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Alex17521 [72]
3 years ago
15

Choose the statement about Masters sports programs that is true.

Physics
2 answers:
likoan [24]3 years ago
7 0
Pretty sure rhe answer to your question is A.
Stels [109]3 years ago
6 0

Answer:

Option A

Explanation:

Master sports programs are an initiative that allows Adults to join their favorite sports.Masters programs allow adults to both exercise and socialize with people who share their love of a sport. It helps the Adults to build a camaraderie and enjoy the sports they love the most. This way they can sustain the passion for the sports.

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A coil with an inductance of 2.3 H and a resistance of 14 Ω is suddenly connected to an ideal battery with ε = 100 V. At 0.13 s
klemol [59]

Given Information:

Resistance = R = 14 Ω

Inductance = L = 2.3 H

voltage = V = 100 V

time = t = 0.13 s

Required Information:

(a) energy is being stored in the magnetic field

(b) thermal energy is appearing in the resistance

(c) energy is being delivered by the battery?

Answer:

(a) energy is being stored in the magnetic field ≈ 219 watts

(b) thermal energy is appearing in the resistance ≈ 267 watts

(c) energy is being delivered by the battery ≈ 481 watts

Explanation:

The energy stored in the inductor is given by

U = \frac{1}{2} Li^{2}

The rate at which the energy is being stored in the inductor is given by

\frac{dU}{dt} = Li\frac{di}{dt} \: \: \: \: eq. 1

The current through the RL circuit is given by

i = \frac{V}{R} (1-e^{-\frac{t}{ \tau} })

Where τ is the the time constant and is given by

\tau = \frac{L}{R}\\ \tau = \frac{2.3}{14}\\ \tau = 0.16

i = \frac{110}{14} (1-e^{-\frac{t}{ 0.16} })\\i = 7.86(1-e^{-6.25t})\\\frac{di}{dt} = 49.125e^{-6.25t}

Therefore, eq. 1 becomes

\frac{dU}{dt} = (2.3)(7.86(1-e^{-6.25t}))(49.125e^{-6.25t})

At t = 0.13 seconds

\frac{dU}{dt} = (2.3) (4.37) (21.8)\\\frac{dU}{dt} = 219.11 \: watts

(b) thermal energy is appearing in the resistance

The thermal energy is given by

P = i^{2}R\\P = (7.86(1-e^{-6.25t}))^{2} \cdot 14\\P = (4.37)^{2}\cdot 14\\P = 267.35 \: watts

(c) energy is being delivered by the battery?

The energy delivered by battery is

P = Vi\\P = 110\cdot 4.37\\P = 481 \: watts

4 0
3 years ago
Why is this event important to include in the biography?
nadya68 [22]

Answer:

D: It shows that Frida Kahlo used art to cope with her pain.

Explanation:

Within the text given it shows her emotions being lonely, immobile and in pain. But it all shows her asking her father for art which states that art is her sort of relief and happy place.

8 0
2 years ago
Read 2 more answers
In the ground state of hydrogen, according to the Bohr model, an electron orbits 5.3 x 10-11 m from the nucleus. It undergoes a
Readme [11.4K]

Answer:

Explanation:

Given

radius of electron(r)=5.3\times 10^{-11} m

centripetal acceleration (a_c)=9\times 10^22 m/s^2

we know

a_c=\frac{v^2}{r}

v=\sqrt{r\times a_c}

v=\sqrt{5.3\times 10^{-11}\times 9\times 10^{22}}

v=\sqrt{47.7\times 10^11}

v=21.84\times 10^5 m/s

(b)For n=10

r=100\times 5.3\times 10^{-11} m\approx 5.3\times 10^{-9} m

a_c=10^4\times 9\times 10^{22} m/s^2

a_c=9\times 10^{26} m/s^2

v=\sqrt{r\times a_c}

v=\sqrt{9\times 10^{26}\times 5.3\times 10^{-9}}

v=21.84\times 10^8 m/s

8 0
3 years ago
A weight of 1400 pounds is suspended from two cables as shown in the figure. What is the tension in the left cable? _________ po
juin [17]

Answer:

Following are the solution to this question:

Explanation:

Law:

\to \theta= 180^{\circ}- 50^{\circ}- 25^{\circ}

      = 180^{\circ}- 75^{\circ}\\\\= 105^{\circ}

\to \frac{T_{L}}{\sin (90+50)}= \frac{T_{R}}{\sin (25+90)}=\frac{1400}{\sin (105)}

\to T_L=931.65 \ pounds \\\\ \to T_R=1313.59 \ pounds \\\\

4 0
3 years ago
A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh
Elanso [62]

Answer:

The value is  v  =  2.3359 *10^{4} \ m/s

Explanation:

From the question we are told that

  The  initial speed is u =  2.05 *10^{4} \  m/s

 Generally the total energy possessed by the space probe when on earth is mathematically represented as

             T__{E}} =  KE__{i}} +  KE__{e}}

Here  KE_i is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

          KE_i =   \frac{1}{2}  *  m  *  u^2

=>       KE_i =   \frac{1}{2}  *  m  *  (2.05 *10^{4})^2

=>       KE_i =  2.101 *10^{8} \ \ m \ \ J

And  KE_e is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

       KE_e =  \frac{1}{2}  *  m *  v_e^2

Here v_e is the escape velocity from earth which has a value v_e =  11.2 *10^{3} \  m/s

=>    KE_e =  \frac{1}{2}  *  m *  (11.3 *10^{3})^2

=>    KE_e =  6.272 *10^{7} \  \  m  \ \   J

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

        KE_p =  \frac{1}{2}  *  m *  v^2

Generally from the law energy conservation we have that

        T__{E}} =  KE_p

So

       2.101 *10^{8}  m  +  6.272 *10^{7}  m  =   \frac{1}{2}  *  m *  v^2

=>     5.4564 *10^{8} =   v^2

=>     v =  \sqrt{5.4564 *10^{8}}

=>     v  =  2.3359 *10^{4} \ m/s

4 0
2 years ago
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