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2 years ago

Define the fundamental difference between kinematics and dynamics. .

1 answer:

5 0

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)

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A man with mass 81 kg lies on the floor. what is the normal force on the man?

butalik [34]

Answer:

Forces come in pairs, so the force of gravity (9.8 N) with the mans weight (794N) on the earth is counteracted with the normal force ( the force of the earth back on the man) which is the same

Explanation:

3 0

3 years ago

An ore sample weighs 17.50 N in air. When the sample is suspended by a light cord and totally immersed in water, the tension in

valkas [14]

Answer:

Volume of the sample: approximately $\rm 0.6422 \; L = 6.422 \times 10^{-4} \; m^{3}$ .

Average density of the sample: approximately $\rm 2.77\; g \cdot cm^{3} = 2.778 \times 10^{3}\; kg \cdot m^{3}$ .

Assumption:

$\rm g = 9.81\; N \cdot kg^{-1}$ .
$\rho(\text{water}) = \rm 1.000\times 10^{3}\; kg \cdot m^{-3}$ .
Volume of the cord is negligible.

Explanation:

<h3>Total volume of the sample</h3>

The size of the buoyant force is equal to $\rm 17.50 - 11.20 = 6.30\; N$ .

That's also equal to the weight (weight, $m \cdot g$ ) of water that the object displaces. To find the mass of water displaced from its weight, divide weight with $g$ .

$\displaystyle m = \frac{m\cdot g}{g} = \rm \frac{6.30\; N}{9.81\; N \cdot kg^{-1}} \approx 0.642\; kg$ .

Assume that the density of water is $\rho(\text{water}) = \rm 1.000\times 10^{3}\; kg \cdot m^{-3}$ . To the volume of water displaced from its mass, divide mass with density $\rho(\text{water})$ .

$\displaystyle V(\text{water displaced}) = \frac{m}{\rho} = \rm \frac{0.642\; kg}{1.000\times 10^{3}\; kg \cdot m^{-3}} \approx 6.42201 \times 10^{-4}\; m^{3}$ .

Assume that the volume of the cord is negligible. Since the sample is fully-immersed in water, its volume should be the same as the volume of water it displaces.

$V(\text{sample}) = V(\text{water displaced}) \approx \rm 6.422\times 10^{-4}\; m^{3}$ .

<h3>Average Density of the sample</h3>

Average density is equal to mass over volume.

To find the mass of the sample from its weight, divide with $g$ .

$\displaystyle m = \frac{m \cdot g}{g} = \rm \frac{17.50\; N}{9.81\; N \cdot kg^{-1}} \approx 1.78389 \; kg$ .

The volume of the sample is found in the previous part.

Divide mass with volume to find the average density.

$\displaystyle \rho(\text{sample, average}) = \frac{m}{V} = \rm \frac{1.78389\; kg}{6.42201 \times 10^{-4}\; m^{3}} \approx 2.778\; kg \cdot m^{-3}$ .

3 0

3 years ago

PY85

aliya0001 [1]

Answer:

460 g

Explanation:

Heat lost by the warm water = heat gained by the cold water

-mCΔT = mCΔT

-m (4.184 J/g/K) (37°C − 85°C) = (1000 g) (4.184 J/g/K) (37°C − 15°C)

-m (37°C − 85°C) = (1000 g) (37°C − 15°C)

-m (-48°C) = (1000 g) (22°C)

m = 458 g

Rounded to two significant figures, you need a mass of 460 g of water.

3 0

2 years ago

A clay ball with a mass of 0.48kg has an initial speed of 4.08m/s it strikes a 3.04kg clay ball at rest and the two balls stick

Rom4ik [11]

Final velocity = 0, thus final kinetic energy is 0
Initial kinetic energy:
0.5mv²
= 0.5 x 0.48 x 4.08²
= 4.0 J
Decrease in kinetic energy = 4 - 0 = 4 Joules

5 0

3 years ago

Elements with atomic numbers 58 through 71 are part of what series?

maria [59]

Elements with atomic numbers from 58 through 71 are part of the

<span>lanthanide</span> series <span />

3 0

3 years ago

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