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Charra [1.4K]
2 years ago
14

The distance between adjacent nodes in a standing wave pattern is 25.0 cm. What is the

Physics
1 answer:
Novay_Z [31]2 years ago
4 0

Answer:

Answer:

Speed of the wave in the string will be 3.2 m/sec

Explanation:

We have given frequency in the string fixed at both ends is 80 Hz

Distance between adjacent antipodes is 20 cm

We know that distance between two adjacent anti nodes is equal to half of the wavelength

So \frac{\lambda }{2}=20cm

2

λ

=20cm

\lambda =40cmλ=40cm

We have to find the speed of the wave in the string

Speed is equal to v=\lambda f=0.04\times 80=3.2m/secv=λf=0.04×80=3.2m/sec

So speed of the wave in the string will be 3.2 m/sec

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Whats the answer???????????
Leviafan [203]

Answer: Less than 4 ohms

Explanation:

We have three resistors with the following resistance:

R_{1}=4\Omega

R_{2}=6\Omega

R_{3}=8\Omega

Now, when the resistors are connected in parallel, the total resistance R is calculated as follows:

\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}

Isolating R:

R=\frac{R_{1}R_{2}R_{3}}{R_{3}(R_{1}+R_{2})+R_{1}R_{2}}

Rewriting with th known values:

R=\frac{(4\Omega)(6\Omega)(8\Omega)}{8\Omega(4\Omega+6\Omega)+(4\Omega)(6\Omega)}

Finally:

R=1.84 \Omega

Hence, the correct option is less than 4 ohms.

4 0
3 years ago
Do you think there might be a point in space between Earth and the Moon where the gravity of each would pull on an object equall
Firlakuza [10]

Answer: Yes.

Explanation:

Assuming Earth and Moon are isolated is space, it is possible to have a point where Earth and Moon will pull at an object with equal force.

That point will be closer to the Moon than the Earth because Moon's gravitational field strength is weaker than Earth's gravitational field strength.

7 0
3 years ago
An ant carries a morsel of food 4.26 meters along a straight path to his nest. He then turns around and follows the path back to
Basile [38]
The displacement is zero. The most important concept to understand is the difference between displacement and total distance traveled. Total distance traveled would be tracking the length of the entire path the ant walked for the whole time (4.26m x 2). Displacement is how far from a designated origin (here, the food source) the ant ended up at the end of the time. Mathematically, the ant walked 4.26m from food source to nest (+4.26m) and then walked from the nest to food source (-4.26m), so the net displacement is [+4.26] + [-4.26] = 0m.
6 0
3 years ago
awhite billiard ball with mass mw = 1.47 kg is moving directly to the right with a speed of v = 3.01 m/s and collides elasticall
pochemuha

Answer:

speed of white ball is 1.13 m/s and speed of black ball is 2.78 m/s

initial kinetic energy = final kinetic energy

KE = 6.66 J

Explanation:

Since there is no external force on the system of two balls so here total momentum of two balls initially must be equal to the total momentum of two balls after collision

So we will have

momentum conservation along x direction

m_1v_{1i} + m_2v_{2i} = m_1v_{1x} + m_2v_{2x}

now plug in all values in it

1.47 \times 3.01 + 0 = 1.47 v_1cos68 + 1.47 v_2cos22

so we have

3.01 = 0.375v_1 + 0.927v_2

similarly in Y direction we have

m_1v_{1i} + m_2v_{2i} = m_1v_{1y} + m_2v_{2y}

now plug in all values in it

0 + 0 = 1.47 v_1sin68 - 1.47 v_2sin22

so we have

0 = 0.927v_1 - 0.375v_2

v_2 = 2.47 v_1

now from 1st equation we have

3.01 = 0.375 v_1 + 0.927(2.47 v_1)

v_1 = 1.13 m/s

v_2 = 2.78 m/s

so speed of white ball is 1.13 m/s and speed of black ball is 2.78 m/s

Also we know that since this is an elastic collision so here kinetic energy is always conserved to

initial kinetic energy = final kinetic energy

KE = \frac{1}{2}(1.47)(3.01^2)

KE = 6.66 J

5 0
3 years ago
a 5 charge is locataed 1.25 m to the left of a -3 charge. What is the magnitude and direction of the electrostatic force on the
Alika [10]

Answer:

The force is 86.5×10^9 N towards the negative charge (to the right)

Explanation:

The electrostatic force on the charges is given by Coulomb's law;

F= Kq1q2/r^2

This an inverse square law.

F= electrostatic force on the charges

K= constant of Coulomb's law

q1 and q2= magnitude of the charges

Since K= 9.0×10^9Nm^2C^2

F= 9.0×10^9 × 5 × 3/(1.25)^2 = 135×10^9/1.56

F= 86.5×10^9 N

The force is 86.5×10^9 N towards the negative charge.

5 0
3 years ago
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