All categories

2 years ago

Match the object to the type of motion it's displaying. which letter goes with which number?

Physics

1 answer:

makkiz [27]2 years ago

8 0

Hello There!!

I think the best answer choice is:

A.  $Linear$ $motion$

B.  $Rotational$ $motion$

D. $Vibrational$ $motion$

4. $Runners$ $sprinting$ $across$ $a$ $straight$ $track$

C. $Curvilinear$ $motion$

3. $Music$ $being\\\\\\$ $played$ $on$ $a$ $stringed$ $instrument,$ $such$ $as$ $a$ $guitar$

2.  $A$ $car$ $driving$ $up$ $a$ $winding$ $mountain$ $road$

1.  $The$ $propeller$ $of$ $an$ $aircraft$ $in$ $midair$

I'm not sure this is 100% right....

P.S Tell me if this is wrong....

$Hope This Helps$ $Sangwoo!!$

# $Be$ $Bold$

# $Always$ $Brainly$

You might be interested in

Is it possible that a speed of 254 and a speed of 100 could be the same speed?

EastWind [94]

Not if both speeds are in the same units.

However, if the 254 is 'centimeters per time' and the 100 is 'inches per time',
then the speeds are equal.

4 0

2 years ago

You start pushing a suitcase full of clothes in the horizontal direction with a force of 25.0 newtons. The weight of the suitcas

sasho [114]

Answer:

Distance traveled will be 5.6307 m

Explanation:

Time t = 3 sec

We have given force F = 25 N

We know that force is given by F = ma

So ma = 25 -----------eqn 1

Weight is given by W = 196 N

We know that weight is given by W = mg

So mg = 196 -----------------eqn 2

From equation 1 and equation 2 $\frac{a}{g}=\frac{25}{196}$

$a=1.2512m/sec^2$

Initial velocity is given as 0 so u = 0 m/sec

From second equation of motion $s=ut+\frac{1}{2}at^2=0\times 3+\frac{1}{2}\times 1.2512\times 3^2=5.6307m$

7 0

3 years ago

A. How does the speed of light in a vacuum change when observed from two different frames of reference?

anyanavicka [17]

a) It is absolute, so it does not change.

b) Inertial ones.

c) Inside the train the time will slow down relatively to the outside clock. So if one travel at nearly the speed if light for 2 hours on his clock, for outdoor observers it will look like 3 hours.

4 0

2 years ago

Read 2 more answers

A child slides down a hill on a toboggan with an acceleration of 1.8 m/s^2. If she starts at rest, how far has she traveled in :

DENIUS [597]

Explanation:

It is given that,

The acceleration of the toboggan, $a=1.8\ m/s^2$

Initial speed of the toboggan, u = 0

We need to find the distance covered by the toboggan. Using the second equation of motion as :

$s=ut+\dfrac{1}{2}at^2$

At t = 1 s

$s=\dfrac{1}{2}\times 1.8\times 1^2$

$s_1=0.9\ m$

At t = 2 s

$s=\dfrac{1}{2}\times 1.8\times 2^2$

$s_2=3.6\ m$

At t = 3 s

$s=\dfrac{1}{2}\times 1.8\times 3^2$

$s_3=8.1\ m$

Hence, this is the required solution.

4 0

3 years ago

A glass flask whose volume is 1000 cm^3 at a temperature of 1.00Â°C is completely filled with mercury at the same temperature. W

Marat540 [252]

Answer:

the coefficient of volume expansion of the glass is $\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}$

Explanation:

Given that:

Initial volume of the glass flask = 1000 cm³ = 10⁻³ m³

temperature of the glass flask and mercury= 1.00° C

After heat is applied ; the final temperature = 52.00° C

Temperature change ΔT = 52.00° C - 1.00° C = 51.00° C

Volume of the mercury overflow = 8.50 cm^3 = 8.50 × 10⁻⁶ m³

the coefficient of volume expansion of mercury is 1.80 × 10⁻⁴ / K

The increase in the volume of the mercury = 10⁻³ m³ × 51.00 × 1.80 × 10⁻⁴

The increase in the volume of the mercury = $9.18*10^{-6} \ m^3$

Increase in volume of the glass = 10⁻³ × 51.00 × $\beta _{glass}$

Now; the mercury overflow = Increase in volume of the mercury - increase in the volume of the flask

the mercury overflow = $(9.18*10^{-6} - 51.00* \beta_{glass}*10^{-3})\ m^3$

$8.50*10^{-6} = (9.18*10^{-6} -51.00* \beta_{glass}* 10^{-3} )\ m^3$

$8.50*10^{-6} - 9.18*10^{-6} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3$

$-6.8*10^{-7} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3$

$6.8*10^{-7} = ( 51.00* \beta_{glass}* 10^{-3} )\ m^3$

$\dfrac{6.8*10^{-7}}{51.00 * 10^{-3}}= ( \beta_{glass} )$

$\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}$

Thus; the coefficient of volume expansion of the glass is $\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}$

5 0

2 years ago