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nata0808 [166]
3 years ago
7

The atomic number of magnesium is 12. This means that its nucleus must contain

Physics
2 answers:
Lena [83]3 years ago
3 0
12 protons and 12 electrons

Marina86 [1]3 years ago
3 0
The atomic of magnesium is 12 this means that its nucleus 12 protons and no electrons are  not in the nucleus  
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in a hydraulic garage the small piston has a radius of 5 cm and the large piston has radius of 15 cm what force must be applied
viktelen [127]

The force applied to small piston = 2.2 x 10³ N

<h3>Further explanation</h3>

Given

a radius of 5 cm and 15 cm

weight 20000 N

Required

Force applied

Solution

Pascal Law :

F₁/A₁=F₂/A₂

A₁ = π.5²

A₂ = π.15²

F₁/ π.5² cm² = 20000/π.15² cm²

F₁ = 2222.22 N⇒2.2 x 10³ N

5 0
3 years ago
A point charge q1 = is located at the center of a thick conducting spherical shell of inner radius a = 2.2 cm and outer radius b
Yanka [14]

Answer:

Ex(P) = -3.602 x 10^6 N/C

Explanation:

  • given q1 = -5.7μC
  • q2 = 2.6 μC = the net charge on the conducting shell
  • inner radius of conducting shell = a = 2.2 cm =0.022m
  • outer radius of conducting shell = b = 4.5 cm = 0.045m

1) To get Ex(P), the value of the x-component of the electric field at point P, located a distance 8.8 cm along the x-axis from q1 ;

Ex(P) = k(q1+q2)/r^2

= 9 x 10^9 (-5.7 + 2.6) x 10^-6 /0.088^2

Ex(P) = -27.9 x 10^3/ 0.007744

Ex(P) = -3.602 x 10^6 N/C

7 0
4 years ago
A 40-gram block of copper at 95*C is placed in 105 g of water at an unknown temperature. After equilibrium is reached, the final
REY [17]

Answer:

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Explanation:

5 0
3 years ago
A test rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward
dusya [7]
I: x = 1/2at²
II: v = at

I+II: x = 1/2a(v²/a²) = 1/2 v²/a

a = 1/2 v²/x

v = 30
x = 49


4 0
3 years ago
Read 2 more answers
One afternoon, a couple walks three-fourths of the way around a circular lake, the radius of which is 2.51 km. they start at the
Marysya12 [62]
(a) Since the lake has a circular shape, the distance they traveled is exactly 3/4 of a circumference. The radius of the lake is 2.51 km, and the circumference is given by 2\pi r, therefore the distance covered is
d= \frac{3}{4} (2 \pi r) = \frac{3}{2} \pi (2.51 km)=11.82 km

(b) We can consider the lake to be on a xy-plane with the origin of the axes being at the center of the lake. In this system of coordinates, the starting point of the motion is at the west side of the lake, so at coordinates (-2.51 km,0). The final point is after 3/4 of circumference, therefore at the north side, at coordinates (0, 2.51 km).
So we can calculate the magnitude of the displacement as
d= \sqrt{(x_f-x_i)^2+(y_f-y_i)^2} = \sqrt{(0-(-2.51))^2+(2.51-0)^2}=
=3.55 km

(c) Considering only the initial and final point of the motion, the couple moved 2.51 km north (on the x-axis) and 2.51 km east (on the y-axis). Therefore, we can calculate the angle of the displacement with respect to the east direction:
\tan \alpha =  \frac{\Delta y}{\Delta x} = \frac{2.51 km}{2.51 km}=1
from which
\alpha=45^{\circ}
3 0
3 years ago
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