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3 years ago

What quantity is scalar

Physics

2 answers:

daser333 [38]3 years ago

5 0

A scalar quantity is a measurement of a quantity, like temperature, or mass.

Anni [7]3 years ago

4 0

Mass, temperature, weight, cost, speed, loudness, distance, hardness, value.

Anything without a direction

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Please help u guys acellus sucks

34kurt

Answer:

6.0cm

Explanation:

Given

focal length = 15.0cm

object distance = 10.0cm

Required

Image distance v

Using the formula

1/f = 1/u + 1/v

1/15 = 1/10+1/v

1/v = 1/15 + 1/10

1/v = 2+3/30

1/v = 5/30

v = 30/5

v = 6.0cm

Hence the image distance is 6.0cm

7 0

3 years ago

When asked to describe resistance you say, "It is an object's ability to

Vesnalui [34]

Accept the flow of current and is measured in ohms

8 0

3 years ago

Read 2 more answers

How large must the coefficient of static friction be between the tires and the road if a car is to round a level curve of radius

Korvikt [17]

Answer:

<h2>The coefficient of static friction will be 0.7</h2>

Explanation:

Given data

the radius of curve= 90m

speed v= 90 km/h to m/s = (90*100)/60*60= 25 m/s

we know that the expression for the centripetal force acting on the car

$Fc= \frac{mv^2}{r}$ -------1

we also know that the expression for the frictional force between road and tire.

Ff= μmg--------2

Equating equation 1 and 2 we have

μmg= mv^2/r

μ= v^2/gr

substituting the values of speed and radius we have (assuming g= 9.81m/s^2)

μ= 25^2/9.81*90

μ= 625/882.9

μ= 0.7

4 0

3 years ago

A radar station sends out a 250000 Hz sound wave at a speed of 340 m/s. The sound wave bounces off a weather ballon and returns

Eddi Din [679]

Answers:

a)The balloon is 68 m away of the radar station

b) The direction of the balloon is towards the radar station

Explanation:

We can solve this problem with the Doppler shift equation:

$f'=\frac{V+V_{o}}{V-V_{s}} f$ (1)

Where:

$f=250,000 Hz$ is the actual frequency of the sound wave

$f'=240,000 Hz$ is the "observed" frequency

$V=340 m/s$ is the velocity of sound

$V_{o}=0 m/s$ is the velocity of the observer, which is stationary

$V_{s}$ is the velocity of the source, which is the balloon

Isolating $V_{s}$ :

$V_{s}=\frac{V(f'-f)}{f'}$ (2)

$V_{s}=\frac{340 m/s(240,000 Hz-250,000 Hz)}{240,000 Hz}$ (3)

$V_{s}=-14.16 m/s$ (4) This is the velocity of the balloon, note the negative sign indicates the direction of motion of the balloon: It is moving towards the radar station.

Now that we have the velocity of the balloon (hence its speed, the positive value) and the time ( $t=4.8 s$ ) given as data, we can find the distance:

$d=V_{s}t$ (5)

$d=(14.16 m/s)(4.8 s)$ (6)

Finally:

$d=68 m$ (8) This is the distance of the balloon from the radar station

6 0

3 years ago

A rotating object has an angular acceleration of α = 0 rad/s2. Which one or more of the following three statements is consistent

Murrr4er [49]

Answer:

A,B and C

Explanation:

Statement A

At all times, angular velocity is $\omega = 0\,{\rm{rad/s}$

Angular acceleration is the rate of change in angular velocity with respect to time.

Angular velocity and angular acceleration are related by

${\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t$

Which when re-arranged becomes

$\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}$

There’s no change in angular velocity anytime when the angular velocity is $\omega = 0\,{\rm{rad/s}}$

The equation can be modified as follows:

$\begin{array}{c}\\\alpha = \frac{{0\,{\rm{rad/s}} - 0\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}$

Therefore, the angular acceleration becomes zero hence statement A is valid.

Statement B

Angular acceleration is the rate of change in angular velocity with respect to time.

Angular velocity and angular acceleration are related by

${\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t$

Which when re-arranged becomes

$\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}$

There’s no change in angular velocity anytime when the angular velocity is $\omega = 10\,{\rm{rad/s}}$ .The final and initial velocities remain the same.

The equation can be modified as follows:

$\begin{array}{c}\\\alpha = \frac{{10\,{\rm{rad/s}} - 10\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}$

Therefore, the angular acceleration becomes zero and statement B is valid

Statement C

Angular velocity is defined as the change in the angular position with respect to time.

Angular velocity and angular displacement are related by

$\theta = \omega t$

Which can also be modified as:

${\theta _{\rm{f}}} - {\theta _{\rm{i}}}$

Note that the final position is ${\theta _{\rm{f}}}$ and initial position is ${\theta _{\rm{i}}}$

Modifying the equation to find the angular velocity we obtain

$\omega = \frac{{{\theta _{\rm{f}}} - {\theta _{\rm{i}}}}}{t}$

When the angular displacement has the same value at all times, the equation becomes

$\begin{array}{c}\\\omega = \frac{{{\theta _{\rm{i}}} - {\theta _{\rm{i}}}}}{t}\\\\ = 0\\\end{array}$

The angular velocity becomes zero.

Angular acceleration and angular velocity are related by

${\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t$

The expression above can be rearranged as follows:

$\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}$

At all times, the angular velocity is $\omega = 0\,{\rm{rad/s}}$ hence initial and final velocities remain the same

We obtain

$\begin{array}{c}\\\alpha = \frac{{0\,{\rm{rad/s}} - 0\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}$

Therefore, the angular acceleration becomes zero and statement C is valid.

Therefore, statements A,B and C are consistent .

4 0

4 years ago