All categories

3 years ago

For a large atom, which of the following would give the most stable nuclear configuration?

Physics

2 answers:

mrs_skeptik [129]3 years ago

6 0

Answer:D

Explanation:

Hope it helps

seraphim [82]3 years ago

4 0

My answer is D but I will still try to word it out again

You might be interested in

If a girl carries groceries up a flight of stairs, is she doing work on the groceries? Explain.

nikdorinn [45]

Answer:

Explanation:

Yes she is doing work. With or without the groceries, she is still doing work. She does more work with the groceries than without because Work is defined by F which is defined by mass. The mass increases with the groceries.

The work done is against the force of gravity.

4 0

3 years ago

What does the earths core do

cestrela7 [59]

Answer: Earth scientists have theorized that the Earth's core is responsible for the planet's magnetic field as well as plate tectonics.

Explanation:

6 0

3 years ago

You attach a meter stick to an oak tree, such that the top of the meter stick is 2.27 meters above the ground. later, an acorn f

Alexandra [31]

The acorn was at a height of <u>4.15 m</u> from the ground before it drops.

The acorn takes a time t to fall through a distance h₁, which is the length of the scale. When the acorn reaches the top of the scale, its velocity is u.

Calculate the speed of the acorn at the top of the scale, using the equation of motion,

$s=ut+ \frac{1}{2} at^2$

Since the acorn falls freely under gravity, its acceleration is equal to the acceleration due to gravity g.

Substitute 2.27 m for s (=h₁), 0.301 s for t and 9.8 m/s² for a (=g).

$s=ut+ \frac{1}{2} at^2\\ (2.27 m)=u(0.301s)+\frac{1}{2}(9.8m/s^2)(0.301s)^2\\ u=\frac{1.8261m}{0.301s} =6.067m/s$

If the acorn starts from rest and reaches a speed of 6.067 m/s at the top of the scale, it would have fallen a distance h₂ to achieve this speed.

Use the equation of motion,

$v^2=u^2+2as$

Substitute 6.067 m/s for v, 0 m/s for u, 9.8 m/s² for a (=g) and h₂ for s.

$v^2=u^2+2as\\ (6.067m/s)^2=(0m/s)^2+2(9.8m/s^2)h_2\\ h_2=\frac{(6.067m/s)^2}{2(9.8m/s^2)} =1.878 m$

The height h above the ground at which the acorn was is given by,

$h=h_1+h_2=(2.27 m)+(1.878 m)=4.148 m$

The acorn was at a height <u>4.15m</u> from the ground before dropping down.

3 0

2 years ago

A group of students collected the data shown below while attempting to measure the coefficient of static friction (of course, it

anzhelika [568]

Answer:

0.130

Explanation:

From the given data, the coefficient of static friction for each trial are:

1. 0.053

2. 0.081

3. 0.118

4. 0.149

5. 0.180

6. 0.198

The sum of the coefficient of static friction = 0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198

= 0.779

So that;

the average coefficient of static friction = $\frac{sum of coefficient of static friction}{number of trials}$

= $\frac{0.779}{6}$

= 0.12983

The average coefficient of static friction is 0.130

8 0

2 years ago

C) A sample of substance of volume 10 cm3 was brought back to Earth from the Moon.

postnew [5]

Answer:

8125

Explanation:

$P=0,13 N\\a=1,6 N/kg\\m=P/a=0,13/1,6=0,08125\\V=10(cm)^{3} =10^{-5} meters^{3} \\p=m/V=0,08125/10^{-5} =0,08125*100000=8125kg/meters^{3}$

3 0

3 years ago