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4 years ago

PLZ HELP

Physics

1 answer:

nalin [4]4 years ago

6 0

Answer: the wavelenght is 9.8 meters

Explanation:

We can use the relationship:

Velocity = wavelenght*frequency.

Initially we have:

wavelenght = 4.9m

velocity = 9.8m/s

then:

9.8m/s = 4.9m*f

f = 9.8m/s/4.9m = 2*1/s

now, if the velocity is doubled and the frequency remains the same, we have:

2*9.8m/s = wavelenght*2*1/s

wavelenght = (2*9.8m/s)*(1/2)s = 9.8 m

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A family leaves home at 1:00pm and arrives at their vacation spot 200 miles away at 5:00pm on the same day. What is their averag

Shkiper50 [21]

The answer is 40 because you just have to do 200 divided by 5

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3 years ago

A roofer drops a nail that hits the ground traveling at 26 m/s. How fast was the nail traveling 1 second before it hits the grou

ELEN [110]

This problem can be solved using a kinematic equation. For this case, the following equation is useful:

v_final = v_initial + at

where,
v_final = final velocity of the nail
v_initial = initial velocity of the nail
a = acceleration due to gravity = 9.8 m/s^2
t = time

First, we determine the time it takes for the nail to hit the ground. We know that the initial velocity is 0 m/s since the nail was only dropped. It has a final velocity of 26 m/s. We substitute these values to the equation and solve for t:

26 = 0 + 9.8*t
t = 26/9.8 = 2.6531 s

The problem asks the velocity of the nail at t = 1 second. We then subtract 1 second from the total time 2.6531 with v_final as unknown.

v_final = 0 + 9.8(2.6531-1) = 16.2004 m/s.

Thus, the nail was traveling at a speed of 16. 2004 m/s, 1 second before it hit the ground.

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4 years ago

A tennis player receives a shot with the ball (0.060 0 kg) traveling horizontally at 56.0 m/s and returns the shot with the ball

Fofino [41]

Explanation:

It is given that,

Mass of the ball, m = 0.06 kg

Initial speed of the ball, u = 56 m/s

Final speed of the ball, v = -34.5 m/s (opposite direction)

(a) Let J is the impulse delivered to the ball by the racquet. It is equal to the change in momentum of the object as :

$J=m(v-u)$

$J=0.06\ kg(-34.5\ m/s-56\ m/s)$

J = -5.43 kg-m/s

(b) The work done by the racquet on thee ball is equal to the change in kinetic energy as :

$W=\dfrac{1}{2}m(v^2-u^2)$

$W=\dfrac{1}{2}\times 0.06((-34.5)^2-(56)^2)$

W = -58.372 Joules

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3 years ago

A rock is thrown horizontally from a high building at 33.8 m/s. What is the magnitude of its velocity 4.25 s later?

Alex17521 [72]

<h2>Answer:53.63 $ms^{-2}$ </h2>

Explanation:

The equations of motion used in this question is $v=u+at$

When a object is projected horizontally from a sufficiently height,the x-component of acceleration remains zero because there is no force that drags the object in x direction.

But,due to gravity,the object accelerates downward at a rate of $9.8ms^{-2}$ .

In X-Direction,

Given that initial velocity= $u_{x}$ = $33.8ms^{-1}$

Using $v=u+at$ ,

$v_{x}=33.8+(0)4.25=33.8ms^{-1}$

In Y-Direction,

Given that initial velocity= $u_{x}$ = $0ms^{-1}$

Using $v=u+at$ ,

$v_{y}=0+(9.8)4.25=41.65ms^{-1}$

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}$

$v=\sqrt{1142.44+1734.72}=\sqrt{2877.163}=53.63ms^{-1}$

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4 years ago

What is 1 joule work?

larisa86 [58]

$\large\color{pink}{{★ƛƝƧƜЄƦ ↦}}$

The amount of work done when a the force of 1 Newton displaces a body through a distance of 1m in the direction of the force applied is said to be as 1joule work

Hope it helps ~

$\sf{\:мѕнαcкεя\: ♪...}$

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2 years ago