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Alenkasestr [34]
3 years ago
14

Christopher conducts an experiment in which he tests how much sugar dissolves at different temperatures of water. One step requi

res him to use water with a temperature of 32.5°C. What is this temperature in kelvins? Record your answer to the nearest tenth. K
Physics
2 answers:
sveticcg [70]3 years ago
5 0

305.5 i just took the quiz

disa [49]3 years ago
4 0

Explanation:

When we have to convert temperature from degree celsius to kelvin then we add 273 into the given degree celsius temperature.

Therefore, convert 32.5^{o}C into kelvin as follows.

         Temperature in kelvin = Temperature in degree celsius + 273

                                               = (32.5 + 273) K

                                               = 305.5 K

Thus, we can conclude that temperature in kelvins is 305.5 K.

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After the first few seconds of a race, a runner runs at the same speed until she approaches the finish line, at which point she
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5 0
3 years ago
f the CD rotates clockwise at 500 rpmrpm (revolutions per minute) while the last song is playing, and then spins down to zero an
vazorg [7]

Answer:

The magnitude of the angular acceleration is  a =  20.14 rad/s^2

Explanation:

From the question we are told that

   The angular speed of CD is  w_{CD} =  500 rpm =  \frac{500  rpm}{\frac{60 \ s }{1 \ min} } * \frac{2 \pi }{ 1 \ rev} = 52.37 rad/s

    time taken to decelerate is t_{CD} =  2.60\ s

    The final angular speed is  w_f= 0 \ rad/s

The angular acceleration is mathematically represented as

         a =  \frac{w_f - w_{CD}}{t}

substituting values

          a =  \frac{0 - 52.37}{2.60}

         a =  - 20.14 rad/s^2

The negative sign show that the CD is decelerating  but the magnitude is

       a =  20.14 rad/s^2

   

3 0
3 years ago
Which of the following objects would have the strongest gravitational pull on a third object positioned at an equal distance fro
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I believe Box B will have a greater gravitational pull because the gravitational pull of an object depends on its mass. The more mass an object has, the greater its gravitational pull will become.

For example, we can take planets. Naturally, they are round because once upon a time there was a larger piece of rock that attracted others. But the size of the rock won't matter, it's the weight that matters. If the rock weighed nothing, the other rocks would just rebound upon contact. But if the rock weighed a lot, then things wouldn't so easily rebound and might actually stick to it.
8 0
3 years ago
A 0.5 m diameter wagon wheel consists of a thin rim having a mass of 7 kg and six spokes, each with a mass of 1.2 kg. 1.2 kg 7 k
Arte-miy333 [17]

Explanation:

It is given that,

Mass of the rim of wheel, m₁ = 7 kg

Mass of one spoke, m₂ = 1.2 kg

Diameter of the wagon, d = 0.5 m

Radius of the wagon, r = 0.25 m

Let I is the the moment of inertia of the wagon wheel for rotation about its axis.

We know that the moment of inertia of the ring is given by :

I_1=m_1r^2

I_1=7\times (0.25)^2=0.437\ kgm^2

The moment of inertia of the rod about one end is given by :

I_2=\dfrac{m_2l^2}{3}

l = r

I_2=\dfrac{m_2r^2}{3}

I_2=\dfrac{1.2\times (0.25)^2}{3}=0.025\ kgm^2

For 6 spokes, I_2=0.025\times 6=0.15\ kgm^2

So, the net moment of inertia of the wagon is :

I=I_1+I_2

I=0.437+0.15=0.587\ kgm^2

So, the moment of inertia of the wagon wheel for rotation about its axis is 0.587\ kgm^2. Hence, this is the required solution.

4 0
3 years ago
A man has 887.5 J of kinetic energy while running with a velocity of 5 m/s. What is his mass?
monitta

Answer:

The mass of the man is 71 kg

Explanation:

Given;

kinetic energy of the man, K.E = 887.5 J

velocity of the man, v = 5 m/s

The mass of the man is calculated as follows;

K.E = ¹/₂mv²

where;

m is the mass of the man

2K.E = mv²

m = 2K.E / v²

m = (2 x 887.5) / (5)²

m = 71 kg

Therefore, the mass of the man is 71 kg

7 0
2 years ago
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