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kondor19780726 [428]
3 years ago
8

Tommy was walking at a rate of 4 miles hour at noon and at 12:30 pm he was walking at a brisk rate of 6 miles hour . Two hours l

ater Tommy was walking at a leisurely rate of 2 miles hour . Which statement about Tommy's accleration is true?
A) Tommy had an accleration of zero the entire time.
B) Tommy had a positive acceleration between noon and 12:30 pm.
C) Tommy had a positive acceleration between 12:30 and 2:30 pm.
D) Tommy had a negative acceleration between noon and 12:30 pm.
Physics
2 answers:
docker41 [41]3 years ago
7 0

Answer: B) Tommy had a positive acceleration between noon and 12:30 pm.

maks197457 [2]3 years ago
3 0

Answer:

B) Tommy had a positive acceleration between noon and 12:30 pm.

Explanation:

Acceleration is defined as the rate of change of velocity:

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time

In the problem,

- At noon, Tommy is walking at a velocity of 4 mi/h

- At 12.30 pm, Tommy is walking at a velocity of 6 mi/h

- A time of half an hour (0.5 h) passed between the two moments

So Tommy's acceleration is

a=\frac{6 mi/h-4 mi/h}{0.5 h}=4 mi/h^2

and the acceleration is positive, since the velocity has increased.

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A ___ is awarded to the opposing team when that team violently or dangerously charges a opponent. ( in soccer)
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Answer:

(C)Direct kick

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Free kick is of two types; direct kick and indirect kick.

Direct kick is awarded when a player outside 18 yards box violently or dangerously charges a opponent.

Indirect kick can be awarded when a goalkeeper inside 18 yards box commits certain offenses.

Therefore, the correction option is (C)Direct kick

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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:

a)

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

b)

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

c)

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}

where

\Delta v is the change in the linear speed

\Delta  t is the time interval

In this problem, the spaceship is moving with constant linear speed equal to

v=8322 m/s

Therefore, its linear speed is not changing, so the change in linear speed is zero:

\Delta v=0

And therefore, the tangential acceleration is zero as well:

a_t=\frac{0}{\Delta t}=0 m/s^2

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