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Radda [10]
3 years ago
12

A type of circuit that does not have any breaks in the loop is called what?

Physics
1 answer:
Korvikt [17]3 years ago
5 0
I think the correct answer is a series circuit. This type of circuit is characterized by having only one path for electric current to flow. All of the components in the circuit share the same current. Hope this answers the question. Have a nice day.
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Why won’t anyone help me please anybody help me I really need help .
irga5000 [103]

Answer:

1➡️ this is the method of decomposition

2➡️ H2 and O2

3➡️ b

sorry if I am wrong

8 0
3 years ago
Two identical 9.10-g metal spheres (small enough to be treated as particles) are hung from separate 300-mm strings attached to t
Musya8 [376]

Answer:

n = 1.266\times 10^{12}

Explanation:

Given data:

mass of sphere is 10 g

Angle between string and vertical axis is \theta = 13 degree

thickness of string  300 mm = 0.3 m

sin\theta =\frac{2}{0.3 m}

r =0.3 sin 13 = 0.067 m

Fe = \frac{ kq_1 q-2}{d^2}

Fe = \frac{kq^2}{(2r)^2} = mg tan\theta

q^2 =  mg tan\theta \frac{(2r)^2}{k}

    = 0.0091 \times 9.8 tan13 \times \frac{(2\times 0.067)^2}{9\times 10^9}

q^2 = 4.10\times 10^{-14}

q = 2.026 \times 10^{-7} C

q = ne

n = \frac{1.6\times 10^{-19}}{2.02\times 10^{-7}}

n = 1.266\times 10^{12}

3 0
3 years ago
Se dispone de dos vasos con agua a 20 grados y queremos calentarlo hasta alcanzar 50 grados. Si el primero contiene 0,5l y el se
BigorU [14]

Answer:

The cup with 0.5L

Explanation:

To know what amount of water you take into account the specific heat of the water. The specific heat of water is:

c_{water}=4186\frac{J}{kg\°C}

Thus, 4186 J of energy are needed to icrease the temperature of 1 kg water in 1°C. Then, more grams of water will need more energy.

You have that one cup has 0.5 L and the other one has 750mL = 0.75L

The second cup of water will need more heat because the amount of water contained in the second cup is greater than in the first cup with 0.5L

4 0
3 years ago
Evaluate the gravitational potential energy between two 5.00-kg spherical steel balls separated by a center-tocenter distance of
navik [9.2K]

Answer:

U = 8.30×10-⁹J

Explanation:

m1 = m2 = 5.00kg masses of the spheres

d = 15.0cm = 15×10-²m

r = 5.10cm = 5.10×10-²m

R = d + r = 15×10-² + 5.10×10-²

R = 20.10 ×10-²m = 0.201m

G = 6.67×10-¹¹Nm²/kg²

U = Gm1×m2/R = potential energybetween the spheres

U = 6.67×10-¹¹×5.00×5.00/0.201

U = 8.30×10-⁹J

7 0
3 years ago
A speeding motorist traveling down a straight highway at 100 km/h passes a parked police car. It takes the police constable 1.0
Lubov Fominskaja [6]

Answer:

t = 7.5 s

Explanation:

The distance traveled by the car at the time of meeting of the two cars must be the same. First, we calculate the distance traveled by the police car. For that we use 2nd equation of motion. Here, we take the time when police car starts to be reference. So,

s₁ = Vi t + (0.5)gt²

where,

s₁ = distance traveled by police car

Vi = Initial Velocity = 0 m/s

t = time taken

Therefore,

s₁ = (0 m/s)(t) + (0.5)(9.8 m/s²)t²

s₁ = 4.9 t²

Now, we calculate the distance traveled by the car. For constant speed and time to be 1 second more than the police car time, due to car starting time, we get:

s₂ = Vt = V(t + 1)

where,

s₂ = distance traveled by car

V = Velocity of car = (100 km/h)(1000 m/1 km)(1 h/ 3600 s) = 27.78 m/s

Therefore,

s₂ = 27.78 t + 27.78

Now, we know that at the time of meeting:

s₁ = s₂

4.9 t² = 27.78 t + 27.78

4.9 t² - 270.78 t - 27.78 = 0

solving the equation and choosing the positive root:

t = 6.5 s

since, we want to know the time from the moment car crossed police car. Therefore, we add 1 second of starting time in this.

t = 6.5 s + 1 s

<u>t = 7.5 s</u>

6 0
4 years ago
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