1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Effectus [21]
3 years ago
10

You are walking towards teh back of the bus that is movind forward with a constant velocity. Descriube your motion relative to t

he bus and relative to a oint on the groun.
Physics
1 answer:
Arisa [49]3 years ago
5 0

Answer:

When walking towards the back of a bus while it moves forward its apparent velocity seems slower than it actually is with respect to the ground.

Explanation:

  • When a someone follows a moving bus from behind by walking then the apparent speed of the bus is slightly slower by the same magnitude as the speed of walking because both are moving in the same direction.
  • This is the concept of relative velocity when the apparent velocity of the same object appears different to different viewers at the same time being on different frame of motion.
  • When the same bus is observed from a point on the road the speed of the bus will appear greater than the former case discussed above because the bus here is moving with respect to a stationary observer.
You might be interested in
Water is being heated in a vertical piston-cylinder device. the piston has a mass of 20 kg and a cross-sectional area of 100 cm2
sesenic [268]
<span>Answer: We have pressure from the weight of the piston + atmospheric pressure acting on the gas pressure from the piston... P = F / area = mass x acceleration due to gravity / area P = (24 kg x 9.80 m/s²) / (90 cm² x (1m/100cm)²) = 26133 N/m² = 2.61x10^4 Pa = 26.1 kPa total pressure acting on the gas = 26+85 kPa = 111 kPa then.. via clausius clapeyron equation.. ln(P1 / P2) = (dHvap / R) x (1 / T2 - 1 / T1) and if I pick say.. the normal boiling point of water for P2, T2.. then... P1 = 111 kPa P2 = 101.325 kPa dHvap = 40680 J/mole R = 8.314 J/moleK T2 = 373.15 K T1 = ?? ---> T1 = 102.46°</span>
4 0
3 years ago
A 2.2 ohm resistor has 0.042 A of current in it. What is the potential difference across the resistor?
Mrrafil [7]
V=IxR
v =0.042x2.2
potential difference =0.0924volts
7 0
3 years ago
What is a good way to study guys for a science test Im in need because I have a quiz tmrw
nikklg [1K]

Answer:

Space out your studying. Nate Kornell “definitely did cram” before big tests when he was a student. ...

Practice, practice, practice! ...

Don't just reread books and notes. ...

Test yourself. ...

Mistakes are okay — as long as you learn from them. ...

Mix it up. ...

Use pictures. ...

Find examples.

Explanation:

5 0
3 years ago
Explain why astronomers long ago believed that space must be filled with some kind of substance (the “aether”) instead of the va
professor190 [17]

Answer:Explained

Explanation:

The scientist of that time could not believe that the electromagnetic waves do not require a medium to travel through so they made a space filling substance necessary to travel of electromagnetic waves called aether.

But after the special 0 theory of relativity this idea is discarded and fell out.

6 0
3 years ago
A 6.00 kg ball is dropped from a height of 12.0 m above one end of a uniform bar that pivots at its center. The bar has mass 5.0
Andre45 [30]

Answer:

The height of the other ball go after the collision is 2.304 m.

Explanation:

Given that,

Mass of ball = 6.00 kg

Height = 12.0 m

Mass of bar =5.00 kg

Length = 4.00 m

Suppose we need to calculate how high will the other ball go after the collision

We need to calculate the velocity of ball

Using formula of velocity

v=\sqrt{2gh}

v=\sqrt{2\times9.8\times12.0}

v=15.33\ m/s

We need to calculate the angular momentum

Using formula of angular momentum

l_{before}=mvr

Put the value into the formula

l_{before}=6.00\times15.33\times2.0

l_{before}=183.96\ kgm^2/s

We need to calculate the angular momentum

Using formula of angular momentum

l_{after}=I_{t}\omega

l_{after}=(\dfrac{ml^2}{12}+m_{1}r^2+m_{2}r^2)\omega

Put the value into the formula

l_{after}=(\dfrac{5\times4.00^2}{12}+6.00\times2.0^2+6.00\times2.0^2)\omega

183.96=54.66\omega

\omega=\dfrac{183.96}{54.66}

\omega=3.36\ rad/sec

After collision the ball leaves with velocity

We need to calculate the velocity after collision

Using formula of the velocity

v= r\omega

v=2.0\times3.36

v=6.72\ m/s

We need to calculate the height

Using formula of height

h=\dfrac{v^2}{2g}

Put the value into the formula

h=\dfrac{(6.72)^2}{2\times9.8}

h=2.304\ m

Hence, The height of the other ball go after the collision is 2.304 m.

5 0
3 years ago
Other questions:
  • If you were trying to describe the difference between power and work you could say...?
    9·1 answer
  • A person with mass mp = 76 kg stands on a spinning platform disk with a radius of R = 1.98 m and mass md = 191 kg. The disk is i
    8·1 answer
  • A 2.26 kg book is dropped from a height of 1.5 m. What is its acceleration?
    7·1 answer
  • Use the drop-down menus to order the steps for writing chemical formulas.
    8·1 answer
  • Peggy is an astronaut and volunteers for the first manned mission to Alpha Centauri, the nearest star system to the Solar System
    8·1 answer
  • A vertical scale on a spring balance reads from 0 to 200 \rm{N}. The scale has a length of 10.0 \rm{cm} from the 0 to 200 \rm{N}
    10·1 answer
  • How much current is in a circult that includes a 9.0-volt battery and a bulb with a resistance of 4.0 ohms?
    11·1 answer
  • prepara un escrito en el que resumas las ideas principales sobre la evolución de los modelos atómicos
    7·1 answer
  • A very long time after the circuit is connected, when the capacitor is fully charged, the charge on the capacitor is
    13·1 answer
  • The rotor in a certain electric motor is a flat, rectangular coil with 80 turns of wire and dimensions 2.50 cm by 4.00 cm . The
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!