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2 years ago

What is the difference between speed and velocity?

1 answer:

4 0

Speed does not include a direction, but velocity does.

30 miles per hour north
and
30 miles per hour west

Same speed.
Different velocity.

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The top of the cylinder head is sealed by the

svetlana [45]

The common name is valve cover, so the correct answer is A.

3 0

3 years ago

Read 2 more answers

A long, fine wire is wound into a coil with inductance 5 mH. The coil is connected across the terminals of a battery, and the cu

Effectus [21]

Answer:

Compared with the current in the first coil, the current in the second coil is unchanged.

Explanation:

All coils, inductors, chokes and transformers create a magnetic field around themselves consist of an Inductance in series with a Resistance forming an LR Series Circuit.

The steady state of current in the LR circuit is:

I= V/R (1 - e^-Rt/L)

Where I= current

R= Resistance

V= Voltage

Where R/L is the time constant.

For a conducting wire, it has a very small resistance. The time constant will be in microseconds. The current will be in a steady state after few second. The current is independent on the inductance and dependent on the resistance. The length of wire and the resistance here are the same. Therefore, the current remains unchanged.

5 0

3 years ago

Read 2 more answers

You pull a solid nickel ball with a density of 8.91 g/cm3 and a radius of 1.40 cm upward through a fluid at a constant speed of

Sunny_sXe [5.5K]

Answer:

$P = 1.090\,N$

Explanation:

The constant speed means that ball is not experimenting acceleration. This elements is modelled by using the following equation of equilibrium:

$\Sigma F = P - W + F_{D}$

$\Sigma F = P - \rho \cdot V \cdot g + c\cdot v = 0$

Now, the exerted force is:

$P = \rho \cdot V \cdot g - c\cdot v$

The volume of a sphere is:

$V = \frac{4\cdot \pi}{3}\cdot R^{3}$

$V = \frac{4\cdot \pi}{3}\cdot (0.014\,m)^{3}$

$V = 1.149\times 10^{-5}\,m^{3}$

Lastly, the force is calculated:

$P = (8910\,\frac{kg}{m^{3}} )\cdot (1.149\cdot 10^{-5}\,m^{3})\cdot (9.81\,\frac{m}{s^{2}} )+(0.950\,\frac{kg}{s})\cdot (0.09\,\frac{m}{s} )$

$P = 1.090\,N$

5 0

2 years ago

When you changed from low to high power, how did the change affect the working distance of the lens?

Basile [38]

The working distance gets shorter as the magnification gets bigger. In order to focus, the high-power objective lens must be significantly nearer to the specimen than the low-power lens. Magnification is negatively correlated with working distance.

Magnification change The magnification of a specimen is increased by switching from low power to high power. The magnification of an image is determined by multiplying the magnification of the objective lens by the magnification of the ocular lens, or eyepiece.

The geometry of the optical system connects the magnifying power, or how much the thing being observed seems expanded, and the field of view, or the size of the object that can be seen.

To know more about working distance

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4 0

1 year ago

The period P of a pendulum depend on its length , mass as well as the acceleration due to gravity . Use dimensional analysis to

Savatey [412]

Explanation:

Period P has units of seconds (s).

Length has units of meters (m).

Mass has units of kilograms (kg).

Acceleration has units of meters per second squared (m/s²).

Dimensional analysis:

s = √(m / (m/s²))

Therefore:

P = k √(L/g)

where k is a dimensionless constant.

6 0

2 years ago