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4 years ago

An observer sitting at a bus stop sees the bus drive by at 30 m/s to the east. He also sees a

Physics

1 answer:

Dafna1 [17]4 years ago

8 0

Answer:

32 m/s

Explanation:

The speed of a bus is 30 m/s due East wrt the passenger

He also sees a passenger on the bus walking to the back at 2 m/s.

We need to find the passenger's velocity relative to the bus. As the observer sees that the bus and the passenger are moving in opposite direction. Let v is the relative velocity. So,

v = 30 m/s + 2 m/s

v = 32 m/s

Hence, the passenger's velocity relative to the bus is 32 m/s.

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In what sense can a transformer be considered an electrical lever? What does it multiply? What does it not multiply?

liberstina [14]

Answer:

This is because it steps up or steps down electrical voltage. It multiplies either voltage (if it is a voltage transformer )or current (if it is a current transformer), but it does not multiply electrical power.

Explanation:

A transformer steps up or steps down electrical voltage, by transmitting power at a voltage, V₁ and Current I₁ at one terminal, to a voltage, V₂ and Current I₂ at its other terminals, just like a lever transmits force from one point to another. Since the power transmitted remains the same, (energy per unit time remains constant), I₁V₁ = I₂V₂ ⇒ I₁/I₂ = V₂/V₁ = n (the turns ratio of the transformer). So, the turns ratio will determine if its a step-up or step-down transformer. V₂ = nV₁. So, if V₁ > V₂ it is a step down transformer and if V₁ < V₂ it is a step-up transformer.It multiplies either voltage (if it is a voltage transformer )or current (if it is a current transformer), but it does not multiply electrical power, since P = IV = constant for the transformer.

5 0

3 years ago

How would you explain social control to someone who has no idea about our society?

Natasha2012 [34]

You can try to explain it by using a parallel between their and your societies .

7 0

3 years ago

Sam receives the kicked football on the 3 yd line and runs straight ahead toward the goal line before cutting to the right at th

Pie

Answer:

Distance: 21 yd, displacement: 15 yd, gain in the play: 12 yd

Explanation:

The distance travelled by Sam is just the sum of the length of each part of Sam's motion, regardless of the direction. Initially, Sam run from the 3 yd line to the 15 yd line, so (15-3)=12 yd. Then, he run also 9 yd to the right. Therefore, the total distance is

d = 12 + 9 = 21 yd

The displacement instead is a vector connecting the starting point with the final point of the motion. Sam run first 12 yd straight ahead and then 9 yd to the right; these two motions are perpendicular to each other, so we can find the displacement simply by using Pythagorean's theorem:

$d=\sqrt{12^2+9^2}=15 yd$

Finally, the yards gained by Sam in the play are simply given by the distance covered along the forward-backward direction only. Since Sam only run from the 3 yd line to the 15 yd line along this direction, then the gain in this play was

d = 15 - 3 = 12 yd

7 0

3 years ago

A spaceship negotiates a circular turn of radius 2925 km at a speed of 29960 km/h. (a) What is the magnitude of the angular spee

emmainna [20.7K]

a) 0.0028 rad/s

b) $23.68 m/s^2$

c) $0 m/s^2$

Explanation:

When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by

$\omega = \frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time interval

The angular speed of an object in circular motion can also be written as

$\omega = \frac{v}{r}$ (1)

where

v is the linear speed of the object

r is the radius of the orbit

For the spaceship in this problem we have:

$v=29,960 km/h$ is the linear speed, converted into m/s,

$v=8322 m/s$

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Subsituting into eq(1), we find the angular speed of the spaceship:

$\omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s$

When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the centre of the orbit: this is called centripetal acceleration, or radial acceleration.

The magnitude of the radial acceleration is given by

$a_r=\omega^2 r$

where

$\omega$ is the angular speed

$r$ is the radius of the orbit

For the spaceship in the problem, we have

$\omega=0.0028 rad/s$ is the angular speed

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Substittuing into the equation above, we find the radial acceleration:

$a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2$

When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.

The tangential acceleration is given by

$a_t=\frac{\Delta v}{\Delta t}$