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2 years ago

Vaccine ASAP don't got much time

Physics

2 answers:

lesantik [10]2 years ago

3 0

Answer:

it's D

Explanation:

Verizon [17]2 years ago

3 0

Answer:

um it is d

Explanation:

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How many total electrons are there in a water molecule<br><br>4<br>10<br>2

andre [41]

10 electrons are in a water molecule. :)

3 0

3 years ago

What is the voltage of the power source if a motor with a resistance of 20 2 draws a

Lady_Fox [76]

Answer:

190v I believe

hope this helped a little and if it did pls mark brainiest :)

8 0

3 years ago

A 0.2-stone is attached to a string and swung in a circle of radius 0.6 m on a horizontal and frictionless surface. If the stone

Nikitich [7]

Answer:

2960 N

Explanation:

Convert rev/min to rad/s:

150 rev/min × (2π rad/rev) × (1 min / 60 s) = 50π rad/s

Sum of forces in the centripetal direction:

∑F = ma

T = m v² / r

T = m ω² r

T = (0.2 kg) (50π rad/s)² (0.6 m)

T = 2960 N

3 0

3 years ago

PLEASE HELP PLEASE HELP ME PLEASE HELP ME

klemol [59]

Answer:24

Explanation:

7 0

2 years ago

A m = 94.2 kg object is released from rest at a distance h = 1.15134 R above the Earth’s surface. The acceleration of gravity is

OleMash [197]

Answer:

v= 4055.08m/s

Explanation:

This is a problem that must be addressed through the laws of classical mechanics that concern Potential Gravitational Energy.

We know for definition that,

$U = \frac{GMm}{r}$

We must find the highest point and the lowest point to identify the change in energy, so

Point a)

The problem tells us that an object is dropped at a distance of h = 1.15134R over the earth.

That is to say that the energy of that object is equal to,

$U_1=-\frac{(6.6738 * 10^{-11})(5.98 * 10^{24})(94.2)}{(1.15134)(6.38*10^6)}$

$U_2= - 5.1180*10^9J$

Point B )

We now use the average radius distance from the earth.

$U_2=-\frac{(6.6738 * 10^{-11})(5.98 * 10^{24})(94.2)}{(6.38*10^6)}$

$U_2= -5.8925*10^9J$

Then,

$\Delta U = U_2 - U_1 = -5.1180*10^9J - ( -5.8925*10^9J)$

$\Delta U = 774.5*10^6$

By the law of conservation of energy we know that,

$\Delta U = \frac{1}{2}mv^2$

clearing v,

$v= \sqrt{2 \Delta U/m}$

$v= \sqrt{2*774.5*10^6 /94.2}$

$v= 4055.08m/s$

Therefore the speed of the object when it strikes the Earth’s surface is 4055.08m/s

8 0

3 years ago