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kondor19780726 [428]
3 years ago
5

What is the purpose of the scientific process

Physics
1 answer:
jeka57 [31]3 years ago
8 0
Phrase your problem as a question and phrase your hypothesis as a statement be specific and stating your hypothesis or purpose but don't be overly wordy
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Why does static friction have a maximum value​
Naddik [55]

Answer:

If you push horizontally with a small force, static friction establishes an equal and opposite force that keeps the book at rest. As you push harder, the static friction force increases to match the force. Eventually maximum static friction force is exceeded and the book moves.

Explanation:

3 0
3 years ago
Define the fundamental difference between kinematics and dynamics. .
Tcecarenko [31]

In kynematics you describe the motion of particles using vectors and their change in time. You define a position vector r for a particle, and then define velocity v and acceleration a as

v=\frac{dr}{dt} \\

a=\frac{dv}{dt}

In dynamics Newton's laws predict the acceleration for a given force. Knowing the acceleration, and the kynematical relations defines above, you can solve for the position as a function of time: r(t)

5 0
3 years ago
Determine the potential difference between two charged parallel plates that are 0.50 cm apart and have an electric field strengt
fiasKO [112]
E = \frac{V}{r} \\ V = \frac{E}{r} \\ V = \frac{9.0V/cm}{0.5cm} \\ V = 18V
4 0
3 years ago
A student pulls a 50-newton sled with a force having a magnitude of 15 newtons. What is the magnitude of the force that the sled
Simora [160]

Answer:

Force = 35 N

Explanation:

From Newton's third law of motion, the boy must apply a force greater than the weight of the sled to lift it.

weight of sled = mg

where m is its mass and g the force of gravity on it.

weight of sled = 50 N

Force applied by the boy on the sled = 15 N

Since the force applied on the sled by the boy is lesser than the weight of the sled, then;

Force that the sled exerts on the student = 50 - 15

                                             = 35 N

The force exerted by the sled on the student is 35 N.

5 0
3 years ago
What is the total entropy change if a 0.280 efficiency engine takes 3.78E3 J of heat from a 3.50E2 degC reservoir and exhausts i
babymother [125]

Answer:

\Delta S=1.69J/K

Explanation:

We know,

\eta=1-\frac{T_2}{T_1}=1-\frac{Q_2}{Q_1}      ..............(1)

where,

η = Efficiency of the engine

T₁ = Initial Temperature

T₂ = Final Temperature

Q₁ = Heat available initially

Q₂ = Heat after reaching the temperature T₂

Given:

η =0.280

T₁ = 3.50×10² °C = 350°C = 350+273 = 623K

Q₁ = 3.78 × 10³ J

Substituting the values in the equation (1) we get

0.28=1-\frac{Q_2}{3.78\times 10^{3}}

or

\frac{Q_2}{3.78\times 10^3}=0.72

or

Q_2=3.78\times 10^3\times0.72

⇒ Q_2 =2.721\times 10^3 J

Now,

The entropy change (\Delta S) is given as:

\Delta S=\frac{\Delta Q}{T_1}

or

\Delta S=\frac{Q_1-Q_2}{T_1}

substituting the values in the above equation we get

\Delta S=\frac{3.78\times 10^{3}-2.721\times 10^3 J}{623K}

\Delta S=1.69J/K

7 0
3 years ago
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