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Mamont248 [21]
2 years ago
6

PLEASEE HELPPP IM GONNA FAIL NEED THIS BEFORE 9:30 MIDDLE SCHOOL SCIENCE​

Physics
1 answer:
dezoksy [38]2 years ago
7 0

Answer:

I think your soppose to multiply

Explanation:

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The ratio of the speed of light in a medium to the speed of light in a vacuum
Liula [17]

Answer:D.Refractive Indez

Explanation:

It is usually expressed the other way: the ratio of the speed of light in a vacuum to the speed of light in a medium. In that case, it is called the "index of refraction".

7 0
3 years ago
Read 2 more answers
A 5-g lead bullet traveling in 20°C air at 300 m/s strikes a flat steel plate and stops.
densk [106]

To solve this problem it is necessary to apply the concepts related to the Kinetic Energy and the Energy Produced by the heat loss. In mathematical terms kinetic energy can be described as:

KE = \frac{1}{2} mv^2

Where,

m = Mass

v = Velocity

Replacing we have that the Total Kinetic Energy is

KE = \frac{1}{2} mv^2

KE = \frac{1}{2} (5*10^{-3})(300)^2

KE =  225J

On the other hand the required Energy to heat up t melting point is

Q_1 = mC_p \Delta T

Q_2 = L_f m

Where,

m = Mass

C_p =Specific Heat

\Delta T =Change at temperature

L_f = Latent heat of fussion

Heat required to heat up to melting point,

Q = Q_1+Q_2

Q = mC_p \Delta T+L_f m

Q = 5*0.128*(327-20) + 5*24.7

Q = 310J

The energy required to melt is larger than the kinetic energy. Therefore the heat of fusion of lead would be 327 ° C: The melting point of lead.

4 0
3 years ago
Guys answer with a clear explanation and plzz don't spam.
timama [110]

Answer:

20.7N

Explanation:

There are many students who can not get answers step by step and on time

So there are a wats up group where you can get help step by step and well explained by the trusted experts.

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3 0
3 years ago
A battery has an electric potential of 1.5V and transfers 10.0 C between the two terminals. How much work was done?
bogdanovich [222]

Answer:

15 Joules

Explanation:

work = charge x potential difference

= 10 x 1.5

= 15

8 0
3 years ago
Saturn has an orbital period of 29.46 years. In two or more complete sentences, explain how to calculate the average distance fr
d1i1m1o1n [39]
For astronomical objects, the time period can be calculated using:
T² = (4π²a³)/GM
where T is time in Earth years, a is distance in Astronomical units, M is solar mass (1 for the sun)
Thus,
T² = a³
a = ∛(29.46²)
a = 0.67 AU
1 AU = 1.496 × 10⁸ Km
0.67 * 1.496 × 10⁸ Km
= 1.43 × 10⁹ Km
3 0
3 years ago
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