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Morgarella [4.7K]
3 years ago
9

Which of the following is the best definition of energy?

Physics
1 answer:
Montano1993 [528]3 years ago
8 0
Energy is the capacity to do some type of work
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Based on Archimedes' principle, the greatest buoyant force an object can experience in water is determined by which quantity?
ValentinkaMS [17]

Answer:

B. The object's volume

Explanation:

When an object is immersed in a fluid, it experiences an upward force which is called buoyant force. The magnitude of the buoyant force is given by:

B=\rho_f V_{disp} g

where

\rho_f is the density of the fluid in which the object is immersed

V_{disp} is the volume of the fluid displaced by the object

g is the acceleration due to gravity

When the object is totally immersed in the fluid, V_{disp} corresponds to the volume of the object; when the object is only partially immersed, V_{disp} corresponds only to the volume of the part of the object immersed.

From the formula, we see that the greatest buoyant force is experienced by the object when it is fully immersed. Moreover, we see that the buoyant force depends only on one property of the object: its volume. Therefore, the correct choice is

B. The object's volume

6 0
3 years ago
Which question should Trudy write
Tanzania [10]

A real question, one with a question mark?

6 0
3 years ago
In which part of the ear are the vibrations occurring in a gas, liquid, and solid?
UkoKoshka [18]

Answer:

eardrum is the correct answer

hope this helps ❤️

Explanation:

Sound waves enter the ear canal and cause the eardrum to vibrate. Three small bones transmit these vibrations to the cochlea. This produces electrical signals which pass through the auditory nerve to the brain, where they are interpreted as sound.

5 0
3 years ago
The "steam" above a freshly made cup of instant coffee is really water vapor droplets condensing after evaporating from the hot
KiRa [710]

Answer:

T_{f} = 85.7 ° C

Explanation:

For this exercise we will use the calorimetry heat ratios, let's start with the heat lost by the evaporation of coffee, since it changes from liquid to vapor state

      Q₁ = m L

Where m is the evaporated mass (m = 2.00 103-3kg) and L is 2.26 106 J / kg, where we use the latent heat of the water

     Q₁ = 2.00 10⁻³ 2.26 10⁶

     Q1 = 4.52 10³ J

Now the heat of coffee in the cup, which does not change state is

     Q coffee = M c_{e} ( T_{f} -T_{i})

Since the only form of energy transfer is terminated, the heat transferred is equal to the evaporated heat

    Qc = - Q₁

    M ce (T_{f} -T_{i}) = - Q₁

The coffee dough left in the cup after evaporation is

    M = 250 -2 = 248 g = 0.248 kg

   T_{f} -Ti = -Q1 / M c_{e}

   T_{f} = Ti - Q1 / M c_{e}

Since coffee is essentially water, let's use the specific heat of water,

    c_{e}= 4186 J / kg ºC

Let's calculate

     T_{f} = 90.0 - 4.52 103 / (0.248 4.186 103)

     T_{f} = 90- 4.35

     T_{f} = 85.65 ° C

     T_{f} = 85.7 ° C

5 0
3 years ago
A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)
Phantasy [73]

Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is U  =  \int\limits^T_0 {P(t)} \, dt

Compute U if the bulb remains on for 5h

Answer:

The value is  U  =  7.563 *10^{5} \  J

Explanation:

From the question we are told that

   The power rating of the bulb is P  =  100 \  W

   The resistance is   R =  143 \ \Omega

   The  voltage is  V  =  V_o  sin [2 \pi ft]

   The  energy expanded is U  =  \int\limits^T_0 {P(t)} \, dt

   The  voltage  V_o  =  110 \  V

   The frequency is  f =  60 \  Hz

    The  time considered is  t =  5 \  h  =  18000 \  s

Generally power is mathematically represented as

             P =  \frac{V^2}{ R}

=>          P =  \frac{( 110  sin [2 \pi * 60t])^2}{ 144}

=>           P =  \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}

So  

     U  =  \int\limits^T_0 { \frac{ 110^2*  [sin [120 \pi t])^2}{ 144}} \, dt

=>  U  =  \frac{110^2}{144} \int\limits^T_0 { (   sin^2 [120 \pi t]} \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | T} \atop {0}} \right.

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | 18000} \atop {0}} \right.

U =  \frac{110^2}{144} [\frac{18000}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi (18000))}{240 \pi} ] ]

=>   U  =  7.563 *10^{5} \  J

7 0
3 years ago
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