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g100num [7]
3 years ago
11

A student is working in a lab to determine how time affects impulse. The student keeps the force the same in each trial but chan

ges the impact time. Some data is shown.

Physics
2 answers:
notsponge [240]3 years ago
8 0

Answer:

Trail 3

Explanation:

As we know by Newton's II law that rate of change in momentum is always net force applied on the object.

Here we know that

F = \frac{\Delta P}{\Delta t}

now we also know that change in momentum is known as impulse

so here impulse and force relation is given as

impulse = F\Delta t

so here in order to find the greatest impulse we need to find the product of force and time to be the maximum as per above formula

so we have

trail 1

impulse = 500 (0.25) = 125 Ns

trial 2

impulse = 500(0.15) = 75 Ns

trial 3

impulse = 500(0.45) = 225 Ns

trial 4

impulse = 500(0.33) = 165 Ns

so out of the four trials maximum impulse is for trial 3

pishuonlain [190]3 years ago
4 0

yea some data is shown what is the question dude

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sleet_krkn [62]

Answer:

Explanation:

Distance between plates d = 2 x 10⁻³m

Potential diff applied = 5 x 10³ V

Electric field = Potential diff applied /  d

= 5 x 10³  / 2 x 10⁻³

= 2.5 x 10⁶ V/m

This is less than  breakdown strength for air  3.0×10⁶ V/m

b ) Let the plates be at a separation of d .so

5 x 10³ / d = 3.0×10⁶ ( break down voltage )

d = 5 x 10³  / 3.0×10⁶

= 1.67 x 10⁻³ m

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5 0
3 years ago
A person is standing on a raft; their
krok68 [10]

Answer:

The volume of water displaced by the raft is 0.233 m³

Explanation:

The question relates to Archimedes' principle which states that the buoyant force experienced by an object immersed in a fluid is equal to the weight of (the force of gravity on) the displaced fluid

The given parameters are;

The combined mass of the person and the raft, m = 233 kg

The liquid on which the raft is located = Water

The density of water, \rho _{water} = 1000 kg/m³

Weight = Mass, m × g

Where;

m = The mass of the object

g = The acceleration due to gravity = 9.8 m/s²

Given that the raft is on the surface of the water (floating), the buoyant force is equal to the combined weight of the person and the raft = 233 kg

The combined weight of the person and the raft, W_{combined} = 233 kg × 9.8 m/s² = 2,283.4 N

Therefore;

The buoyant force = 2,283.4 N = The weight of the water displaced

The mass of the water displaced, m_{water}, = 2,283.4 N/(9.8 m/s²) = 233 kg

Density = Mass/Volume

The volume of water displaced by the raft = The mass of the water displaced/(The density of the water) = 233 kg/(1,000 kg/m³) = 0.233 m³.

3 0
2 years ago
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The center of mass of the system is the point at which the total mass of the system could be concentrated without changing the _
Sveta_85 [38]

  The centre of mass of the system is the point at which the total

  mass of system could be concentrated without changing the moment

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Centre of Mass is the point at which the whole mass of the system

is assumed to be concentrated.

 The general formula for the COM is:

              xₙ =  Σmₐxₐ / Σmₐ         where,  a = 1,2,3.........n

  Here the term Σ mₐ xₐ is called the first moment of the system and the

  denominator expression is called total mass of the system.

    Therefore, from this theory we can say that the moment of the system

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  Hence, The centre of mass of the system is the point at which the total

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             brainly.com/question/3454419

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8 0
1 year ago
How much heat is needed to raise the temperature of an empty 2.0 x 101 kg vat made of
scZoUnD [109]

Answer:

Q = 7272 Kilojoules.

Explanation:

<u>Given the following data;</u>

Mass = 2.0*101kg = 202kg

Initial temperature, T1 = 10°C

Final temperature, T2 = 90°C

We know that the specific heat capacity of iron = 450J/kg°C

*To find the quantity of heat*

Heat capacity is given by the formula;

Q = mcdt

Where;

  • Q represents the heat capacity or quantity of heat.
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dt = T2 - T1

dt = 90 - 10

dt = 80°C

Substituting the values into the equation, we have;

Q = 202*450*80

Q = 7272KJ or 7272000 Joules.

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bezimeni [28]

Answer:

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7 0
2 years ago
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