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AfilCa [17]
3 years ago
8

When a planet's orbit takes it closest to the Sun, its called:________.

Physics
1 answer:
andriy [413]3 years ago
3 0

Answer:

perihelion

Explanation:

The point at which a planet is closest to the sun is called perihelion. The farthest point is called aphelion

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What property does matter have that energy does not'?
krok68 [10]
<span>The property that matter has that energy does not is that matter has size, shape and occupies space. Matter also has inertia. Energy does not have any of these.</span>
8 0
3 years ago
Read 2 more answers
On a hot day, the deck of a small ship reaches a temperature of 48
AlekseyPX

The final temperature of the seawater-deck system is 990°C.

<h3>What is heat?</h3>

The increment in temperature adds up the thermal energy into the object. This energy is Heat energy.

The deck of a small ship reaches a temperature Ti= 48.17°C seawater on the deck to cool it down. During the cooling, heat Q =3,710,000 J are transferred to the seawater from the deck. Specific heat of seawater= 3,930 J/kg°C.

Suppose for 1 kg of sea water, the heat transferred from the system is given by

3,710,000 = 1 x 3,930 x (T - 48.17)

T = 990°C  to the nearest tenth.

The final temperature of the seawater-deck system is 990°C.

Learn more about heat.

brainly.com/question/13860901

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6 0
2 years ago
How many electrons circulate each second through the cross section of a conductor, which has a current intensity of 4A.
guajiro [1.7K]

Answer:

2.5×10¹⁹

Explanation:

4 C/s × (1 electron / 1.60×10⁻¹⁹ C) = 2.5×10¹⁹ electrons/second

8 0
4 years ago
A positive point charge Q1 = 2.5 x 10-5 C is fixed at the origin of coordinates, and a negative point charge Q2 = -5.0 x 10-6 C
mario62 [17]

Answer:

3.62 m  and - 1.4 m

Explanation:

Consider a location towards the positive side of x-axis beyond the location of charge Q₂

x = distance of the location from charge Q₂

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

\frac{kQ_{1}}{(2 + x)^{2}}= \frac{kQ_{2}}{x^{2}}

\frac{2.5\times 10^{-5}}{(2 + x)^{2}}= \frac{5 \times 10^{-6}}{x^{2}}

x = 1.62 m

So location is 2 + 1.62 = 3.62 m

Consider a location towards the negative side of x-axis beyond the location of charge Q₁

x = distance of the location from charge Q₁

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

\frac{kQ_{1}}{(x)^{2}}= \frac{kQ_{2}}{ (2 + x)^{2}}

\frac{2.5\times 10^{-5}}{(x)^{2}}= \frac{5 \times 10^{-6}}{(2+x)^{2}}

x = - 1.4 m

6 0
4 years ago
A radioactive isotope of potassium has a half-life of 20 minutes if a 40 g sample of this isotope is allowed to decay for 80 min
FinnZ [79.3K]

There will be 2.5g left.


Explanation:

80/20 = 4 half lives

40/2 = 20

20/2 = 10

10/2=5

5/2 = 2.5


5 0
3 years ago
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