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3 years ago

Salita ng karanasan sa buya1.2.3.4.5.

Physics

1 answer:

sergeinik [125]3 years ago

4 0

Answer:

Explanation:Salita ng karanasan sa buya 1. 2. 3. 4. 5.. Add answer+5 pts. Log in to add comment. jeniahashby18 is waiting for your help. Add your answer

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an ice sheet 5m thick covers a lake that is 20m deep. at what is the temperature of the water at the bottom of the lake?

muminat

Answer:

4°C

Explanation:

Water is densest at 4°C. Since dense water sinks, the bottom of the lake will be 4°C.

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3 years ago

The unit of energy is a derived unit

Tems11 [23]

Explanation:

Hi, there!!

Energy is defined as the capacity or ability to do work. It's SI unit is Joule.

here,

Joule = (kg×m×m)/(s×s)

 = kg×m^2/s^2.

Therefore, the derived unit is kg.m^2 by s^2.

Hope it helps...

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3 years ago

What distance does light travel in water, glass, and diamond during the time that it travels 1.0 m in vacuum? the refractive ind

alexdok [17]

We are given with
distance traveled through vacuum = 1.0 m
refractive index of water = 1.33
refractive index of glass = 1.50
refractive index of diamond = 2.42

distance traveled through water is = 1.0/1.33 = 0.75 m
distance traveled through water is = 1.0/1.50 = 0.67 m
distance traveled through water is = 1.0/2.42 = 0.41 m

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3 years ago

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A 25kg box fell 200m with an acceleration of 5 m/s2. with what force did it hit the floor when it landed?

Marizza181 [45]

According to Newton's Second Law of motion, the net force acting on the object is equal to its mass multiplied by its acceleration. In formula, it is written as

Net Force =mass * acceleration
Net force = 25 kg * 5m/s^2
Net force = 125 Newtons

6 0

3 years ago

What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of −25.0 nc? g?

Daniel [21]

To calculate the force between two negative charges, we use the formula which is given by the Coulomb`s Law as

$F=\frac{kq_{1}q_{2} }{r^{2} }$

Here, $q_{1}$ and $q_{2}$ are the charges on the pith balls, r is the separation between the charges and k is constant and its value is $8.99\times 10^9 N m^2/C^2$ .

Given $q_{1} =q_{2} =-25 nC=-25\times 10^{-9}C$ and $r=8 cm=8\times10^{-2} m$ .

Substituting these values in above formula we get,

$F= 8.99\times 10^9 N m^2/C^2\frac{(-25\times 10^{-9}C)(-25\times 10^{-9}C)}{(8\times10^{-2} m)^2} \\\\\ F= 877929.7\times 10^{-9} N\\\\F=8.8\times10^{-4}N$

Thus, the repulsive force between two pith balls is $8.8\times10^{-4}N$ .

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3 years ago

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Salita ng karanasan sa buya1.2.3.4.5.​

Salita ng karanasan sa buya1.2.3.4.5.