All categories

4 years ago

If ice didn't float, but sank instead, this change wouldn't affect aquatic life. true of false

Physics

2 answers:

ruslelena [56]4 years ago

6 0

Answer:

No it will not (false)

Explanation:

I say this because if anything the ice will sink and melt from a solid to a liquid and that will lead to i guess more water for the aquatic life. (THIS IS A THEORY SHAWTY)

Volgvan4 years ago

5 0

False

the aquatic lif would have to make significant changes to how they hunt and find shelter. Especially in Antartica

You might be interested in

For your senior project, you would like to build a cyclotron that will accelerate protons to 10% of the speed of light. The larg

jeyben [28]

Answer:

<h2>Magnetic field strength in that region is 1.2 T</h2>

Explanation:

As we know by the formula of radius of charge moving in external field is given as

$R = \frac{mv}{qB}$

so we will have

$R = 25 cm$

$m = 1.6 \times 10^{-27} kg$

$q = 1.6 \times 10^{-19} C$

$v = 0.10 \times 3 \times 10^8 m/s$

now we have

$0.25 = \frac{(1.6 \times 10^{-27})(3\times 10^7)}{(1.6 \times 10^{-19})B}$

now we have

$B = 1.2 T$

8 0

3 years ago

Pls Answer Quick Will Mark 5 Stars And brain lists

ra1l [238]

Answer:

Food mixer will convert electrical energy into kinetic energy

Explanation:

Energy conservation means that energy can not be created nor destroyed it can be converted from one form into another

7 0

3 years ago

Please help!!! I have a physics exam tomorrow and I just can't wrap my head around this one!

Ainat [17]

The total electrostatic force on charge A is $28 \mu N$

Explanation:

The magnitude of the electrostatic force between two charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

Here we have three positively charged particles A,B and C, located at the following positions:

$x_A = 0\\x_B = 10 m\\x_C = 20 m$

The magnitudes of the three charges are:

$q_A = q_B = q_C = 0.5 \mu C = 0.5\cdot 10^{-6}C$

The force exerted by B on A is to the left (because the force between two positive charges is repulsive), and the force exerted by C on A is also to the left (also repulsive). Therefore, the net force on A is just the sum of the two forces exerted by charges B and C:

$F_A = F_{BA} + F_{CA} = k\frac{q_B q_A}{(x_B-x_A)^2}+k\frac{q_C q_A}{(x_C-x_A)^2}=\\=(8.99\cdot 10^9) \frac{(0.5\cdot 10^{-6})^2}{(10)^2}+(8.99\cdot 10^9) \frac{(0.5\cdot 10^{-6})^2}{(20)^2}=2.8\cdot 10^{-5} N = 28 \mu N$