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3 years ago

The average kinetic energy of particles in an object is also known as _________.

1 answer:

7 0

C
Because the definition of temperature is "the average kinetic energy within a substance," temperate is a measure of heat so heat cannot be the answer. And the Thermal energy (or heat energy measured in joules) is the total kinetic energy within a substance.

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What is the mechanical advantage of the level shown below?

snow_lady [41]

the correct answer is B. 1.27

Mechanical advantage of a lever is simply the ratio of the effort arm to the load arm.Effort arm is the distance from the pivot to the point of application of force while load arm is the distance of the lord from the pivot.

therefore, in this question, the effort arm is 0.28m while the load arm is 0.22 m. MA is calculated as follows: MA=effort arm/load arm

=0.28m/0.22m=1.27

5 0

3 years ago

Describe how newtons third law is related to fluid pressure.

Hatshy [7]

Fish swimming forward in the water, the water gets pushed backward because the fish moving forward is forcing the water to move backward, the motion forward and backward are the same, they are opposite and equal.

3 0

3 years ago

6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres

sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

$2ah=v_f^2-v_0^2$

Where:

$\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}$

If we assume the initial velocity to be 0 we get:

$2ah=v_f^2$

The acceleration is the acceleration due to gravity:

$2gh=v_f^2$

Now, we take the square root to both sides:

$\sqrt{2gh}=v_f$

Now, we substitute the values:

$\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f$

solving the operations:

$280\frac{m}{s}=v$

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

$F_d=\frac{1}{2}C\rho_{air}Av^2$

Where:

$\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}$

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced:

$F_d=mg$

Now, we determine the mass of the raindrop using the following formula:

$m=\rho_{water}V$

Where:

$\begin{gathered} \rho_{water}=\text{ density of water} \\ V=\text{ volume} \end{gathered}$

The volume is the volume of a sphere, therefore:

$m=\rho_{water}(\frac{4}{3}\pi r^3)$

Since the diameter of the raindrop is 3 millimeters, the radius is 1.5 mm or 0.0015 meters. Substituting we get:

$m=(0.98\times10^3\frac{kg}{m^3})(\frac{4}{3}\pi(0.0015m)^3)$

Solving the operations:

$m=1.39\times10^{-5}kg$

Now, we substitute the values in the formula for the drag force:

$F_d=(1.39\times10^{-5}kg)(9.8\frac{m}{s^2})$

Solving the operations:

$F_d=1.36\times10^{-4}N$

Now, we substitute in the formula:

$1.36\times10^{-4}N=\frac{1}{2}C\rho_{air}Av^2$

Now, we solve for the velocity:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}C\rho_{air}A}=v^2$

Now, we substitute the values. We will use the area of a circle:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^3})(\pi r^2)}=v^2$

Substituting the radius:

$\frac{1.36\cdot10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^{3}})(\pi(0.0015m)^2)}=v^2$

Solving the operations:

$70.67\frac{m^2}{s^2}=v^2$

Now, we take the square root to both sides:

$\begin{gathered} \sqrt{70.67\frac{m^2}{s^2}}=v \\ \\ 8.4\frac{m}{s}=v \\ \end{gathered}$

Therefore, the velocity is 8.4 m/s

7 0

1 year ago

g typical coffee mug holds up to 12 oz (340 grams) of water. How much heat energy is required to warm enough water from 23⁰C to

Kryger [21]

Answer:

Explanation:

Heat required to raise the temperature

= mass x specific heat x rise in temperature

= .34 x 4200 x ( 95 - 23 )

= 102816 J .

1 kWh = 1000 x 60 x 60 J

= 3600000 J

102816 J = 102816 / 3600000

= .02856 kWh .

8 0

3 years ago

Question 1 (13 marks) A charge Q is located on the top-left corner of a square. Charges of 2Q, 3Q and 4Q are located on the othe

Colt1911 [192]

Answer:

Explanation:

First of all we shall calculate electric field near charge 2Q .

electric field due to charge Q = K x Q / (5 x 10⁻² )²

E₁ = KQ / 25 x 10⁻⁴ = KQ x 10⁴ / 25 . It is acting along positive x axis

E₁ = KQ x 10⁴ i / 25

Similarly electric field due to charge 3Q near 2Q

= 3KQ x 10⁴ i / 25 . It is acting along y-axis

E₂ = 3KQ x 10⁴ j / 25

Similarly electric field due to charge 4Q near 2Q

= 4KQ x 10⁴ j / (25 x 2 )

= 2 KQ x 10⁴ / 25 . It is acting acting along north east direction

unit vector in north east direction = ( i + j )/ √2

So E₃ can be represented by

E₃ = 2 KQ x 10⁴ ( i + j ) / 25 x √2

Total field = KQ x 10⁴ i / 25 + 3KQ x 10⁴ j / 25 + 2 KQ x 10⁴ ( i + j ) / ( 25 x √2 )

= KQ x 10⁴ [ i + 3 j + √2 i + √2 j ) / 25

= 400 KQ ( 2.414 i + 4.414 j ) N / C

Force on 2Q = Field x charge = 400 KQ ( 2.414 i + 4.414 j ) x 2Q N

= 800 KQ² ( 2.414 i + 4.414 j ) N

= 800 x 9 x 10⁹ x ( 2.5 x 10⁻⁶ )² x 2.414 x ( i + 2 j ) N

= 108.63 ( i + 2 j ) N .

Magnitude of this force

= 108.63 x √5

= 243 N approx .

4 0

3 years ago