When a carousel is in motion, the movement of the carousel horse can best be described as?

Lesechka [4]

What is Newton's second law of motion?

Newton's second law of motion states that the rate of change of momentum is directly proportional to the applied force and the direction of change of momentum takes place towards direction of applied force.

<h3> $\bullet$ What do you come to know from Newton's second law of motion?</h3>

1. Concept of Momentum.

2. Measurement of Force.

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<h3 /><h3 /><h3 /><h3> $\bf Hope \: it \: helps \: you$ .</h3>

7 0

2 years ago

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Before 1960, people believed that the maximum attainable coefficient of static friction for an automobile tire on a roadway was

True [87]

This problem is looking for the minimum value of μs that is necessary to achieve the record time. To solve this problem:

Assuming the front wheels are off the ground for the entire ¼ mile = 402.3 m, the acceleration a = µs·9.8 m/s².

For a constant acceleration, distance = 402.3

m = 1/2at^2 = 804.6 m / (4.43 s)^2 = a = µs·9.8 m/s^2

µs = 804.6 m / (4.43s)^2 / 9.8 m/s^2 = 4.18

5 0

2 years ago

A 27.0-m steel wire and a 48.0-m copper wire are attached end to end and stretched to a tension of 145 N. Both wires have a radi

algol13

Answer:

The time taken by the wave to travel along the combination of two wires is 458 ms.

Explanation:

Given that,

Length of steel wire= 27.0 m

Length of copper wire = 48.0 m

Tension = 145 N

Radius of both wires = 0.450 mm

Density of steel wire $\rho_{s}= 7.86\times10^{3}\ kg/m^{3}$

Density of copper wire $\rho_{c}=8.92\times10^{3}\ kg/m^3$

We need to calculate the linear density of steel wire

Using formula of linear density

$\mu_{s}=\rho_{s}A$

$\mu_{s}=\rho_{s}\times\pi r^2$

Put the value into the formula

$\mu_{s}=7.86\times10^{3}\times\pi\times(0.450\times10^{-3})^2$

$\mu_{s}=5.00\times10^{-3}\ kg/m$

We need to calculate the linear density of copper wire

Using formula of linear density

$\mu_{c}=\rho_{s}A$

$\mu_{c}=\rho_{s}\times\pi r^2$

Put the value into the formula

$\mu_{c}=8.92\times10^{3}\times\pi\times(0.450\times10^{-3})^2$

$\mu_{c}=5.67\times10^{-3}\ kg/m$

We need to calculate the velocity of the wave along the steel wire

Using formula of velocity

$v_{s}=\sqrt{\dfrac{T}{\mu_{s}}}$

$v_{s}=\sqrt{\dfrac{145}{5.00\times10^{-3}}}$

$v_{s}=170.3\ m/s$

We need to calculate the velocity of the wave along the steel wire

Using formula of velocity

$v_{c}=\sqrt{\dfrac{T}{\mu_{c}}}$

$v_{c}=\sqrt{\dfrac{145}{5.67\times10^{-3}}}$

$v_{c}=159.9\ m/s$

We need to calculate the time taken by the wave to travel along the combination of two wires

$t=t_{s}+t_{c}$

$t=\dfrac{l_{s}}{v_{s}}+\dfrac{l_{c}}{v_{c}}$

Put the value into the formula

$t=\dfrac{27.0}{170.3}+\dfrac{48.0}{159.9}$

$t=0.458\ sec$

$t=458\ ms$

Hence, The time taken by the wave to travel along the combination of two wires is 458 ms.

4 0

3 years ago

A photon ionizes a hydrogen atom from the ground state. The liberated electron 11. recombines with a proton into the first excit

anygoal [31]

Answer:

a) 23.2 e V

b) energy of the original photon is 36.8 eV

Explanation:

given,

energy at ground level = -13.6 e V

energy at first exited state = - 3.4 e V

A photon of energy ionized from ground state and electron of energy K is released.

h ν₁ - 13.6 = K

K combine with photon in first exited state giving out photon of energy

$h\nu_2 =\dfrac{hc}{\lambda}=\dfrac{12400}{466}$

= 26.6 e V

h c = 6.626 × 10⁻³⁴ × 3 × 10⁸ = 12400 e V A°

K + ( 3.4 ) = 26.6 e V

a) energy of free electron

K = 26.6 - 3.4 = 23.2 e V

b) energy of the original photon

h ν₁ - 13.6 = K

h ν₁ = 23.2 + 13.6

= 36.8 e V

energy of the original photon is 36.8 eV

3 0

3 years ago

Which of the following quantities is inversely proportional to the gravitational pull between two objects?

Contact [7]

Answer:

C

Explanation:

Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces

I hope this helps a little bit

7 0

2 years ago

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