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givi [52]
3 years ago
10

Can we use position as another word for phase?

Physics
1 answer:
Yuri [45]3 years ago
6 0

Answer:

yes

Explanation:

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Suman with a mass of 45kg climbs in 3m heigh ladder in 10 seconds. Calculate her power. Plz help me​
attashe74 [19]

Answer:

work done=fdcosø

f=mg=450N

d=+3m

ø=180

450×3×cos(180°) power=work done/

time taken

1350×-1 p=135watts

7 0
3 years ago
Determine the kinetic energy of a 2000 kg roller coaster car that is moving at the speed of 10 ms
kondaur [170]

Answer:

\boxed {\boxed {\sf 100,000 \ Joules}}

Explanation:

Kinetic energy is energy due to motion. The formula is half the product of mass and velocity squared.

E_k= \frac{1}{2} mv^2

The mass of the roller coaster car is 2000 kilograms and the car is moving 10 meters per second.

  • m= 2000 kg
  • s= 10 m/s

Substitute these values into the formula.

E_k= \frac{1}{2} (2000 \ kg ) \times (10 \ m/s)^2

Solve the exponent.

  • (10 m/s)²= 10 m/s * 10 m/s= 100 m²/s²

E_k= \frac{1}{2} (2000 \ kg ) \times (100 \ m^2/s^2)

Multiply the first two numbers together.

E_k= 1000 \ kg  \times (100 \ m^2/s^2)

Multiply again.

E_k= 100,000 \ kg*m^2/s^2

  • 1 kilogram square meter per square second is equal to 1 Joule.
  • Our answer of 100,000 kg*m²/s² is equal to 100,000 Joules.

E_k= 100,000 \ J

The roller coaster car has <u>100,000 Joules</u> of kinetic energy.

3 0
3 years ago
Which of the following elements most likely has the highest boiling point?
JulsSmile [24]

Answer:

Explanationalluminiim

4 0
3 years ago
Read 2 more answers
According to newton's third law of motion, when a hammer strikes and exerts force to push it into a piece of wood, the nail
Darya [45]
According to newton's third law of motion, when a hammer strikes and exerts force to push it into a piece of wood, the nail <span>C. exerts an equal or opposite force on the hammer. The third law of motion states that every action has an equal BUT opposite reaction. This means that the nail exerts the same force the hammer exerts on it.</span>
6 0
4 years ago
Read 2 more answers
A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end.
Veronika [31]

Answer:

A) ω = 6v/19L

B) K2/K1 = 3/19

Explanation:

Mr = Mass of rod

Mb = Mass of bullet = Mr/4

Ir = (1/3)(Mr)L²

Ib = MbRb²

Radius of rotation of bullet Rb = L/2

A) From conservation of angular momentum,

L1 = L2

(Mb)v(L/2) = (Ir+ Ib)ω2

Where Ir is moment of inertia of rod while Ib is moment of inertia of bullet.

(Mr/4)(vL/2) = [(1/3)(Mr)L² + (Mr/4)(L/2)²]ω2

(MrvL/8) = [((Mr)L²/3) + (MrL²/16)]ω2

Divide each term by Mr;

vL/8 = (L²/3 + L²/16)ω2

vL/8 = (19L²/48)ω2

Divide both sides by L to obtain;

v/8 = (19L/48)ω2

Thus;

ω2 = 48v/(19x8L) = 6v/19L

B) K1 = K1b + K1r

K1 = (1/2)(Mb)v² + Ir(w1²)

= (1/2)(Mr/4)v² + (1/3)(Mr)L²(0²)

= (1/8)(Mr)v²

K2 = (1/2)(Isys)(ω2²)

I(sys) is (Ir+ Ib). This gives us;

Isys = (19L²Mr/48)

K2 =(1/2)(19L²Mr/48)(6v/19L)²

= (1/2)(36v²Mr/(48x19)) = 3v²Mr/152

Thus, the ratio, K2/K1 =

[3v²Mr/152] / (1/8)(Mr)v² = 24/152 = 3/19

3 0
3 years ago
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