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4 years ago

Ii) The distance travelled by an object in time

Physics

1 answer:

dsp734 years ago

5 0

Answer:

v = (5.2 ± 0.2) 10⁻² m / s

Explanation:

This is an exercise in kinematics of uniform motion. The formula for the average speed is

v = x / t

let's calculate

v = 5.2 / 100

v = 0.052 m / s

v = 5.2 10-2 m / s

we have to find the uncertainty of this quantity we propagate the error using derivatives

Δv = dv /dx Δx + dv /dt Δt

taking everything in the most unfavorable case, I mean all positive

let's make the derivatives

dv / dx = 1 / t

dv / dt = -x / t²

we substitute

Δv = 1 /t Δx + x /t² Δt

let's calculate the error

Δv = 1/100 0.1 + 5.2 / 100² 1

Δv = 1 10⁻³ + 5.2 10⁻⁴

Δv = 1.52 10⁻³ m/s

the error must be given with a significant figure

Dv = 2 10⁻³ m / s

the answer is

v = (5.2 ± 0.2) 10⁻² m / s

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Answer: Heat energy is transferred from warmer objects to cooler objects.

Explanation:

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3 years ago

Question 4. A tuning fork ‘A’ produces 6 beats/sec with another fork ‘B’ of un-known frequency. On

8090 [49]

Clever problem.

We know that the beat frequency is the DIFFERENCE between the frequencies of the two tuning forks. So if Fork-A is 256 Hz and the beat is 6 Hz, then Fork-B has to be EITHER 250 Hz OR 262 Hz. But which one is it ?

Well, loading Fork-B with wax increases its mass and makes it vibrate SLOWER, and when that happens, the beat drops to 5 Hz. That means that when Fork-B slowed down, its frequency got CLOSER to the frequency of Fork-A ... their DIFFERENCE dropped from 6 Hz to 5 Hz.

If slowing down Fork-B pushed it CLOSER to the frequency of Fork-A, then its natural frequency must be ABOVE Fork-A.

The natural frequency of Fork-B, after it gets cleaned up and returns to its normal condition, is 262 Hz. While it was loaded with wax, it was 261 Hz.

4 0

4 years ago

Part A A uniform solid disk of radius 1.60 m and mass 2.30 kg rolls without slipping to the bottom of an inclined plane. If the

kifflom [539]

Answer:

Height will be 3.8971 m

Explanation:

We have given that radius of the solid r = 1.60 m

Mass of the solid disk m = 2.30 kg

Angular velocity $\omega =4.46rad/sec$

Moment of inertia is given by $I=\frac{1}{2}mr^2$

Transnational Kinetic energy is given by $KE=\frac{1}{2}mv^2$ as we know that v = $v=\omega r$

So $KE=\frac{1}{2}m(\omega r)^2$

Rotational kinetic energy is given by $KE_{ROTATIONAL}=\frac{1}{2}I\omega ^2=\frac{1}{2}\left ( \frac{1}{2}mr^2 \right )\omega ^2=\frac{1}{4}m(r\omega )^2$

Potential energy is given by mgh

According to energy conservation

$mgh=\frac{1}{2}m(\omega r)^2+\frac{1}{4}m(\omega r)^2$

$h=\frac{3r^2\omega ^2}{4g}=\frac{3\times 1.60^2\times 4.46^2}{4\times 9.8}=3.8971m$

3 0

3 years ago

What is solar wind i ask that you help me

Novay_Z [31]

My answer -

the corona, the sun's outer layer, reaches temperatures of up to 2 million degrees Fahrenheit (1.1 million Celsius). At this level, the sun's gravity can't hold on to the rapidly moving particles, and it streams away from the star.

The sun's activity shifts over the course of its 11-year cycle, with sun spot numbers, radiation levels, and ejected material changing over time. These alterations affect the properties of the solar wind, including its magnetic field properties, velocity, temperature and density. The wind also differs based on where on the sun it comes from and how quickly that portion is rotating.

The velocity of the solar wind is higher over coronal holes, reaching speeds of up to 500 miles (800 kilometers) per second. The temperature and density over coronal holes are low, and the magnetic field is weak, so the field lines are open to space. These holes occur at the poles and low latitudes, and reach their largest when activity on the sun is at its minimum. Temperatures in the fast wind can reach up to 1 million degrees F (800,000 C).

At the coronal streamer belt around the equator, the solar wind travels more slowly, at around 200 miles (300 km) per second. Temperatures in the slow wind reach up to 2.9 million F (1.6 million C).

p.s

Glad to help you and if you need anything else on brainly let me know so I can elp you again have an AWESOME!!! :^)

4 0

3 years ago

A planet has a surface temperature of 95.0 K and an atmospheric pressure of 1.60 atm. If 4.87 L of atmosphere has a mass of 28.6

mr Goodwill [35]

Answer:

M=28.88 gm/mol

Explanation:

Given that

T= 95 K

P= 1.6 atm

V= 4.87 L

m = 28.6 g

R=0.08206L atm .mol .K

We know that gas equation for ideal gas

P V = n R T

P=Pressure , V=Volume ,n=Moles,T= Temperature ,R=gas constant

Now by putting the values

P V = n R T

1.6 x 4.87 = n x 0.08206 x 95

n=0.99 moles

We know that number of moles given as

$n=\dfrac{m}{M}$

M=Molar mass

$n=\dfrac{m}{M}$

$0.99=\dfrac{28.6}{M}$

M=28.88 gm/mol

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3 years ago