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Natalka [10]
2 years ago
6

The bottom of the inner curve of a hook is called

Physics
1 answer:
padilas [110]2 years ago
8 0
Slip hook or blunt hook
You might be interested in
A 4.0-mF capacitor initially charged to 50 V and a 6.0-mF capacitor charged to 30 V are connected to each other with the positiv
Juli2301 [7.4K]

Answer:

<em>The final charge on the 6.0 mF capacitor would be 12 mC</em>

Explanation:

The initial charge on 4 mF capacitor  = 4 mf  x 50 V = 200 mC

The initial Charge on 6 mF capacitor  = 6 mf x 30 V =180 mC

Since the negative ends are joined together  the total charge on both capacity would be;

q = q_{1} -q_{2}

q = 200 - 180

q = 20 mC

In order to find the final charge on the 6.0 mF capacitor we have to find the combined voltage

q = (4 x V) + (6 x V)

20 = 10 V

V = 2 V

For the final charge on 6.0 mF;

q = CV

q = 6.0 mF x 2 V

q =  12 mC

Therefore the final charge on the 6.0 mF capacitor would be 12 mC

5 0
2 years ago
Read 2 more answers
"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas i
asambeis [7]

This question is incomplete, the complete question is;

The Figure shows a container that is sealed at the top by a moveable piston, Inside the container is an ideal gas at 1.00 atm. 20.0°C and 1.00 L.

"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant?"

Answer:

the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant

Explanation:

Given that;

P₁ = 1.00 atm

P₂ = ?

V₁ = 1 L

V₂ = 1.60 L

the temperature of the gas is kept constant

we know that;

P₁V₁ = P₂V₂

so we substitute

1 × 1 = P₂ × 1.60

P₂ = 1 / 1.60

P₂ = 0.625 atm

Therefore the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant

5 0
1 year ago
You have a suspect on the run in a dry, wooded area; they’ve ditched the gun but are hiding from you. how long do you have to fi
erma4kov [3.2K]

Answer:

C. a few days

Explanation:

The evidence that we will have to investigate is the gunshot residue on the body of the suspect. When a person fires a gun there will be a deposition of material from the components of the chemicals present in the gun. This deposit is called gunshot residue and consists of particles of the propellant or explosive primer.

Experimentation has shown that the period of time that is person will test possible for gunshot residue is 5.27 days.

7 0
1 year ago
Study the diagram and calculate the effort required to balance the load​
Lisa [10]

Answer:

Solution given;

load =600N

effort=?

load distance: 0.6m

effort distance;2.6m

we have

load *load distance :effort *effort distance

0.6*600=effort *2.6

360/2.6=effort

effort:138.46N

<u>R</u><u>e</u><u>q</u><u>u</u><u>i</u><u>r</u><u>e</u><u>d</u><u> </u><u>e</u><u>f</u><u>f</u><u>o</u><u>r</u><u>t</u><u> </u><u>i</u><u>s</u><u> </u><u>1</u><u>3</u><u>8</u><u>.</u><u>5</u><u>N</u><u>.</u>

6 0
1 year ago
Two point charges totaling 8 μC exert a repulsive force of 0.15 N on one another when separated by 0.5 m. What is the charge on
nexus9112 [7]

Answer:

B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C

Explanation:

From Coulomb's Law the electrostatic repulsive force is given by the following formula:

F = kq₁q₂/r²

where,

F = Repulsive Force = 0.15 N

k = Coulomb's Constant = 9 x 10⁹ N.m²/C²

q₁ = Magnitude of 1st Charge = ?

q₂ = Magnitude of 2nd Charge = ?

r = Distance between Charges = 0.5 m

Therefore,

0.15 N = (9 x 10⁹ N.m²/C²)q₁q₂/(0.5 m)²

q₁q₂ = (0.15 N)(0.5 m)²/(9 x 10⁹ N.m²/C²)

q₁q₂ = 4.17 x 10⁻¹²

q₁ = (4.17 x 10⁻¹²)/q₂   -------------------- equation (1)

The sum of charges is given as:

q₁ + q₂ = 8 μC

q₁ + q₂ = 8 x 10⁻⁶

using equation (1):

(4.17 x 10⁻¹²)/q₂ + q₂ = 8 x 10⁻⁶

(4.17 x 10⁻¹²) + q₂² = 8 x 10⁻⁶ q₂

q₂² - (8 x 10⁻⁶) q₂ + (4.17 x 10⁻¹²) = 0

Solving this quadratic equation:

q₂ = 7.4 x 10⁻⁶ C   (OR)   q₂ = 0.56 x 10⁻⁶ C

q₂ = 7.4 μC   (OR)   q₂ = 0.6 μC

Therefore,

q₁ = (4.17 x 10⁻¹² C)/(7.4 x 10⁻⁶ C)

q₁ = 0.6μC

Now, if we solve with q₂ = 0.6 μC, we will get q₁ = 7.4 μC.

Therefore, the correct option will be:

<u>B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C</u>

8 0
1 year ago
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