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3 years ago

Why is pluto not a planet?

Physics

2 answers:

Veseljchak [2.6K]3 years ago

8 0

Pluto is not a planet because it's a dwarf planet it's to small to be a planet.

Hopefully this helped =)!

mihalych1998 [28]3 years ago

3 0

Pluto revolves around a star instead of our SUN. So, it's an "Exo-planet" rather than a Planet. It has declared in 2006 by the scientists.

Hope this helps!

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An airplane flies in a loop (a circular path in a vertical plane) of radius 160 m . The pilot's head always points toward the ce

notka56 [123]

Answer:

a) 39.6 m/s b) 4123 N

Explanation:

a) At the top of the loop, all of the forces point downwards (force of gravity and normal force).

Fnet=ma

ma=m(v^2/R) (centripetal acceleration)

mg=m(v^2/R)

m cancels out (this is why pilot feels weightless) so,

g=(v^2/R)

9.8 m/s^2 = v^2/160 m

v^2=1568 m^2/s^2

v=39.6 m/s

b) At the bottom of the loop, the normal force and the force of gravity point in opposite directions. The normal force is the weight felt.

Convert 300 km/hr to m/s

300 km/hr=83.3 m/s

Convert pilot's weight into mass:

760 N = 77.55 kg

Fnet=ma

n-mg=m(v^2/R)

n=(77.55 kg)(((83.3 m/s)^2)/160 m)+(77.55 kg)(9.8 m/s^2)

n=3363.2 N+760 N=4123 N

5 0

3 years ago

A toaster using a Nichrome heating element operates on 120 V. When it is switched on at 28 ∘С, the heating element carries an in

sukhopar [10]

Answer:

The final temperature of the element = 262.67°C

The power dissipated in the heating element initially = 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A = 147.60 W

Explanation:

Our given parameters include;

A Nichrome heating element operates on 120 V.

Voltage (V) = 120V

Initial Current (I₁) = 1.36 A

Initial Temperature (T₁) = 28°C

Final Current (I₂) = 1.23 A

Final Temperature (T₂) = unknown ????

Temperature dependencies of resistance is given by:

$R_{T(2)}=R_1[1+\alpha (T_2-T_1)]$ ---------------------- (1)

in which R₁ is the resistance at temperature T₁

$R_{T(2)$ is the resistance at temperature T₂

Given that V= IR

R = $\frac{V}{I}$

Therefore, the resistance at temperature 28°C is;

$R_{28}= \frac{120V}{1.36A}$

= 88.24Ω

$R_{T(2)$ = $\frac{120V}{1.23A}$

= 97.56Ω

From (1) above;

$R_{T(2)}=R_1[1+\alpha (T_2-T_1)]$

97.56 = 88.24 [ 1 + 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)]

$\frac{97.56}{88.24}= 1+(4.5*10^{-4})(T-28^0C)$

1.1056 - 1 = 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)

0.1056 = 4.5×10⁻⁴(T₂-28°C)

$\frac{0.1056}{4.5*10^{-4}}= T-28^0C$

T - 28° C = 234.67

T = 234.67 + 28° C

T = 262.67 ° C

(b)

What is the power dissipated in the heating element initially and when the current reaches 1.23 A

The power dissipated in the heating element initially can be calculated as:

P = I²₁R₂₈

P = (1.36A)²(88.24Ω)

P = 163.209 W

P ≅ 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A can be calculated as:

$P= I^2_2R_{T^0C$

P = (1.23)²(97.56Ω)

P = 147.598524

P ≅ 147.60 W

6 0

3 years ago

What process occurs when all of the energy from light waves is transferred to a medium?

nadya68 [22]

Absorption occurs when <span>all of the energy from light waves is transferred to a medium! </span>

7 0

3 years ago

Read 2 more answers

9. Una jeringa contiene cloro gaseoso, que ocupa un volumen de 95 mL a una presión de 0,96 atm. ¿Qué presión debemos ejercer en

masha68 [24]

Answer:

2.61 atm

Ley de Boyle

Explanation:

$P_1$ = Presión inicial = 0.96 atm

$P_2$ = Presión final

$V_1$ = Volumen inicial = 95 mL

$V_2$ = Volumen final = 35 mL

En este problema usaremos la ley de Boyle.

$\dfrac{P_1}{P_2}=\dfrac{V_2}{V_1}\\\Rightarrow P_2=\dfrac{P_1V_1}{V_2}\\\Rightarrow P_2=\dfrac{0.96\times 95}{35}\\\Rightarrow P_1=2.61\ \text{atm}$

La presión ejercida sobre el émbolo para reducir su volumen es de 2.61 atm.

4 0

3 years ago

Một dây dẫn đặt trong không khí có dòng điện I = 12A chạy qua, được gấp thành hình vuông cạnh a = 10cm. Xác định vectơ cường độ

disa [49]

Answer:

The net magnetic field ta the center of square is $1.36\times10^{-4} T$ .

Explanation:

Current, I = 12 A , side ,a = 10 cm = 0.1 m

Let the magnetic field due to the one side is B.

The magnetic field is given by

$B = \frac{\mu o}{4\pi}\times \frac{I}{r}\times \left (Sin A +Sin B \right )\\\\B = 10^{-7}\times \frac{12}{0.05}\times \left ( sin 45 + sin 45 \right )\\\\B = 3.4\times 10^{-5} T$

Net magnetic field at the center of the square is

B' = 4 B

$B'= 4\times 3.4\times 10^{-5}\\\\B' = 1.36\times10^{-4} T$

4 0

3 years ago