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podryga [215]
2 years ago
10

Mateo makes a table of the symbols that represent the parts of an electric circuit. What label should Mateo place in the marked

cell? Uses electrical energy Provides electrical energy Measures electric current Controls electric current
Physics
2 answers:
deff fn [24]2 years ago
7 0
The correct option is PROVIDE ELECTRICAL ENERGY.
The marked cell in the given electric circuit is the battery and the function of the battery in electric cells is to provide electrical energy that will drive the electrical current in the electric circuit. A battery in an electric circuit is thus a source of energy for the circuit. 
Vikki [24]2 years ago
4 0

Answer:

B Provide electrical energy

Explanation:

Took the tests

You might be interested in
Consider a space shuttle which has a mass of about 1.0 x 105 kg and circles the Earth at an altitude of about 200.0 km. Calculat
kodGreya [7K]

Answer:

1.6675×10^-16N

Explanation:

The force of gravity that the space shuttle experiences is expressed as;

g = GM/r²

G is the gravitational constant

M is the mass = 1.0 x 10^5 kg

r is the altitude = 200km = 200,000m

Substitute into the formula

g = 6.67×10^-11 × 1.0×10^5/(2×10^5)²

g = 6.67×10^-6/4×10^10

g = 1.6675×10^{-6-10}

g = 1.6675×10^-16N

Hence the force of gravity experienced by the shuttle is 1.6675×10^-16N

7 0
2 years ago
A siphon pumps water from a large reservoir to a lower tank that is initially empty. The tank also has a rounded orifice 20 ft b
trasher [3.6K]

Answer:

height of the water rise in tank is 10ft

Explanation:

Apply the bernoulli's equation between the reservoir surface (1) and siphon exit (2)

\frac{P_1}{pg} + \frac{V^2_1}{2g} + z_1= \frac{P_2}{pg} + \frac{V_2^2}{2g} +z_2

\frac{P_1}{pg} + \frac{V^2_1}{2g} +( z_1-z_2)= \frac{P_2}{pg} + \frac{V_2^2}{2g}-------(1)

substitute P_a_t_m for P_1, (P_a_t_m +pgh) for P_2

0ft/s for V₁, 20ft for (z₁ - z₂) and 32.2ft/s² for g in eqn (1)

\frac{P_1}{pg} + \frac{V^2_1}{2g} +( z_1-z_2)= \frac{P_2}{pg} + \frac{V_2^2}{2g}

\frac{P_1}{pg} + \frac{0^2_1}{2g} +( 20)= \frac{(P_a_t_m+pgh)}{pg} +\frac{V^2_2}{2\times32.2} \\\\V_2 = \sqrt{64.4(20-h)}

Applying bernoulli's equation between tank surface (3) and orifice exit (4)

\frac{P_3}{pg} + \frac{V^2_3}{2g} + z_3= \frac{P_4}{pg} + \frac{V_4^2}{2g} +z_4

substitute

P_a_t_m for P_3, P_a_t_m for P_4

0ft/s for V₃, h for z₃, 0ft for z₄, 32,2ft/s² for g

\frac{P_a_t_m}{pg} + \frac{0^2}{2g} +h=\frac{P_a_t_m}{pg} + \frac{V_4^2}{2\times32.2} +0\\\\V_4 =\sqrt{64.4h}

At equillibrium Fow rate at point 2 is equal to flow rate at point 4

Q₂ = Q₄

A₂V₂ = A₃V₃

The diameter of the orifice and the siphon are equal , hence there area should be the same

substitute A₂ for A₃

\sqrt{64.4(20-h)} for V₂

\sqrt{64.4h} for V₄

A₂V₂ = A₃V₃

A_2\sqrt{64.4(20-h)} = A_2\sqrt{64.4h}\\\\20-h=h\\\\h= 10ft

Therefore ,height of the water rise in tank is 10ft

3 0
2 years ago
On an amusement park ride, passengers are seated in a horizontal circle of radius 7.5 m. The seats begin from rest and are unifo
timofeeve [1]

Answer:

a = 0.5 m/s²

Explanation:

Applying the definition of angular acceleration, as the rate of change of the angular acceleration, and as the seats begin from rest, we can get the value of the angular acceleration, as follows:

ωf = ω₀ + α*t

⇒ ωf = α*t ⇒ α = \frac{wf}{t} = \frac{1.4 rad/s}{21 s} = 0.067 rad/s2

The angular velocity, and the linear speed, are related by the following expression:

v = ω*r

Applying the definition of linear acceleration (tangential acceleration in this case) and angular acceleration, we can find a similar relationship between the tangential and angular acceleration, as follows:

a = α*r⇒ a = 0.067 rad/sec²*7.5 m = 0.5 m/s²

3 0
2 years ago
At 5000-kg freight car runs into 10000-kg freight car at rest. they couple upon collision and move with a speed of 2 m/s. what w
aliina [53]

Solution for the problem is:

Total momentum before collision is always equal to total momentum after collision. So note that:
Momentum of car A = 5000 x Xm/s 
Momentum of car A + B = 15,000 x 2m/s 

So combining the two, will give us the equation:
15,000/5,000 = 3 
3 x 2 =6m/s

6 0
2 years ago
Two narrow slits spaced 100 microns apart are exposed to light of 600 nm. At what angle does the first minimum (dark space) occu
kumpel [21]

Answer:

The angle is   \theta  =  0.1719^o

Explanation:

From the question we are told that

   The  distance of separation is  d =  100 * 10^{-6} \  m

    The  wavelength of light is  \lambda  =  600 nm =  600 *10^{-9} \  m

   

Generally the condition for destructive interference is mathematically represented as

         dsin(\theta )  =[m  +  \frac{1}{2} ]\lambda

Here  m is the order of maxima,  first minimum (dark space) m = 0

 So  

      100 *10^{-6 } *  sin(\theta )  =[0  +  \frac{1}{2} ]600 *10^{-9}

=>   \theta  =  sin^{-1} [0.003]

=>   \theta  =  0.1719^o

     

7 0
2 years ago
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