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3 years ago

What two things do you need to know to describe the velocity of an object?

Physics

2 answers:

mezya [45]3 years ago

4 0

By definition, we have that the speed of an object is given by:

$v = \frac{d}{t}$

Where,

d: distance [in units of length: meters, feet, miles]

t: time [in units of time: minutes, seconds, hours]

Therefore, knowing the distance traveled, and the time to travel this distance, we can know the speed of an object.

Then, since velocity is a vector, then we need the direction of the vector.

Therefore, the velocity vector can be written as:

$v = |v|(cos(\alpha) i + sin(\alpha) j)$

Answer:

two things you need to know to describe the velocity of an object are:

1) Magnitude (speed)

2) direction

pickupchik [31]3 years ago

3 0

The distance it traveled and the time that it took to travel that distance

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The pressure exerted by 15m of liquid is 1500pa.The acceleration due to gravity g=10m/s^2.Calculate the density liquid.

Natali [406]

Answer:

1500 divided by 150(15m x 10m/s^2) = 10

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3 years ago

You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th

PolarNik [594]

<u>Answer:</u> The Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

$Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}$

where,

Y = Young's Modulus

F = force exerted by the weight = $m\times g$

m = mass of the ball = 10 kg

g = acceleration due to gravity = $9.81m/s^2$

l = length of wire = 2.6 m

A = area of cross section = $\pi r^2$

r = radius of the wire = $\frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m$ (Conversion factor: 1 m = 1000 mm)

$\Delta l$ = change in length = 1.99 mm = $1.99\times 10^{-3}m$

Putting values in above equation, we get:

$Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2$

Hence, the Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

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4 years ago

A parallel plate capacitor has a capacitance of 6.78 μF and it is connected to a 12V battery. What is the charge on the capacito

AfilCa [17]

Answer:

8.136×10⁻⁵ J

Explanation:

Applying,

Q = Cv................ Equation 1

Where Q = Charge on the capacitor, v = voltage of the battery, C = capacitance of the capacitor.

From the question,

Given: C = 6.78μF = 6.78×10⁻⁶ F, v = 12 V

Substitute these values into equation 1

Q = (6.78×10⁻⁶ )(12)

Q = 8.136×10⁻⁵ J

Hence the charge on the capacitor is 8.136×10⁻⁵ J

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3 years ago

Position vs Time

Annette [7]

Answer:

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3 years ago

What is the best use of an atomic model to explain the charge of the particles in Thomson’s beams?

kodGreya [7K]

Answer:

The interpretation of the subject in question is characterized in the discussion section below.

Explanation:

The atom seems to be the smallest particle of negatively charged electrons as well as positive. The negative particulates are comparatively small, as well as distant from both the considerably high beneficial particles. When the negative objects traveled away from the desired ones those who formed an intangible beam which was electrically charged.
Thomson would use a closed glass globe with just a single positive and another negative electrode with extraordinarily low present pressure. He was forced to submit those other gases to quite a voltage level, and also that the rise in popularity of emission levels, which have been called cathode rays, must have been observed.
As the negatives shifted away from those in the positives, an imaginary electrically conductive beam was formed. Not only that but the negative particles have been completely circumvented due to the extreme distance seen between positively and negatively particle size.

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3 years ago