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3 years ago

Help pls it’s urgent

Physics

2 answers:

Igoryamba3 years ago

3 0

The wave length becomes shorter and the pitch becomes higher

Svetllana [295]3 years ago

3 0

The wavelength becomes shorter and the frequency gets higher.

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A ball of putty with a mass of 0.250 kg is thrown with a velocity of +7.50 m/s directly toward a nearby wall. The putty sticks t

nlexa [21]

The impulse experienced by an object is the force. • time.
The momentum change of an object is the mass. • velocity change.
The impulse equals the momentum change.

Your impulse is 1.875 I hope this helps

8 0

3 years ago

For a projectile, which of the following quantities are constant during the flight: x, y, vx, vy, v, ax, ay? Check all that appl

geniusboy [140]

Answer:

C. vx

F. ax

G. ay

Explanation:

The projectile moves along a curved path towards the ground, so its x and y positions change.

Since there is no applied force in the x-direction, the acceleration in the x-direction is zero. This means, ax and vx do not change.

The projectile is under the influence of gratify, which is directed towards the ground. So its velocity increases, because its y-component of the velocity increases.

However, the y-component of its acceleration is constant, since it is gravitational acceleration.

5 0

3 years ago

A wave with a period of 0.008 second has a frequency of

rodikova [14]

D. 125 Hz
$f = \frac{1}{t} = \frac{1}{0.008} \\ = 125hz$

4 0

4 years ago

A wave with a frequency of 1200 Hz propagates along a wire that is under a tension of 800 N. Its wavelength is 39.1 cm. What wil

mash [69]

Answer:

The wavelength will be 33.9 cm

Explanation:

Given;

frequency of the wave, F = 1200 Hz

Tension on the wire, T = 800 N

wavelength, λ = 39.1 cm

$F = \frac{ \sqrt{\frac{T}{\mu} }}{\lambda}$

Where;

F is the frequency of the wave

T is tension on the string

μ is mass per unit length of the string

λ is wavelength

$\sqrt{\frac{T}{\mu} } = F \lambda\\\\\frac{T}{\mu} = F^2\lambda^2\\\\\mu = \frac{T}{F^2\lambda^2} \\\\\frac{T_1}{F^2\lambda _1^2} = \frac{T_2}{F^2\lambda _2^2} \\\\\frac{T_1}{\lambda _1^2} = \frac{T_2}{\lambda _2^2}\\\\T_1 \lambda _2^2 = T_2\lambda _1^2\\\\$

when the tension is decreased to 600 N, that is T₂ = 600 N

$T_1 \lambda _2^2 = T_2\lambda _1^2\\\\\lambda _2^2 = \frac{T_2\lambda _1^2}{T_1} \\\\\lambda _2 = \sqrt{\frac{T_2\lambda _1^2}{T_1}} \\\\\lambda _2 = \sqrt{\frac{600* 0.391^2}{800}}\\\\\lambda _2 = \sqrt{0.11466} \\\\\lambda _2 =0.339 \ m\\\\\lambda _2 =33.9 \ cm$