All categories

3 years ago

When ___ attacks the surface of a metal, it becomes tarnished.

Physics

2 answers:

alexandr1967 [171]3 years ago

6 0

Answer:corrosion (i believe)

Explanation:

Tcecarenko [31]3 years ago

5 0

Answer

Mix baking soda and salt with hot water and cover everything with it. The proportions are not crucial, but about 1 tablespoon of salt and 1 tablespoon of baking soda to 3 dl water should do the trick. Lightly tarnished objects should clean up in a few minutes, and you just rinse them of and dry them.

Explanation:

You might be interested in

2 Select the correct answer. Javed suffers from a severe food allergy. Which device should Javed always carry? A. inhaler B. epi

Slav-nsk [51]

The best answer would be:

B. Epi pen

Here is more about it:

An epi pen injects epinephrine into the body to counteract the effects of an allergic reaction. Epinephrine helps narrow blood vessels and open up the airways. This can be used for food allergies, insect bites, and some other allergens. It is very effective and immediately works, but it is short-lived, so seeking medical attention after is still needed.

4 0

3 years ago

An observer at the top of a 462-ft cliff measures the angle of depression from the top of the cliff to a point on the ground to

Lana71 [14]

complete question:

An observer at the top of a 462-ft cliff measures the angle of depression from the top of the cliff to a point on the ground to be 5°. What is the distance from the base of the cliff to the point on the ground? Round to the nearest foot

Answer:

a ≈ 5281 ft

Explanation:

The observer at the top of a 462 ft cliff measures the angle of depression from the top of the cliff to a point on the ground to be 5°.

The angle of depression form the top of the cliff = 5°

The 5° is outside the triangle formed . To find the angle in the triangle we have to subtract 5° from 90°. 90° - 5° = 85° Note sum of an angle on a right angle is 90°.

using SOHCAHTOA principle we can solve for the distance from the base of the cliff to the point on the ground(a)

tan 85° = opposite / adjacent

tan 85° = a / 462

cross multiply

462 × tan 85° = a

a = 11.4300523 × 462

a = 5280.66 ft

a ≈ 5281 ft

5 0

3 years ago

We have three identical metallic spheres A, B, C. Initially sphere A is charged with charge Q, while B and C are neutral. First,

larisa [96]

Answer:

The final charges of each sphere are: q_A = 3/8 Q , q_B = 3/8 Q , q_C = 3/4 Q

Explanation:

This problem asks for the final charge of each sphere, for this we must use that the charge is distributed evenly over a metal surface.

Let's start Sphere A makes contact with sphere B, whereby each one ends with half of the initial charge, at this point

q_A = Q / 2

q_B = Q / 2

Now sphere A touches sphere C, ending with half the charge

q_A = ½ (Q / 2) = ¼ Q

q_B = ¼ Q

Now the sphere A that has Q / 4 of the initial charge is put in contact with the sphere B that has Q / 2 of the initial charge, the total charge is the sum of the charge

q = Q / 4 + Q / 2 = ¾ Q

This is the charge distributed between the two spheres, sphere A is 3/8 Q and sphere B is 3/8 Q

q_A = 3/8 Q

q_B = 3/8 Q

The final charges of each sphere are:

q_A = 3/8 Q

q_B = 3/8 Q

q_C = 3/4 Q

7 0

3 years ago

A major artery with a cross sectional area of 1.00cm^2 branches into 18 smaller arteries, each with an average cross sectional a

dybincka [34]

Here we can say that rate of flow must be constant

so here we will have

$A_1v_1 = 18 A_2v_2$

now we know that

$A_1 = 1 cm^2$

$A_2 = 0.4 cm^2$

now from above equation

$1 cm^2 v_1 = 18(0.400 cm^2)v_2$

$\frac{v_2}{v_1} = \frac{1}{18\times 0.4}$

$\frac{v_2}{v_1} = 0.14$

so velocity will reduce by factor 0.14

3 0

3 years ago

An 85,000 kg stunt plane performs a loop-the-loop, flying in a 260-m-diameter vertical circle. at the point where the plane is f

konstantin123 [22]

A) When the plane is flying straight down, there are three forces acting on it:
- the centripetal force $F=m \frac{v^2}{r}$ , directed toward the center of the circle (so, horizontally)
- the weight of the plane: $W=mg$ , downward, so vertically
- a third force, given by the propulsion of the plane, which is accelerating it towards the ground (because the problem says that the plane has an acceleration of a=12 m/s^2 towards the ground)

The radius of the circle is $r= \frac{260 m}{2} = 130 m$ , so the centripetal force acting on the plane is
$F_c=m \frac{v^2}{r} = \frac{(85000 kg)(55 m/s)^2}{130 m}=1.98 \cdot 10^6 N$
On the vertical axis, we have two forces: the weight
$W=mg=(85000 kg)(9.81 m/s^2)=8.34 \cdot 10^5 N$
and the other force F given by the propulsion. Since we know that their sum should generate an acceleration equal to $a=12 m/s^2$ , we can find the magnitude of this other force F by using Newton's second law:
$F+mg=ma$
$F=m(a-g)=(85000kg)(12 m/s^2-9.81 m/s^2)=1.86 \cdot 10^5 N$

So, the net force acting on the plane will be the resultant of the centripetal force (acting in the horizontal direction) and the two forces W and F (acting in the vertical direction):
$R= \sqrt{(F_c^2+(W+F)^2}=$
$= \sqrt{(1.98\cdot 10^6N)^2+(8.34 \cdot 10^5N+1.86 \cdot 10^5 N)^2} =2.23 \cdot 10^6 N$

(b) The tangent of the angle with respect to the horizontal is the ratio between the sum of the forces in the vertical direction (taken with negative sign, since they are directed downward) and the forces acting in the horizontal direction, so:
$\tan \theta = \frac{-(W+F)}{F_c}= -0.5$
And so, the angle is
$\theta = \arctan (-0.5)=-26.8 ^{\circ}$

7 0

3 years ago

When ___ attacks the surface of a metal, it becomes tarnished.​

When ___ attacks the surface of a metal, it becomes tarnished.