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olya-2409 [2.1K]

3 years ago

14

Which of the following pairings are more likely to be held together with the strong nuclear force?

1 answer:

aivan3 [116]3 years ago

8 0

Which of the following pairings are more likely to be held together with the strong nuclear force

Explanation:

1.What does a strong nuclear force do in an atom? It repels electrons from other electrons. It repels protons from other protons. It attracts protons and neutrons.

2.The chain reaction requires both the release of neutrons from fissile isotopes undergoing nuclear fission and the subsequent absorption of some of these neutrons in fissile isotopes.

3.The strong nuclear force holds most ordinary matter together because it confines quarks into hadron particles such as the proton and neutron. In addition, the strong force binds these neutrons and protons to create atomic nuclei.

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0.0010 kg pellet is fired at a speed of 52.2 m/s at a motionless 3.3 kg piece of balsa wood. When the pellet

Paraphin [41]

The formula for momentum is p = m*v

The conservation of momentum suggests:

m*vi = m*vf (initial mass times initial velocity = final mass times final velocity or initial momentum = final momentum)

(0.0010)(52.2) = (0.0010 + 3.3)vf

vf = (0.0010)(52.2)/(0.0010 + 3.3) = 0.0522/3.301 ≈ 0.01581 m/s

To the nearest thousandth ≈ .016 m/s

7 0

3 years ago

A very elastic rubber ball is dropped from a certain height and hits the floor with a (2pts) downward speed v. Since it is so el

artcher [175]

Answer:

3.True. The magnitude of momentum is the same

Explanation:

Let's propose the solution of the problem

The initial moment is

p₀ = m v

The final moment

$p_{f}$ = m (-v)

p₀ = - $p_{f}$

Now we can review the claims

1. False. We see that the moment module is the same, but its direction changes

2. False. The impulse is a vector

3.True. The magnitude of momentum is the same

7 0

3 years ago

At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107

dangina [55]

Answer:

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

Explanation:

The bulk modulus is represented by the following differential equation:

$K = - V\cdot \frac{dP}{dV}$

Where:

$K$ - Bulk module, measured in pascals.

$V$ - Sample volume, measured in cubic meters.

$P$ - Local pressure, measured in pascals.

Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:

$-\frac{K \,dV}{V} = dP$

This resultant expression is solved by definite integration and algebraic handling:

$-K\int\limits^{V_{f}}_{V_{o}} {\frac{dV}{V} } = \int\limits^{P_{f}}_{P_{o}}\, dP$

$-K\cdot \ln \left |\frac{V_{f}}{V_{o}} \right| = P_{f} - P_{o}$

$\ln \left| \frac{V_{f}}{V_{o}} \right| = \frac{P_{o}-P_{f}}{K}$

$\frac{V_{f}}{V_{o}} = e^{\frac{P_{o}-P_{f}}{K} }$

The final volume is predicted by:

$V_{f} = V_{o}\cdot e^{\frac{P_{o}-P_{f}}{K} }$

If $V_{o} = 1\,m^{3}$ , $P_{o} - P_{f} = -10132500\,Pa$ and $K = 2.3\times 10^{9}\,Pa$ , then:

$V_{f} = (1\,m^{3}) \cdot e^{\frac{-10.1325\times 10^{6}\,Pa}{2.3 \times 10^{9}\,Pa} }$

$V_{f} \approx 0.996\,m^{3}$

Change in volume due to increasure on pressure is:

$\Delta V = V_{o} - V_{f}$

$\Delta V = 1\,m^{3} - 0.996\,m^{3}$

$\Delta V = 0.004\,m^{3}$

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

8 0

3 years ago

A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider

shutvik [7]

Answer:

4.0 m/s

Explanation:

The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.

Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is

$d=v_x t$

where here we have

d = 3.0 m is the horizontal distance covered

vx is the horizontal velocity

t = 1.3 s is the duration of the fall

Solving for vx,

$v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s$

Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

$y(t) = h + v_y t - \frac{1}{2}gt^2$

where

h = 4.0 m is the initial height

vy is the initial vertical velocity

We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy

$0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s$

So now we can find the magnitude of the initial velocity:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.3 m/s)^2+(3.3 m/s)^2}=4.0 m/s$

4 0

3 years ago

Draw vectors to represent these motions:<br>moving 3 cm to the right<br>moving 2 cm to the left

Ira Lisetskai [31]

Vector. Oh yeeeeeeeeahhh

7 0

2 years ago