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garri49 [273]
3 years ago
14

Jupiter and the other jovian planets are sometimes called "gas giants." in what sense is this term misleading?

Physics
1 answer:
shepuryov [24]3 years ago
6 0
DEFINITION
The light stripes are regions of high clouds, and the dark stripes are regions where we can see down to deeper, darker clouds.

GOES WITH
Which of the following best why we see horizontal "stripes" in photographs of Jupiter and Saturn?
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Why is water considered a good solvent?
dem82 [27]

Answer:

D: Water is polar, so it can form attractions to and pull apart molecules of many types of substances.

Explanation:

Water is an excellent solvent simply because its polar water molecules usually form hydrogen bonds with ions and polar molecules, thereby allowing the ionic and polar covalent compounds to easily disperse easily in water.

Thus, the correct option is; D

7 0
3 years ago
Read 2 more answers
If a 5kg ball is traveling at 20 m/s and it’s stopped in 4s, what’s the impulse on the ball?
Anna [14]
FT=MV

F(4)=(5)(20)

F= 100/4

F= 25 N
4 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
A-delta fibers : A) are small, myelinated fibers. B) transmit pain signals at a slower rate than C-fibers. C) typically transmit
daser333 [38]

Answer:

A

Explanation:

Aδ fibers carry cold, pressure, and acute pain signals, and because they are thin (2 to 5 μm in diameter) and myelinated, they send impulses faster than unmyelinated C fibers, but more slowly than other, more thickly myelinated group A nerve fibers. Their conduction velocities are moderate.

7 0
3 years ago
Can you respond this two questions, please? :
Andrews [41]

Where's the diagram for question 1?

6 0
3 years ago
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