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OverLord2011 [107]
3 years ago
6

Which contributes most to our average dose of background radiation?

Physics
1 answer:
Anuta_ua [19.1K]3 years ago
6 0
C) Natural and Artificial sources
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A 150 g egg is dropped from 3.0 meters. The egg is moving at 4.4 m/s right before it hits the ground. The egg comes to a stop in
algol [13]

Answer: Magnitude of the force exerted on the egg by the ground is 9.2N

Explanation:

Given the following :

Mass of egg (m) = 150g = 0.15kg

Height(h) from which egg is dropped = 3m

velocity of egg before hitting the ground (u) = 4.4m/s

Final velocity of egg (V) = 0

Time taken (t) = 0.072s

Magnitude of the force exerted on the egg by the ground can be found by applying Newton's 2nd law:

Momentum = mass × velocity

From Newton's second law:

Force = mass × change in Velocity with time ;

That is

F = m * ΔV / t

Inputting our values

F = 0.15 * (4.4 - 0) / 0.072

F = 0.15 × (4.4 / 0.072)

F = 0.15 × 61.11

F = 9.16N

F = 9.2N

8 0
2 years ago
A block has two strings attached to it on opposite ends. One string has a force of 5 N,
juin [17]

Unless you have a diagram to include or any other additional info, I'll assume the block is being pulled by two opposing forces along the horizontal surface.

Horizontally, the block is under the influence of

• one rope pulling in one direction with magnitude 15 N,

• the other rope pulling in the opposite direction with mag. 5 N, and

• friction, opposing the direction of the block's motion, with mag. 3 N.

It stands to reason that the block is accelerating in the direction of the larger pulling force.

(A) By Newton's second law, we have

15 N + (-5 N) + (-3 N) = <em>m</em> (1 m/s²)

where <em>m</em> is the mass of the block. Solve for <em>m</em> :

7 N = <em>m</em> (1 m/s²)

<em>m</em> = (7 N) / (1 m/s²)

<em>m</em> = 7 kg

(B) The friction force is proportional to the normal force, so that if <em>f</em> is the mag. of friction and <em>n</em> is the mag. of the normal force, then <em>f</em> = <em>µ</em> <em>n</em> where <em>µ</em> is the coefficient of friction.

The block does not bounce up and down, so its vertical forces are balanced, which means the normal force and the block's weight (mag. <em>w</em>) cancel out:

<em>n</em> + (-<em>w</em>) = 0

<em>n</em> = <em>w</em>

<em>n</em> = <em>m</em> <em>g</em>

where <em>g</em> = 9.8 m/s² is the mag. of the acceleration due to gravity.

<em>n</em> = (7 kg) (9.8 m/s²)

<em>n</em> = 68.6 N

Then

3 N = <em>µ</em> (68.6 N)

<em>µ</em> = (3 N) / (68.6 N)

<em>µ</em> ≈ 0.044

4 0
2 years ago
Interactive Solution 6.39 presents a model for solving this problem. A slingshot fires a pebble from the top of a building at a
mariarad [96]

(a) 29.8 m/s

To solve this problem, we start by analyze the vertical motion first. This is a free fall motion, so we can use the following suvat equation:

v_y^2 - u_y^2 = 2as

where, taking upward as positive direction:

v_y is the final vertical velocity

u_y = 0 is the initial vertical velocity (zero because the pebble is launched horizontally)

a=g=-9.8 m/s^2 is the acceleration of gravity

s = -25.0 m is the displacement

Solving for vy,

v_y = \sqrt{u^2+2as}=\sqrt{0+2(-9.8)(-25)}=-22.1 m/s (downward, so we take the negative solution)

The pebble also have a horizontal component of the velocity, which remains constant during the whole motion, so it is

v_x = 20.0 m/s

So, the final speed of the pebble as it strikes the ground is

v=\sqrt{v_x^2+v_y^2}=\sqrt{20.0^2+(-22.1)^2}=29.8 m/s

(b) 29.8 m/s

In this case, the pebble is launched straight up, so its initial vertical velocity is

u_y = 20.0 m/s

So we can find the final vertical velocity using the same suvat equation as before:

v_y^2 - u_y^2 = 2as

v_y = \sqrt{u^2+2as}=\sqrt{(20.0)^2+2(-9.8)(-25)}=-29.8 m/s (downward, so we take the negative solution)

The horizontal speed instead is zero, since the pebble is initially launched vertically, so the final speed is just equal to the magnitude of the vertical velocity:

v = 29.8 m/s

(c) 29.8 m/s

This case is similarly to the previous one: the only difference here is that the pebble is launched straight down instead than up, therefore

u_y = -20.0 m/s

Using again the same suvat equation:

v_y^2 - u_y^2 = 2as

v_y = \sqrt{u^2+2as}=\sqrt{(-20.0)^2+2(-9.8)(-25)}=-29.8 m/s (downward, so we take the negative solution)

As before, the horizontal speed instead is zero, since the pebble is initially launched vertically, so the final speed is just equal to the magnitude of the vertical velocity:

v = 29.8 m/s

We notice that the final value of the speed is always the same in all the three parts, so it does not depend on the direction of launching. This is due to the law of conservation of energy: in fact, the initial mechanical energy of the pebble (kinetic+potential) is the same in all three cases (because the height h does not change, and the speed v does not change either), and the kinetic energy gained during the fall is also the same (since the pebble falls the same distance in all 3 cases), therefore the final speed must also be the same.

7 0
3 years ago
Use the table of electric force between objects in two different interactions to answer the question. Interaction Charge on Obje
forsale [732]

Answer:?that’s a lot

Explanation:

6 0
2 years ago
Which trait is determined entirely by heredity ?
quester [9]
Shape of the nose is entirely hereditary, as genes have nothing to do with any of the other attributes.
7 0
3 years ago
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