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chubhunter [2.5K]
3 years ago
6

From inside to out, describe the components of an atom. What makes one element's atoms differ from another element's atoms?

Physics
1 answer:
Vesna [10]3 years ago
4 0
From the outside is it’s shell, and on the inside is the nucleus (on the middle) which is made or protons (positive charge) and neutrons (neutral charge which is negate and positive) and Lastly there are electrons (negative charge) that orbit the nucleus like planets orbiting the Sun. The shell gets made as the electrons orbit the nucleus.
How do atoms differ from eachother?- their atoms might have different elections, neutrons, and/ or protons inside of it. Plus the location of the elections might be different too.
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A 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the cable for the following case
Murljashka [212]

Answer:

The answer is below

Explanation:

A 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the

cable for the following cases:

a. The load moves downward at a constant velocity

b. The load accelerates downward at a rate 0.4 m/s??

C. The load accelerates upward at a rate 0.4 m/s??

Solution:

Acceleration due to gravity (g) = 10 m/s²

a) Given that the mass of the crane (m) is 140 kg. If the load moves downward, the tension (T) is given by:

mg - T = ma

Since the load has a constant velocity, hence acceleration (a) = 0. Therefore:

mg - T = m(0)

mg - T = 0

T = mg

T = 140(10) = 1400 N

T = 1400 N

b)  If the load moves downward, the tension (T) is given by:

mg - T = ma

T = mg - ma = m(g - a)

T = 140(10 - 0.4) = 140(9.96) = 134.4

T = 134.4 N

c)  If the load moves upward, the tension (T) is given by:

T - mg = ma

T = ma + mg = m(a + g)

T = 140(0.4 + 10) = 140(10.4)

T = 145.6 N

2) To find the distance (s) if the load move from rest (u= 0) and accelerates for 20 seconds (t = 20). We use:

s = ut + (1/2)gt²

s = 0(20) + (1/2)(10)(20)²

s = 2000 m

7 0
3 years ago
A 200. N wagon is to be pulled up a 30 degree incline at constant speed. How large a force parallel to the incline force is need
Sever21 [200]
For the question above, here is the equation to follow:
<span>F = mgsinα = Wsinα
      =200 x 0.5 = 100 N
</span>OR

<span>Sin30 * 200N = 100 N
</span>
The asnwer is 100N. I hope this answer helps.
3 0
3 years ago
Read 2 more answers
calculate the temperature at which the reading on Fahrenheit scale is equal to half the reading of Celsius scale​
RideAnS [48]

Answer:

Therefore, the temperature at which the Fahrenheit scale reading is equal to half of the Celsius scale is −24.6∘C .

3 0
3 years ago
An object is dropped from a​ tower, 400 ft above the ground. The​ object's height above ground x seconds after the fall is ​s(x)
pav-90 [236]

1) 5 s

The vertical position of the object is given by

y(t) = h - \frac{1}{2}gt^2 =  400 - 16 t^2

where

h=400 ft represents the initial height

g = 32 ft/s^2 is the acceleration of gravity

t is the time

We want to find the time t at which the object reaches the ground, so the time t at which

y(t) = 0

By substituting this into the equation, we find

0 = 400 - 16t^2\\t=\sqrt{\frac{400}{16}}=5 s

2) 160 ft/s

The object is released from rest, so the initial velocity is zero

u = 0

The final vertical velocity can be found by using

v^2 - u^2 = 2ah

where

v is the final velocity

a = 32 ft/s^2 is the acceleration of gravity

h = 400 ft is the vertical distance covered

Solving for v, we find

v=\sqrt{u^2 +2ay}=\sqrt{2(32 ft/s)(400 ft)}=160 ft/s

8 0
3 years ago
Which has more momentum, a 2000 lb car moving at 100 km/hr or a 4000 lb truck moving at 50 km/hr ?
BabaBlast [244]

Answer:

The truck and car have the same momentum.

Explanation:

m_1 = Mass of car = 2000\ \text{lb}=2000\times0.45359237\ \text{kg}

v_1 = Velocity of car = \dfrac{100}{3.6}\ \text{m/s}

m_2 = Mass of truck = 4000\times0.45359237\ \text{kg}

v_2 = Velocity of truck = \dfrac{50}{3.6}\ \text{m/s}

Momentum of car

p_1=m_1v_1\\\Rightarrow p_1=2000\times0.45359237\times \dfrac{100}{3.6}\\\Rightarrow p_1=25199.58\ \text{kg m/s}

Momentum of the truck

p_2=m_2v_2\\\Rightarrow p_2=4000\times0.45359237\times \dfrac{50}{3.6}\\\Rightarrow p_2=25199.58\ \text{kg m/s}

Both the truck and car have the same momentum of 25199.58\ \text{kg m/s}.

8 0
3 years ago
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