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2 years ago

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A bob of mass m is suspended from a fixed point with a massless string of length L (i.e., it is a pendulum). You are to investig

ate the motion in which the string moves in a cone with half-angle θ. What tangential speed, v , must the bob have so that it moves in a horizontal circle with the string always making an angle θ from the vertical?Express your answer in terms of some or all of the variables m, L, and θ, as well as the acceleration due to gravity g.v = Lsin(θ) / sqrt(Lcos(θ) / g)v = Lsin(θ) / sqrt(Lcos(θ) / g)

1 answer:

Brrunno [24]2 years ago

7 0

Answer:

b

Explanation:

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What is the atomic number of an Neon atom with 10 protons and 11 neutrons in its nucleus, & 10 Electrons in orbitals around

vesna_86 [32]

Answer:

10

Explanation:

The atomic number is only based on the number of the elements protons. The neutrons and electrons will only change the atomic mass and properties of the element. So the atomic number is 10.

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3 years ago

Read 2 more answers

You walk 100m due north. You then turn and walk 55m due east. You then make another turn and walk 12m due south. What is the res

lord [1]

Answer:

Explanation:

Important here is to know that due north is a 90 degree angle, due east is a 0 degree angle, and due south is a 270 degree angle. Then we find the x and y components of each part of this journey using the sin and cos of the angles multiplied by each magnitude:

$A_x=100cos90\\A_x=0\\B_x=55cos0\\B_x=55\\C_x=12cos270\\C_x=55$

Add them all together to get the x component of the resultant vector, V:

$V_x=55$

Do the same to find the y components of the part of this journey:

$A_y=100sin90\\A_y=100\\B_y=55sin0\\B_y=0\\C_y=12sin270\\C_y=-12$

Add them together to get the y component of the resultant vector, V:

$V_y=88$

One thing of import to note is that both of these components are positive, so the resultant angle lies in QI.

We find the final magnitude:

$V_{mag}=\sqrt{55^2+88^2}$ and, rounding to 2 sig dig's as needed:

$V_{mag}=$ 1.0 × 10² m; now for the direction:

$\theta=tan^{-1}(\frac{88}{55})=$ 58°

7 0

2 years ago

A guitar string is 90 cm long and has a mass of 3.7 g . The distance from the bridge to the support post is L=62cm, and the stri

zvonat [6]

To solve this problem we will apply the concept of frequency in a string from the nodes, the tension, the linear density and the length of the string, that is,

$f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})$

Here

n = Number of node

T = Tension

$\mu$ = Linear density

L = Length

Replacing the values in the frequency and value of n is one for fundamental overtone

$f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})$

$f = \frac{1}{2(62*10^{-2})}(\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}})$

$\mathbf{f = 281.2Hz}$

Similarly plug in 2 for n for first overtone and determine the value of frequency

$f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})$

$f = \frac{2}{2(62*10^{-2})}(\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}})$

$\mathbf{f = 562.4Hz}$

Similarly plug in 3 for n for first overtone and determine the value of frequency

$f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})$

$f = \frac{3}{2(62*10^{-2})}\bigg (\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}} \bigg)$

$\mathbf{f= 843.7Hz}$

4 0

2 years ago

What’s the voltage of a battery in a circuit with resistance of 3 ohms and current of 5 amps?

alexira [117]

Answer: The correct answer is-15 Volts.

Explanation-

Voltage of a battery can be defined as the difference in electric potential that lies between the positive and negative terminals of a battery.

It can be calculated using Ohm's law, which states that the electric potential difference between two points on a circuit is equal to the product of the current that flows between the two points (I) and the total resistance that sis present between the two points. It can be mathematically depicted as-

ΔV = I • R

Putting the value of 'I' and 'R', we get-

ΔV = 5 X 3

= 15 V

8 0

3 years ago

An object of mass 6 kg. is resting on a horizontal surface. A horizontal force

son4ous [18]

Answer:

a) The work done by the applied force is 1500 joules.

b) The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) 300 joules of energy are lost during motion.

Explanation:

a) Since the object has a constant mass, on which a constant horizontal force is exerted. The work done by the force ( $W$ ), measured in joules, is defined by the following expression:

$W = F\cdot \Delta x$ (1)

Where:

$F$ - Force, measured in newtons.

$\Delta x$ - Distance, measured in meters.

If we know that $F = 15\,N$ and $\Delta x = 100\,m$ , then the work done by the force exerted on the object is:

$W = (15\,N)\cdot (100\,m)$

$W = 1500\,J$

The work done by the applied force is 1500 joules.

b) At first we need to calculate the net acceleration of the object ( $a$ ), measured in meters per square second. By assuming a constant acceleration, we use the following kinematic formula:

$\Delta x = v_{o}\cdot t +\frac{1}{2}\cdot a\cdot t^{2}$ (2)

Where $v_{o}$ is the initial velocity of the object, measured in meters per second.

We clear the acceleration within the equation above:

$\frac{1}{2}\cdot a \cdot t^{2} = \Delta x-v_{o}\cdot t$

$a = \frac{2\cdot (\Delta x - v_{o}\cdot t)}{t^{2}}$

If we know that $\Delta x = 100\,m$ , $v_{o} = 0\,\frac{m}{s}$ and $t = 10\,s$ , then the net acceleration experimented by the object is:

$a = \frac{2\cdot \left[100\,m-\left(0\,\frac{m}{s} \right)\cdot (10\,s)\right]}{(10\,s)^{2}}$

$a = 2\,\frac{m}{s^{2}}$

By the 2nd Newton's Law, we construct the following equation of equilibrium under the consideration of a friction force acting against the motion of the object:

$\Sigma F = F - f = m\cdot a$ (3)

Where:

$F$ - External force exerted on the object, measured in newtons.

$f$ - Kinetic friction force, measured in newtons.

If we know that $F = 15\,N$ , $m = 6\,kg$ and $a = 2\,\frac{m}{s^{2}}$ , the kinetic friction force is:

$f = F-m\cdot a$

$f = 15\,N-(6\,kg)\cdot \left(2\,\frac{m}{s^{2}} \right)$

$f = 3\,N$

The work done by friction ( $W'$ ), measured in joules, is:

$W' = f\cdot \Delta x$ (4)

$W' = (3\,N) \cdot (100\,m)$

$W' = 300\,J$

And the net work experimented by the object is:

$\Delta W = 1500\,J - 300\,J$

$\Delta W = 1200\,J$

By the Work-Energy Theorem we understand that change in translational kinetic energy ( $\Delta K$ ), measured in joules, is equal to the change in net work. That is:

$\Delta K = \Delta W$ (5)

If we know that $\Delta W = 1200\,J$ , then the change in translational kinetic energy is:

$\Delta K = 1200\,J$

The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) The energy lost by the object is equal to the work done by the force of friction. Therefore, 300 joules of energy are lost during motion.

7 0

3 years ago