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3 years ago

Electrons contribute___

Physics

2 answers:

dangina [55]3 years ago

4 0

C, electrons are negative, and most of the atom’s mass comes from the nucleus of the atom

coldgirl [10]3 years ago

3 0

C negative, little hope this helps

You might be interested in

What is the gravitational potential energy of a 3 kg ball that is 1 meter above the floor?

Sveta_85 [38]

Answer:

PE=mgh

M= Mass (kg)

G= Gravitational field strength (N/kg)

H= Hight (m)

PE= Gravitational Potential Energy (J)

Explanation:

Gravitational Potential Energy is the energy stored in a object due to its position above the Earth's surface.

7 0

3 years ago

A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr

KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s$

Time the parachutist falls without friction is 3.19 seconds

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s$

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

$v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s$

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2$

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2$

Now the initial velocity of the last half height will be the final velocity of the first half height.

$v=u+at\\\Rightarrow v=at$

Since the height are equal

$\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0$

$t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45$

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2$

Magnitude of acceleration is -43200 ft/s²

5 0

3 years ago

Which of the following experiments could be used to determine the inertial mass of a block? A. Place the block on a rough horizo

valentinak56 [21]

Answer:

D, using a spring scale to exert a force on the block. Measure the acceleration of the block and the applied force

Explanation:

For this you would use the net force equation acceleration=net force * mass however you will want to isolate mass so it would be acceleration/ net force to get mass. Then process of elimination comes to play.

3 0

3 years ago

A distant large asteroid is detected that might pose a threat to Earth. If it were to continue moving in a straight line at cons

Vlada [557]

Answer:

The minimum speed required is 5.7395km/s.

Explanation:

To escape earth, the kinetic energy of the asteroid must be greater or equal to its gravitational potential energy:

$K.E\geq P.E$

$\dfrac{1}{2}mv^2 \geq G\dfrac{Mm}{R}$

where $m$ is the mass of the asteroid, $R= 24,000,000\:m$ is its distance form earth's center, $M = 5.9*10^{24}kg$ is the mass of the earth, and $G = 6.7*10^{-11}m^3/kg\: s^2$ is the gravitational constant.