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lukranit [14]
2 years ago
9

IF YOU GOOD AT SCIENCE PLEASE ANSWER THIS ASAP

Physics
2 answers:
Yuri [45]2 years ago
6 0

Answer:

THE STUDY OF MATTER AND ENERGY

Explanation:

xz_007 [3.2K]2 years ago
4 0

Answer:

I would believe that it would be the last option

Explanation:

Physical science is a type of science that mainly focuses on natural objects that are not alive, such as minerals and rocks.

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How long would it take 2.0x10^20 electrons to pass through a point in a conductor if the current was 10.0A?
inysia [295]

1 coulomb of electric charge is carried by  6.25 x 10^18 electrons

1 Ampere = 1 coulomb per second
10 A = 10 coulombs per second

(2.0 x 10^20 electrons) x (coul / 6.25 x 10^18 electrons) / (10 coul/sec) =

         (2.0 x 10^20) / (6.25 x 10^18 x 10)    sec  =  <em>3.2 seconds</em>


6 0
3 years ago
A lawn roller is rolled across a lawn by a force of 107 N along the direction of the handle, which is 13.5 ◦ above the horizonta
-BARSIC- [3]

Answer:

22.02 m

Explanation:

given,

Force, F = 107 N

angle made with horizontal = 13.5◦

Power develop by the lawn roller = 69.4 W

time = 33 s

distance = ?

Force along horizontal= F cos θ

          = 107 cos 13.5°= 104 N

Power = \dfrac{work\ done}{time}

69.4 = \dfrac{W}{33}

W = 2290.2 J

Work done= Force x displacement

displacement= \dfrac{2290.2}{104}

                      = 22.02 m

6 0
3 years ago
(a) Calculate the absolute pressure at the bottom of a fresh- water lake at a depth of 27.5 m. Assume the density of the water i
ddd [48]

Answer:

a) P = 370.993\,kPa, b) F = 25.948\,kN

Explanation:

a) The absolute pressure at a depth of 27.5 meters is:

P = P_{atm} + P_{man}

P = 101.3\,kPa + \left(1000\,\frac{kg}{m^{3}}\right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (27.5\,m)\cdot \left(\frac{1\,kPa}{1000\,Pa} \right)

P = 370.993\,kPa

b) The force exerted by the water is:

F = (P - P_{atm})\cdot A

F = (370.993\,kPa-101.3\,kPa)\cdot \left(\frac{\pi}{4} \right)\cdot (0.35\,m)^{2}

F = 25.948\,kN

5 0
2 years ago
Read 2 more answers
A 5.20 kg chunk of ice is sliding at 13.5 m/s on the floor of an ice-covered valley when it collides with and sticks to another
Stella [2.4K]

Answer:

The chunk went as high as

2.32m above the valley floor

Explanation:

This type of collision between both ice is an example of inelastic collision, kinetic energy is conserved after the ice stuck together.

Applying the principle of energy conservation for the two ice we have based on the scenery

Momentum before impact = momentum after impact

M1U1+M2U2=(M1+M2)V

Given data

Mass of ice 1 M1= 5.20kg

Mass of ice 2 M2= 5.20kg

velocity of ice 1 before impact U1= 13.5 m/s

velocity of ice 2 before impact U2= 0m/s

Velocity of both ice after impact V=?

Inputting our data into the energy conservation formula to solve for V

5.2*13.5+5.2*0=(10.4)V

70.2+0=10.4V

V=70.2/10.4

V=6.75m/s

Therefore the common velocity of both ice is 6.75m/s

Now after impact the chunk slide up a hill to solve for the height it climbs

Let us use the equation of motion

v²=u²-2gh

The negative sign indicates that the chunk moved against gravity

And assuming g=9.81m/s

Initial velocity of the chunk u=0m/s

Substituting we have

6.75²= 0²-2*9.81*h

45.56=19.62h

h=45.56/19.62

h=2.32m

5 0
3 years ago
an object at rest will remain at rest and an object in motion stays in straight-line motion unless acted upon by a_____ or unbal
LekaFEV [45]
Equal is the answer your looking for
5 0
3 years ago
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