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Amanda [17]
2 years ago
14

Listed in the Item Bank are some important labels for sections of the image below. To find out more information about labels, so

me have more details available when you click on them. Drag and drop each label to the corresponding area it identifies in the image.
ITEM BANK: Move to Bottom
BoilingGas/VaporLiquidMeltingSolidTemperatureTime
0
Physics
1 answer:
gizmo_the_mogwai [7]2 years ago
5 0

Answer:reactant,active site,enzyme below,substrate,products

Explanation:

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Parallel to the handle, what is the force of friction acting on the suitcase?
mrs_skeptik [129]
 <span>Lets call F the friction force which will act horizontally backwards. 

As you are travelling at a constant velosity horizontally there is no overall resultant force in this direction. 

ie. the force you pull with will be equal to the friction force resisting you. (you will initially have to have pulled with a greater force than the friction to get the suitcase moving) 

the value of your force pulling is 60 cos26.9 (horizontally) - you should have learnt about resolving forces. 

this must be equal to F 

so 

F=60cos26.9 
F=53.5N

hope this helps you
please mark this as brainliest answer</span>
5 0
3 years ago
Volleyball was invented in 1905.
Lady_Fox [76]

Answer:

REALLY??

Explanation:

8 0
2 years ago
Read 2 more answers
An unstrained horizontal spring has a length of 0.26 m and a spring constant of 180 N/m. Two small charged objects are attached
MakcuM [25]

Answer:

a)

two like charges always repel each other while two unlike charges attract each other. Since the spring stretches by 0.039 m, the charges have the same sign. both charges are positive(+) or Negative (-)

b)

both q1 and q1 are 8.35 × 10⁻⁶ C or -8.35 × 10⁻⁶ C

Explanation:

Given that;

L = 0.26 m

k = 180 N/m

x = 0.039 m

a)

we know that two like charges always repel each other while two unlike charges attract each other. Since the spring stretches by 0.039 m, the charges have the same sign.

b)

Spring force F = kx

F = 180 × 0.039

F = 7.02 N

Now, Electrostatic force F = Keq²/r²

where r = L + x = ( 0.26 + 0.039 )

we know that proportionality constant in electrostatics equations Ke = 9×10⁹ kg⋅m3⋅s−2⋅C−2

so from the equation; F = Keq²/r²

Fr² = Keq²

q = √ ( Fr² / Ke )

we substitute

q = √ ( 7.02 N × ( 0.26 + 0.039 )² / 9×10⁹ )

q = √ ( 7.02 N × ( 0.26 + 0.039 )² / 9×10⁹ )

q =  √ (0.627595 / 9×10⁹)

q = √(6.97 × 10⁻¹¹)

q = 8.35 × 10⁻⁶ C

Therefore both q1 and q1 are 8.35 × 10⁻⁶ C or -8.35 × 10⁻⁶ C

5 0
2 years ago
For two objects with masses 100 kg and 500 kg to be 20 meters apart , find the gravity between them
Vikki [24]

Answer:

8,3375*10^-9 N

Explanation:

4 0
3 years ago
Two protons are released from rest when they are 0.720 nm apart. For related problem-solving tips and strategies, you may want t
Gnesinka [82]

Answer:

a) Speed of the electrons at maximum speed = (1.384 × 10⁴) m/s

The maximum speed occurs at the point where all of the initial potential energy is converted into kinetic energy.

b) Maximum acceleration of the protons = (2.660 × 10¹⁷) m/s²

The maximum acceleration occurs at the minimum distance apart for the two protons.

Explanation:

The maximum speed occurs when all the potential energy of the protons has been converted to kinetic energy.

The potential energy between the two protons at the instant of release is given by

U = (kq₁q₂/r)

k = Coulomb' s constant = (8.988 × 10⁹) Nm²/C²

q₁ = q₂ = charge on a proton = q = (1.602 × 10⁻¹⁹) C

r = separation between the two protons = 0.72 nm = (7.2 × 10⁻¹⁰) m

U = (kq²/r) = [(8.988 × 10⁹) × (1.602 × 10⁻¹⁹)²] ÷ (7.2 × 10⁻¹⁰) = (3.204 × 10⁻¹⁹) N/m or Joules

At the maximum speeds, the two protons will not possess any potential Energy, only kinetic energy.

The sum of kinetic and potential energies is always constant for the system

(Initial Kinetic Energy) + (Initial Potential Energy) = (Kinetic Energy at maximum speed) + (Potential Energy at maximum speed)

Initial Kinetic Energy of the system = 0 J (Since both protons were intially at rest)

Initial Potential Energy = (3.204 × 10⁻¹⁹) J

Kinetic Energy at maximum speed = Sum of the kinetic energies of the protons at this point = (½mv²) + (½mv²) = (mv²) J (Since theu are both protons, they have the same mass and the same speed at maximum speed)

Potential Energy at maximum speed = 0 J

0 + (3.204 × 10⁻¹⁹) = mv² + 0

mv² = (3.204 × 10⁻¹⁹)

m = mass of a proton = (1.673 × 10⁻²⁷) kg

v = speed of each of the protons at maximum speed = ?

v = √[(3.204 × 10⁻¹⁹) ÷ m]

v = √[(3.204 × 10⁻¹⁹) ÷ (1.673 × 10⁻²⁷)]

v = √(1.915 × 10⁸) = 13,838.8 m/s = (1.384 × 10⁴) m/s

b) Since the two protons repel each other and force of repulsion reduces as the dI stance between the protons increases, the maximum acceleration occurs at the minimum distance apart for the two protons.

Force of repulsion acting on each proton is given through Coulomb's law as

F = (kq₁q₂/r²)

And the force acting on each proton is obtainable using Newton's law that

F = ma

So, the acceleration of each proton at any time is obtainable through a relation of these 2 formulas.

ma = (kq₁q₂/r²)

a = (kq₁q₂/r²m)

k = Coulomb' s constant = (8.988 × 10⁹) Nm²/C²

q₁ = q₂ = charge on a proton = q = (1.602 × 10⁻¹⁹) C

r = separation between the two protons = 0.72 nm = (7.2 × 10⁻¹⁰) m

m = mass of a proton = (1.673 × 10⁻²⁷) kg

a = [(8.988 × 10⁹) × (1.602 × 10⁻¹⁹)²] ÷ [(7.2 × 10⁻¹⁰)² × (1.673 × 10⁻²⁷)]

a = (2.660 × 10¹⁷) m/s²

Hope this Helps!!!

5 0
2 years ago
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