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3 years ago

Which answer is right? thanks if u help me! :)

Physics

2 answers:

WINSTONCH [101]3 years ago

8 0

Answer: I believe the answer is 7.5 x 10^(14)Hz or 7.5E14Hz

Explanation: the formula for frequency is speed of the wave divided by the wavelength

f= c/wavelength

f- frequency

c-speed

(3 x 10^(8)) x (400 x 10^(-9))= 750,000,000,000,000 or 7.5E14 or 7.5 x10^(14)

enyata [817]3 years ago

7 0

Answer:7.5 x 10* 14

Explanation:

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valentina_108 [34]

To solve this problem it is necessary to apply the concepts related to the kinetic energy expressed in terms of simple harmonic movement, as well as the concepts related to angular velocity and acceleration and linear acceleration and velocity.

By definition we know that the angular velocity of a body can be described as a function of mass and spring constant as

$\omega = \sqrt{\frac{k}{m}}$

Where,

k = Spring constant

m = mass

From the given values the angular velocity would be

$\omega = \sqrt{\frac{277}{0.4}}$

$\omega = 26.31rad/s$

The kinetic energy on its part is expressed as

$E = \frac{1}{2} m\omega^2A^2$

Where,

A = Amplitude

$\omega$ = Angular Velocity

$m = Mass$

PART A) Replacing previously given values the energy in the system would be

$E = \frac{1}{2} m\omega^2A^2$

$E = \frac{1}{2} (0.4)(26.31)^2(3*10^{-2})^2$

$E= 0.1245J$

PART B) Through the amplitude and angular velocity it is possible to know the linear velocity by means of the relation

$v = A\omega$

$v = (3*10^{-2})(26.31)$

$v = 0.7893m/s$

PART C) Finally, the relationship between linear acceleration and angular velocity is subject to

$a = A\omega^2$

$a = (3*10^{-2})(26.31)^2$

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3 years ago

In her hand, a softball pitcher swings a ball of mass 0.245 kg around a vertical circular path of radius 59.8 cm before releasin

GuDViN [60]

Answer:

The velocity is $v_b = 20.17 \ m/s$

Explanation:

From the question we are told that

The mass of the ball is $m = 0.245 \ kg$

The radius is $r = 59.8 \ cm = 0.598 \ m$

The force is $F = 30.9 \ N$

The speed of the ball is $v = 16.0 \ m/s.$

Generally the kinetic energy at the top of the circle is mathematically represented as

$K_t = \frac{1}{2} * m * v^2$

=> $K_t = \frac{1}{2} * 0.245 * 16.0 ^2$

=> $K_t = 31.36 \ J$

Generally the work done by the force applied on the ball from the top to the bottom is mathematically represented as

$W = F * d$

Here d is the length of a semi - circular arc which is mathematically represented as

$d = \pi * r$

$W = 30.9 * 0.598$

$W = 18.48 \ J$

Generally the kinetic energy at the bottom is mathematically represented as

$K_b = \frac{1}{2} * m * v_b^2$

=> $K_b = \frac{1}{2} * 0.245 * v_b^2$

=> $K_b = 0.1225 * v_b^2$

From the law of energy conservation

$K_t + W =K_b$

=> $31.36+ 18.48 = 0.1225 * v_b^2$

=> $v_b = 20.17 \ m/s$

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3 years ago

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Goshia [24]

Explanation:

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3 years ago

A sled slides along a horizontal surface on which the coefficient of kinetic friction is 0.25. Its velocity at point A is 8.4 m/

Bad White [126]

Answer:

1.06 secs

Explanation:

Initial speed of sled, u = 8.4 m/s

Final speed of sled, v = 5.8 m/s

Coefficient of kinetic friction, μ = 0.25

Using the impulse momentum theory, we know that the impulse applied to the sled is equal to change in momentum of the sled:

FΔt = mv - mu

where m = mass of the object

Δt = time interval

F = force applied

The force applied on the sled is the frictional force, which is given as:

F = -μmg

where g = acceleration due to gravity

Therefore:

-μmgΔt = mv - mu

-μmgΔt = m(v - u)

-μgΔt = v - u

Making Δt subject of formula:

Δt = (v - u) / -μg

Δt = (5.8 - 8.4) / (-0.25 * 9.8)

Δt = -2.6/ -2.45

Δt = 1.06 secs

It took the sled 1.06 secs to travel from A to B.

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