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den301095 [7]
2 years ago
14

Which hrase describes what atoms are made of? OA. A cloud of protons and neutrons surrounded by a nucleus of electrons OB. A nuc

leus of neutrons and electrons surrounded by a cloud of protons OC. A nucleus of protons and neutrons surrounded by a cloud of electrons OD. A nucleus of protons and electrons surrounded by a cloud of neutrons ​
Physics
1 answer:
ExtremeBDS [4]2 years ago
7 0

Answer:

(OC) a nucleus of protons and neutrons surrounded by a cloud of electrons - modern description of an atom - where the nucleus may be hundreds of times smaller in diameter than the cloud of electrons

You might be interested in
Problem 3: Thermal expansionThe steel rod has the length 2 m and cross-section area 200 cm2at the room temperature 20◦C. Weapply
zalisa [80]

Answer:

a) 2.00024 m

b) 0.036%

c) 436.67°C

Explanation:

Given

Initial length = L₀ = 2 m

Initial cross sectional Area = A₀ = 200 cm² = 0.02 m²

We can obtain initial volume = V₀ = A₀L₀ = 0.02 × 2 = 0.04 m³

Initial Temperature = T₀ = 20°C

Coefficient of linear expansivity = α = (2 × 10⁻⁶) (°C)⁻¹

a) New length of the rod after heating to 80°C

Linear expansion is given as

ΔL = L₀ × α ×ΔT

ΔL = 2 × 2 × 10⁻⁶ × (80 - 20) = 0.00024 m = 0.24 mm

New length = old length + expansion = 2 + 0.00024 = 2.00024 m

b) The percentage of the volume change of the rod.

Volume expansion is given by

ΔV = V₀ × (3α) × ΔT

Volume expansivity ≈ 3 × (linear expansivity)

ΔV = 0.04 × (3×2×10⁻⁶) × (80 - 20) = 0.0000144 m³

Percentage change in volume = 100% × (ΔV/V₀) = 100% × (0.0000144/0.04) = 0.036%

c) The maximal temperature we can allow if the volume should not increase by more than half percent.

For a half percent increase in volume, the corresponding change in volume needs to be first calculated.

Percentage change in volume = 100% × (ΔV/V₀)

0.5 = 100% × (ΔV/0.04)

(ΔV/0.04) = 0.005

ΔV = 0.0002 m³

Then we now investigate the corresponding temperature that causes this.

ΔV = V₀ × (3α) × ΔT

0.0002 = 0.04 × (3×2×10⁻⁶) × ΔT

ΔT = (0.0002)/(0.04 × 3 × 2 × 10⁻⁶) = 416.67°C

Maximal temperature = T₀ + ΔT = 20 + 416.67 = 436.67°C

4 0
2 years ago
In a circus act, when the tumbling lever is measured, the measurements show that the effort arm is 8 meters long and the resista
Yuki888 [10]

The statement the mechanical advantage of the lever is 16 is False.

<h3>Mechanical Advantage</h3>

Mechanical advantage MA = d/D where

  • d = distance moved by effort and
  • D = distance moved by load

Given that the effort arm is 8 meters long, d = distance moved by effort = 8 m.

Also, given that the resistance arm is 2 meters long, D = distance moved by load = 2 m.

<h3>Calculating the mechanical advantage</h3>

So, substituting the values of the variables into the equation, we have

MA = d/D

MA = 8 m/2 m

MA = 4

Since MA = 4, so, the stament is False.

So, the statement the mechanical advantage of the lever is 16 is False.

Learn more about mechanical advantage here:

brainly.com/question/13779480

3 0
2 years ago
The skateboarder has a mass of 100 kg. When traveling downward from E to D, he reaches a velocity of 11 m/s. Calculate his kinet
patriot [66]

6050 J is the kinetic energy at D

<u>Explanation:</u>

In physics, the object's kinetic energy (K.E) defined as the energy it possesses during movement. It can be defined as the required work to accelerate a certain body weight in order to rest at a certain speed. When the body receives this energy as it speeds up (accelerates), it retains this energy unless speed varies. The equation is given as,

           K . E=\frac{1}{2} \times m \times v^{2}

Where,

m - mass of an object

v - velocity of the object

Here,

Given data:

m  = 100 kg

v = 11 m/s

By substituting the given values in the above equation, we get

            K . E=\frac{1}{2} \times 100 \times(11)^{2}=\frac{1}{2} \times 100 \times 121=\frac{12100}{2}=6050\ \mathrm{J}

6 0
3 years ago
Consider three scenarios in which a particular box moves downward under the pull of gravity :
luda_lava [24]

Answer:

1. True  WA > WB > WC

Explanation:

In this exercise they give work for several different configurations and ask that we show the relationship between them, the best way to do this is to calculate each work separately.

A) Work is the product of force by distance and the cosine of the angle between them

    WA = W h cos 0

   WA = mg h

B) On a ramp without rubbing

     Sin30 = h / L

     L = h / sin 30

     WB = F d cos θ  

     WB = F L cos 30

     WB = mf (h / sin30) cos 30

     WB = mg h ctan 30

C) Ramp with rubbing

    W sin 30 - fr = ma

   N- Wcos30 = 0

   W sin 30 - μ W cos 30 = ma

    F = W (sin30 - μ cos30)

   WC = mg (sin30 - μ cos30) h / sin30

   Wc = mg (1 - μ ctan30) h

When we review the affirmation it is the work where there is rubbing is the smallest and the work where it comes in free fall at the maximum

Let's review the claims

1. True The work of gravity is the greatest and the work where there is friction is the least

2 False. The job where there is friction is the least

3 False work with rubbing is the least

4 False work with rubbing is the least

5 0
3 years ago
If an object is thrown in an upward direction from the top of a building 1.6 x 10^2 ft. high at an initial velocity of 21.82 mi/
Goshia [24]

If an object is thrown in an upward direction from the top of a building 1.60 x 102 ft. high at an initial velocity of 21.82 mi/h, what is its final velocity when it hits the ground? (Disregard wind resistance. Round answer to nearest whole number and do not reflect negative direction in your answer.)


this question is troubling me i guessed 96 ft/s

can someone help me out and explain it thanks so much!!!!!!



7 0
3 years ago
Read 2 more answers
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