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Vesna [10]
3 years ago
13

A thermometer is placed in water in order to measure the water’s temperature. What would cause the liquid in the thermometer to

drop?
Physics
2 answers:
sineoko [7]3 years ago
3 0
The temperature of the water getting colder would cause the liquid in the thermometer to drop due to less heat being transferred from the water to the liquid, so the liquid molecules are closer than when they have high energy.
alexandr402 [8]3 years ago
3 0

Answer:

The kinetic energy of the water molecules decreases.

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an auditorium measures 30.0 m ✕ 15.0 m ✕ 5.0 m. the density of air is 1.20 kg/m3. (a) what is the volume of the room in cubic fe
Fynjy0 [20]

dimension = 30.0 m ✕ 15.0 m ✕ 5.0 m.

density = 1.20 kg/m3

(a)volume = lenght * breadth * height

      = 30 * 15 * 5

      = 2250 metre cube = 2.25 cubic meter

(b)   mass of air = density * volume

        mass of air = 1.2 * 2250

mass of air = 2700kg

weight  = mass * 9.8

             = 2700 * 9.8

             = 26,460 N

  • The definition of Density is the amount of matter in a given space, or volume
  • Density = mass/volume
  • units for density kg/m^3
  • Density of water 1g/ml
  • Salt water is denser that is why  don't sink as easily.

To know more about density  visit : brainly.com/question/15164682

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5 0
1 year ago
Would u rather/ be in avengers or x-men
vlabodo [156]
X men those mutants are amazing
6 0
2 years ago
A uniform disk with mass 35.2 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stati
Sergio [31]

Answer:

a) v = 1.01 m/s

b) a = 5.6 m/s²

Explanation:

a)

  • If the disk is initially at rest, and it is applied a constant force tangential to the rim, we can apply the following expression (that resembles Newton's 2nd law, applying to rigid bodies instead of point masses) as follows:

       \tau = I * \alpha  (1)

  • Where τ is the external torque applied to the body, I is the rotational inertia of the body regarding the axis of rotation, and α is the angular acceleration as a consequence of the torque.
  • Since the force is applied tangentially to the rim of the disk, it's perpendicular to the radius, so the torque can be calculated simply as follows:
  • τ = F*r (2)
  • For a solid uniform disk, the rotational inertia regarding an axle passing through its center  is just I = m*r²/2 (3).
  • Replacing (2) and (3) in (1), we can solve for α, as follows:

       \alpha = \frac{2*F}{m*r} = \frac{2*34.5N}{35.2kg*0.2m} = 9.8 rad/s2 (4)

  • Since the angular acceleration is constant, we can use the following kinematic equation:

        \omega_{f}^{2}  - \omega_{o}^{2} = 2*\Delta \theta * \alpha (5)

  • Prior to solve it, we need to convert the angle rotated from revs to radians, as follows:

       0.2 rev*\frac{2*\pi rad}{1 rev} = 1.3 rad (6)

  • Replacing (6) in (5), taking into account that ω₀ = 0 (due to the disk starts from rest), we can solve for ωf, as follows:

       \omega_{f} = \sqrt{2*\alpha *\Delta\theta} = \sqrt{2*1.3rad*9.8rad/s2} = 5.1 rad/sec (7)

  • Now, we know that there exists a fixed relationship the tangential speed and the angular speed, as follows:

        v = \omega * r (8)

  • where r is the radius of the circular movement. If we want to know the tangential speed of a point located on the rim of  the disk, r becomes the radius of the disk, 0.200 m.
  • Replacing this value and (7) in (8), we get:

       v= 5.1 rad/sec* 0.2 m = 1.01 m/s (9)

b)    

  • There exists a fixed relationship between the tangential and the angular acceleration in a circular movement, as follows:

       a_{t} = \alpha * r (9)

  • where r is the radius of the circular movement. In this case the point is located on the rim of the disk, so r becomes the radius of the disk.
  • Replacing this value and (4), in (9), we get:

       a_{t}  = 9.8 rad/s2 * 0.200 m = 1.96 m/s2 (10)

  • Now, the resultant acceleration of a point of the rim, in magnitude, is the vector sum of the tangential acceleration and the radial acceleration.
  • The radial acceleration is just the centripetal acceleration, that can be expressed as follows:

       a_{c} = \omega^{2} * r  (11)

  • Since we are asked to get the acceleration after the disk has rotated 0.2 rev, and we have just got the value of the angular speed after rotating this same angle, we can replace (7) in (11).
  • Since the point is located on the rim of the disk, r becomes simply the radius of the disk,, 0.200 m.
  • Replacing this value and (7) in (11) we get:

       a_{c} = \omega^{2} * r   = (5.1 rad/sec)^{2} * 0.200 m = 5.2 m/s2 (12)

  • The magnitude of the resultant acceleration will be simply the vector sum of the tangential and the radial acceleration.
  • Since both are perpendicular each other, we can find the resultant acceleration applying the Pythagorean Theorem to both perpendicular components, as follows:

       a = \sqrt{a_{t} ^{2} + a_{c} ^{2} } = \sqrt{(1.96m/s2)^{2} +(5.2m/s2)^{2} } = 5.6 m/s2 (13)

6 0
2 years ago
A man runs at an average speed of 5.0m/s how long will it take him to run 5.2km on a perfectly straight line
Anni [7]

The time taken is 1040 s.

<h3>What is speed?</h3>

The term speed refers to the rate at which the distance changes per unit time. This is why we define speed as the ratio of the distance to time for a body that is moving along a straight line.

Now;

We must first convert the distance to meters;

distance = 5.2km  or 5200m

Speed = distance/time

time = distance/speed

time =  5200m/5 m/s

time = 1040 s

Learn more about speed:brainly.com/question/28224010

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8 0
1 year ago
A Cessna aircraft has a lift off speed of 147 km/h. What minimum constant acceleration does this require if the aircraft is to b
netineya [11]

Answer:

The acceleration is  a =51945 \ km/h^2

Explanation:

From the question we are told that

   The lift up speed is  v  = 147 \  km/h

    The distance covered for the take off run is s =  208 m = 0.208 \ km

Generally from kinematic equation we have that

      v^2 = u^2 + 2as

Here u is the initial  speed of the aircraft with value 0 m/ s give that the aircraft started from rest

So  

    147^2 = 0^2 + 2* a* 0.208

=>  a =51945 \ km/h^2

6 0
2 years ago
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