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Arte-miy333 [17]
3 years ago
5

In which of the following eons did a significant amount of oxygen gas exist in Earth's atmosphere?

Physics
1 answer:
Llana [10]3 years ago
6 0
C that is the answer
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NEED ANSWER ASAP!!! Angela has a bucket of mass 2 kg tied to a string. She places a drinking glass of mass 0.5 kg in the bucket.
Schach [20]

a. The free-body diagram for the glass when it is at the top of the circle is attached below.

b. The equation for the net force on the glass at the top of the circle in terms of w, Fn, m, v, and r is mg x g + N -  mg x Vtop² /R =0

c. The glass will fall out of the bucket if the normal force between the glass and bucket equals zero. The speed with which she spin the bucket to prevent this from happening is 3.83 m/s.

d. The string will break if the tension on it is more than 100 N. The range of speeds can  prevent the string from breaking is 3.83< Vtop<4.99 m/s

<h3>What is Net force?</h3>

When two or more forces are acting on the system of objects, then the to attain equilibrium, net force must be zero.

Given, Angela has a bucket of mass 2 kg tied to a string. She places a drinking glass of mass 0.5 kg in the bucket. She spins the bucket in a vertical circle of radius 1.5 m. She must swing the bucket to keep the glass from falling out.

a. The free body diagram of the bucket and glass is attached below.

b. Bucket will undergo centrifugal force

Fb = mVtop² /R

From the equilibrium of forces, we have

For bucket,

T +mb xg - N =  mb x Vtop² /R..............(1)

For glass,

mg x g + N =  mg x Vtop² /R..............(2)

Thus, this is the net force equation on the glass.

c. On adding both the equations. we have

T + (mb + mg) xg = (mb + mg) Vtop² /R

Substituting the values, T = 0 and from the question, we get

0 + (2+0.5) 9.81 = (2+0.5)(Vtop²/0.5)

Vtop = 3.83 m/s

Thus, the speed of spin to prevent glass from falling out is 3.83 m/s

d. The string will break if the tension on it is more than 100 N

100 + (2+0.5) 9.81 = (2+0.5)(Vtop²/0.5)

Vtop = 4.99 m/s

Thus, the range of velocity is  3.83< Vtop<4.99 m/s

Learn more about net force.

brainly.com/question/18031889

#SPJ1

8 0
2 years ago
Which best describes two counteracting forces on an object
Natali5045456 [20]

The correct answer is B two children pulling apart a wishbone

Let me know if you have any questions, and have a nice day!

6 0
3 years ago
Read 2 more answers
Bose-Einstein Condensate can be defined by Group of answer choices slowly vibrating molecules all possibilities listed are corre
deff fn [24]

Answer:

The correct option is;

Absolute zero

Explanation:

A Bose-Einstein condensate is known as the fifth state of matter which is made of a collection of ultra cooled atoms (at almost absolute zero degrees -273.15 °C) such that the there is very slight free energy within the atoms which results in almost no relative motion between the atoms. The atoms then combine forming clumps such that phenomena usually observed at the microscopic level such as wavefunction interference become observable at the microscopic level.

4 0
3 years ago
Help with physics projectile motion​
BlackZzzverrR [31]

Answer:

10.4 m/s

Explanation:

First, find the time it takes for the projectile to fall 6 m.

Given:

y₀ = 6 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

(0 m) = (6 m) + (0 m/s) t + ½ (-9.8 m/s²) t²

t = 1.11 s

Now find the horizontal position of the target after that time:

Given:

x₀ = 6 m

v₀ = 5 m/s

a = 0 m/s²

t = 1.11 s

Find: x

x = x₀ + v₀ t + ½ at²

x = (6 m) + (5 m/s) (1.11 s) + ½ (0 m/s²) (1.11 s)²

x = 11.5 m

Finally, find the launch velocity needed to travel that distance in that time.

Given:

x₀ = 0 m

x = 11.5 m

t = 1.11 s

a = 0 m/s²

Find: v₀

(11.5 m) = (0 m) + v₀ (1.11 s) + ½ (0 m/s²) (1.11 s)²

v₀ = 10.4 m/s

3 0
3 years ago
A viola string with a fundamental frequency of D4 (293 Hz) is generally tuned using a tension of 49.0 N. However, just before a
Alborosie

Answer:

Explanation:

For fundamental frequency in a vibrating string , the formula is

n = 1 / 2L  x  √ ( T /m₁ )

n is frequency , L is length , T is tension and m₁ is mass per unit length .

For first string ,

293 =  1 / 2L  x  √ ( 49 N  /m₁ )

For second string , let mass per unit length be m₂ .

196 =  1 / 2L  x  √ ( 49 N  /m₂ ) ------ ( 1 )

To bring its frequency back to previous one let tension be T

293  =  1 / 2L  x  √ ( T  /m₂ ) ------- ( 2 )

Dividing

293 / 196 = √ ( T  /49  )

1.4948 = √ ( T  /49  )

2.2344 = T  /49

T = 109.48 N .

8 0
2 years ago
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