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2 years ago

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A small object slides along the frictionless loop-the-loop with a diameter of 3 m. what minimum speed must it have at the top of

the loop in order to remain in contact with the loop

1 answer:

otez555 [7]2 years ago

4 0

<span>3.834 m/s. In this problem we need to have a centripetal force that is at least as great as the gravitational attraction the object has. The equation for centripetal force is F = mv^2/r and the equation for gravitational attraction is F = ma Since m is the same in both cases, we can cancel it out and then set the equations equal to each other, so a = v^2/r Substitute the known values (radius is diameter/2) and solve for v 9.8 m/s^2 <= v^2/1.5 m 14.7 m^2/s^2 <= v^2 3.834057903 m/s <= v So the minimum velocity needed is 3.834 m/s.</span>

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If c1=c2=4.00μf and c4=8.00μf, what must the capacitance c3 be if the network is to store 2.70×10−3 j of electrical energy?

Akimi4 [234]

Missing detail in the text: total voltage of the circuit $\Delta V = 46.0 V$
Missing figure: https://www.physicsforums.com/attachments/prob-24-68-jpg.190851/

Solution:

1) The energy stored in a circuit of capacitors is given by
$U= \frac{1}{2} C_{eq} (\Delta V)^2$
where $C_{eq}$ is the equivalent capacitance of the circuit. We can find the value for $C_{eq}$ by using $\Delta V=46.0 V$ and the energy of the system, $U=2.7\cdot 10^{-3} J$
$C_{eq}= \frac{2U}{(\Delta V)^2}=2.55\cdot 10^{-6} F=2.55\mu F$

2) Then, let's calculate the equivalente capacitance of C1 and C2. The two capacitors are in series, so their equivalente capacitance is given by
$\frac{1}{C_{12}}= \frac{1}{C_1}+ \frac{1}{C_2}= \frac{1}{4 \mu F} + \frac{1}{4 \mu F}$
from which we find $C_{12}=2 \mu F$

3) Then let's find $C_{123}$ , the equivalent capacitance of $C_{12}$ and C3. $C_{123}$ is in series with C4, therefore we can write
$\frac{1}{C_{eq}}= \frac{1}{C_{123}}+ \frac{1}{C_4}$
Since we already know $C_4=8 \mu F$ and $C_{eq}=2.55 \mu F$ , we find
$C_{123}=3.70 \mu F$

4) Finally, we can find $C_{3}$ , because it is in parallel with $C_{12}$ , and the equivalent capacitance of the two must be equal to $C_{123}$ :
$C_{123}=C_{12}+C_3$
So, using $C_{123}=3.70 \mu F$ and $C_{12}=2 \mu F$ , we find
$C_3=1.70 \mu F$

7 0

1 year ago

Question 5 At 12:00 pm, a spaceship is at position ⎡⎣324⎤⎦ km ⎣ ⎢ ⎡ 3 2 4 ⎦ ⎥ ⎤ km away from the origin with respect to so

Anettt [7]

Answer:

[1, 6, -2]

Explanation:

Given the following :

Initial Position of spaceship : [3 2 4] km

Velocity of spaceship : [-1 2 - 3] km/hr

Location of ship after two hours have passed :

Distance moved by spaceship :

Velocity × time

[-1 2 -3] × 2 = [-2 4 -6]

Location of ship after two hours :

Initial position + distance moved

[3 2 4] + [-2 4 -6] = [3 + (-2)], [2 + 4], [4 + (-6)]

= [3-2, 2+4, 4-6] = [1, 6, -2]

4 0

2 years ago

Xander's friend Corey has a skateboard that he rides at Newton's Skate Park.

DanielleElmas [232]

Answer:

0.5 m/s2

Explanation:

Step 1:

Data obtained from the question.

Total Mass = 60Kg

Net force = 30N

Acceleration =?

Step 2

Determination of the acceleration.

Force = Mass x Acceleration.

With the above equation, we can easily obtain the acceleration as follow:

30 = 60 x Acceleration

Divide both side by 60

Acceleration = 30/60

Acceleration = 0.5 m/s2

Now, we can thus say that the acceleration at that moment is 0.5 m/s2

8 0

2 years ago

In a distant solar system, a giant planet has

sergeinik [125]

Answer:

mass of the planet: $5.9\,10^{26}\,kg$

Explanation:

When a moon keeps a circular orbit around a planet, it is the force of gravity the one that provides the centripetal force to keep it in its circular trajectory of radius R. So if we can write that in such cases (being the mass of the planet "M" and the mass of the moon "m"), we can form an equation by making the centripetal force on the moon equal the force of gravity (using the Newton's Universal Law of Gravity):

$m\frac{v^2}{R}=G\frac{M\,m}{R^2}$

where we used here the tangential velocity (v) of the moon around the planet. This equation can be further simplified by dividing both sides by "m" and multiplying both sides by the orbital radius R:

$m\frac{v^2}{R}=G\frac{M\,m}{R^2}\\v^2=G\frac{M}{R}$

Notice that the mass of the moon has actually disappeared from the equation, which tells us that the orbiting velocity and period do not depend on the mass of the moon, but on the mass of the actual planet.

We know the orbital radius R ( $5.32\,10^5\,km=5.32\,10^8\,m$ , the value of the Universal Gravitational constant G, and we can estimate the value of the tangential velocity of the moon since we know it period: 36.3 hrs = 388800 seconds.

We know that the moon makes a full circumference ( $2\,\pi\,R$ ) in 388800 seconds, therefore its tangential velocity is:

$v=\frac{2\,\pi\,5.32\,10^8}{388800} \frac{m}{s} \\v=8.6\,10^3\,\frac{m}{s}$

where we rounded the velocity to one decimal.

Notice that we have converted all units to the SI system, so when using the formula to solve for the mass of the planet, the answer comes directly in kg.

Now we use this value for the tangential velocity to estimate the mass of the planet in the first equation we made and simplified:

$v^2=G\frac{M}{R}\\M=\frac{v^2\,R}{G} \\M=\frac{(8.6\,10^3)^2\,5.32\,10^8}{6.67\,10^{-11}}kg\\M=5.9\,10^{26}\,kg$

8 0

2 years ago

A heat recovery system (HRS) is used to conserve heat from the surroundings and supply it to the Mars Rover. The HRS fluid loop

Annette [7]

Answer: Volume of Freon = 12.23. $10^{3}$ L

Explanation: First, the temperatures have to be in the same unit, so all the tmeperatures is transformed into Kelvin:

T₁ = 192K

T₂ = $\frac{5}{9}$ . (F - 32)+273.15

T₂ = $\frac{5}{9}$ . (- 65 - 32)+273.15

T₂ = 219.26K.

Second, BTU is an unit of energy. It can be transformed into Joules by the relation 1BTU = 1055.06J.

Q = 1055.06 . 46.9 = 49482.314J

The letter Q represents Heat and is calculated as Q = m.Cp.ΔT, where:

m is mass of the element;

Cp is the heat capacity specific for each element;

ΔT is the difference between the final and initial temperature;

Third, it will be needed the properties of the Freon:

Freon (CF2Cl2) = 120.91g/mol; Cp = 74J/mol.K; ρ = 1.49kg/m³

Fourth, we calculate the mass of Freon necessary for the remove of 46.9BTU of energy from the system:

Q = m.Cp.ΔT

m= $\frac{Q}{c(T-T_{0} )}$

m = $\frac{49482.314}{74(219.26-192)}$

m = 18228.214g or m=18.228Kg

Fifth, using the density of Freon, Volume can be found

ρ = $\frac{m}{V}$

V = m / ρ

V = 18.228 / 1.49

V = 12.23m³

As SI for density is Kg/m³, the volume found is in m³.

Cubic meters is related to Liters as the following: 1m³=1000l.

So, volume of Freon = 12.23. $10^{3}$ L

The volume of Freon needed is 12.23. $10^{3}$ L.

7 0

1 year ago