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Pachacha [2.7K]
3 years ago
12

Why isnt this technology used in coal plants

Physics
1 answer:
n200080 [17]3 years ago
5 0
Can cause atmospheric problems.

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You attach a 1.10 kg block to a horizontal spring that is fixed at one end. You pull the block until the spring is stretched by
dangina [55]

Answer:

6.3 m/s

Explanation:

m = mass of the block = 1.10 kg

k = spring constant of the spring

x = stretch in the spring = 0.2 m

t = time taken by block to come to zero speed first time = 0.100 s

T = Time period of oscillation

Time period of oscillation is given as

T = 2t

T = 2 (0.1)

T = 0.2 s

Time period is also given as

T = 2\pi \sqrt{\frac{m}{k}}

0.2 = 2(3.14) \sqrt{\frac{1.10}{k}}

k = 1084.6 N/m

v = maximum speed of the block

using conservation of energy

Maximum kinetic energy = Maximum spring potential energy

(0.5) m v² = (0.5) k x²

m v² = k x²

(1.10) v² = (1084.6) (0.2)²

v = 6.3 m/s

6 0
3 years ago
a firework rock is burning at the rate of 14g/s all the products of combustion are ejected in one direction at the speed of 210m
Naddika [18.5K]

SO MY NAME IS ROHITH AND I LIKE MM

Explanation:

5 0
3 years ago
What are power tools​
butalik [34]

Answer:

Tools that require electricty like drills

Explanation:

8 0
3 years ago
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after an investigation kuri determines that her hypothesis was wrong what is the best thing for kuri to do next​
aksik [14]
: You could make slight changes to the process or consider wheather the experiment was carried out correctly
7 0
3 years ago
Read 2 more answers
A point charge q1=+5.00nC is at the fixed position x=0, y=0, z=0. You find that you must do 8.10×10−6J of work to bring a second
Maru [420]

The value of the second charge is 1.2 nC.

<h3>Electric potential</h3>

The work done in moving the charge from infinity to the given position is calculated as follows;

W = Eq₂

E = W/q₂

<h3>Magnitude of second charge</h3>

The magnitude of the second charge is determined by applying Coulomb's law.

E = \frac{kq_2}{r^2} \\\\\frac{kq_2}{r^2} = \frac{W}{q_2} \\\\kq_2^2 = Wr^2\\\\q_2^2 = \frac{Wr^2}{k} \\\\q_2 = \sqrt{\frac{Wr^2}{k} } \\\\q_2 =  \sqrt{\frac{(8.1 \times 10^{-6}) \times (0.04)^2}{9\times 10^9} } \\\\q_2 = 1.2 \times 10^{-9} \ C\\\\q_2 = 1.2 \ nC

Thus, the  value of the second charge is 1.2 nC.

Learn more about electric potential here: brainly.com/question/14306881

7 0
2 years ago
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