [-9.00 × 10⁻⁹/2.00 × 10⁻²] = -4050 V = -4.050 kV. The potential difference V₁ - V₂ = 2700 - (-4050) V = 6750 V = 6.75 kV. Since the potential difference is positive, V₁ is higher. So, the inner shell is at a higher potential.

8 0

3 years ago

Electrons are BLANK not Blank -lonic Bond

DerKrebs [107]

Answer:

electrons are NEGATIVE not POSITIVE ionic bond

Explanation:

4 0

3 years ago