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Alexxandr [17]
3 years ago
8

What is the basic building block of all matter

Physics
2 answers:
Natalka [10]3 years ago
7 0
I think that is atom. 
Igoryamba3 years ago
6 0
An Atom is correct hope u pass your class
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Select all the correct answers.
sdas [7]
I believe there are two correct answers, and those answers are A and D
5 0
3 years ago
Which method of testing substances is a sure way to identify a chemical reaction?
Sergeu [11.5K]

Answer:

B

Explanation:

That's the answer. Hope it helped!

7 0
2 years ago
interval (from lowering the body to his feet taking off) lasts for 0.3 s and the mass of the player is 90 kg. Ignore air resista
steposvetlana [31]

Answer:

The force applied to the surface is 9 kilo Newton.

Explanation:

While jumping on the surface the player applies the force that is equal to its weight on the surface.

The mass of the player is given as 90 kg.

Force applied by the player = weight of the player

Force applied by the player = m × g

Where m is the mass of the player and g is acceleration due to gravity

Force applied by the player = 90 × 9.8

Force applied by the player = 882 Newton

Expressing your answer to one significant figure, we get

Force applied by the player =0. 9 kilo Newton

The force applied to the surface is 0.9 kilo Newton.

8 0
3 years ago
The PE of the box that is on a 2.0 m high self is 1600 J. What is the power expelled to lift the box to this height in 10.0 seco
Pani-rosa [81]

Answer:

  160 W

Explanation:

Power is the ratio of work to time:

  (1600 J)/(10 s) = 160 J/s = 160 W

7 0
2 years ago
A car travels on a straight, level road. (a) Starting from rest, the car is going 38 ft/s (26 mi/h) at the end of 4.0 s. What is
lbvjy [14]

Answer:

a)9.5\frac{ft}{s^2}\\ b) 12.66\frac{ft}{s^2}

Explanation:

A body has acceleration when there is a change in the velocity vector, either in magnitude or direction. In this case we only have a change in magnitude. The average acceleration represents the speed variation that takes place in a given time interval.

a)

a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{38\frac{ft}{s}-0}{4 s- 0}=9.5\frac{ft}{s^2}\\

b)

a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{76\frac{ft}{s}-38\frac{ft}{s}}{7 s- 4s}\\a_{avg}=\frac{38\frac{ft}{s}}{3s}=12.66\frac{ft}{s^2}

8 0
3 years ago
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