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3 years ago

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

Physics

1 answer:

makkiz [27]3 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....

$Hope This Helps$ $Sangwoo!!$

# $Be$ $Bold$

# $Always$ $Brainly$

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An ice cube at 0c was dropped into 30.0 g of water in a cup at 45.0c. at the instant that all of the ice was melted, the tempera

Ede4ka [16]

The amount of heat given by the water to the block of ice can be calculated by using
$Q=m_w C_{sw} \Delta T_w$
where
$m_w = 30 g$ is the mass of the water
$C_{sw}=4.18 J/(g ^{\circ}C)$ is the specific heat capacity of water
$\Delta T_w = 45.0^{\circ}-19.5^{\circ}C = 20.5^{\circ}C$ is the variation of temperature of the water.

Using these numbers, we find
$Q=(30 g)(4.18 J/(g^{\circ}C))(20.5^{\circ}C)=2571 J$

This is the amount of heat released by the water, but this is exactly equal to the amount of heat absorbed by the ice, used to melt it into water according to the formula:
$Q = m_i L_f$
where $m_i$ is the mass of the ice while $L_f =334 J/g$ is the specific latent heat of fusion of the ice.
Re-arranging this formula and using the heat Q that we found previously, we can calculate the mass of the ice:
$m_i = \frac{Q}{L_f}= \frac{2571 J}{334 J/g} =7.7 g$

3 0

3 years ago

What are the uses of evaporative salts?

Makovka662 [10]

to preserve foods, dye fabric, and DE-ice roads i hopes this helps

4 0

3 years ago

A 73.9 kg weight-watcher wishes to climb a

Tresset [83]

The height to which the weight-watcher must climb to work off the equivalent 991 (food) Calories is 0.59 Km

<h3>How to determine the energy. </h3>

1 food calorie = 103 calories

Therefore,

991 food calories = 991 × 103

991 food calories = 102073 calories

Multiply by 4.2 to express in joule (J)

991 food calories = 102073 × 4.2

991 food calories = 428706.6 J

<h3>How to determine the height </h3>

Energy (E) = 428706.6 J
Mass (m) = 73.9 kg
Acceleration due to gravity (g) = 9.8 m/s²

Height (h) =?

E = mgh

Divide both side by mg

h = E / mg

h = 428706.6 / (73.9 × 9.8)

h = 591.95 m

Divide by 1000 to express in km

h = 591.95 / 1000

h = 0.59 Km

Learn more about energy:

brainly.com/question/10703928

7 0

2 years ago

The head of a rattlesnake can accelerate at 49 m/s2 in striking a victim. If a car could do as well, how long would it take to r

Anettt [7]

<h2>Time taken is 0.459 seconds</h2>

Explanation:

We have equation of motion v = u + at

Initial velocity, u = 0 m/s

Final velocity, v = 81 km/hr = 22.5 m/s

Time, t = ?

Acceleration, a = 49 m/s²

Substituting

v = u + at

22.5 = 0 + 49 x t

t = 0.459 seconds

Time taken is 0.459 seconds

3 0

3 years ago

What quantity of heat is needed to convert 1 kg of ice at -13 degrees C to steam at 100 degrees C?

Effectus [21]

Answer:

Heat energy needed = 3036.17 kJ

Explanation:

We have

heat of fusion of water = 334 J/g

heat of vaporization of water = 2257 J/g

specific heat of ice = 2.09 J/g·°C

specific heat of water = 4.18 J/g·°C

specific heat of steam = 2.09 J/g·°C

Here wee need to convert 1 kg ice from -13°C to vapor at 100°C

First the ice changes to -13°C from 0°C , then it changes to water, then its temperature increases from 0°C to 100°C, then it changes to steam.

Mass of water = 1000 g

Heat energy required to change ice temperature from -13°C to 0°C

H₁ = mcΔT = 1000 x 2.09 x 13 = 27.17 kJ

Heat energy required to change ice from 0°C to water at 0°C

H₂ = mL = 1000 x 334 = 334 kJ

Heat energy required to change water temperature from 0°C to 100°C

H₃ = mcΔT = 1000 x 4.18 x 100 = 418 kJ

Heat energy required to change water from 100°C to steam at 100°C

H₄ = mL = 1000 x 2257 = 2257 kJ

Total heat energy required

H = H₁ + H₂ + H₃ + H₄ = 27.17 + 334 + 418 +2257 = 3036.17 kJ

Heat energy needed = 3036.17 kJ

5 0

3 years ago