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2 years ago

The graphs display velocity data. Velocity is on the y-axis (m/s), while time is on the x-axis (s). Based on the graphs,

Physics

1 answer:

EastWind [94]2 years ago

3 0

Answer:

0 1 2 3 4 . because the particle moved with a constant acceleration

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1) Halving the distance (i.s., decreasing by a factor of two) between two charged objects will cause the electrical force betwee

Vedmedyk [2.9K]

Answer:

Explanation:

For an electric force, F the formula:

F = kQq/r^2

Given:

r2 = 1/2 × r1

F1 × r1 = k

F1 × r1 = F2 × r2

F2 = (F1 × r1^2)/(0.5 × r1)^2

= (F1 × r1^2)/0.25r1^2

= 4 × F1.

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3 years ago

An electric immersion heater is put at the bottom of a large tank of water. The water next to the heater becomes warm

djyliett [7]

Answer:

when the water is heated with immersion heater, the water becomes less dense due to which the warm water rises up and the cooler water fills it's space.

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3 years ago

An airplane was 300 km [N] of Toronto airport, 3 hours later the airplane 600 km [S]

Alecsey [184]

Answer:

Explanation:

a) 300 + 600 = 900 km S

b) 900/3 = 300 km/hr S

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3 years ago

A 6.0-cm-diameter horizontal pipe gradually narrows to 4.0 cm. When water flows through this pipe at a certain rate, the gauge p

denpristay [2]

Answer:

a n c

Explanation:

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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

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