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3 years ago

How does increasing the tension of a spring affect a wave on the spring?

Physics

2 answers:

Liono4ka [1.6K]3 years ago

8 0

More energy is being used in the movement

Luden [163]3 years ago

7 0

For plato the answer is, a. wave velocity increases

posting this for future users!

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Will give brainliest! how does an engineer use physical science?

pentagon [3]

Answer: gravity, circuits

Explanation:

3 0

3 years ago

NASA scientists suggest using rotating cylindrical spacecraft to replicate gravity while in a weightless environment. Consider s

dusya [7]

Answer:

Explanation:

Given

diameter of spacecraft $d=148\ m$

radius $r=74\ m$

Force of gravity $F_g$ =mg

where m =mass of object

g=acceleration due to gravity on earth

Suppose v is the speed at which spacecraft is rotating so a net centripetal acceleration is acting on spacecraft which is given by

$F_c=\frac{mv^2}{r}$

$F_c=F_g$

$\frac{mv^2}{r}=mg$

$\frac{v^2}{r}=g$

$v=\sqrt{gr}$

$v=\sqrt{1450.4}$

$v=38.08\ m/s$

8 0

3 years ago

Your cat "Ms." (mass 7.00 {\rm kg}) is trying to make it to the top of a frictionless ramp 2.00 {\rm m} long and inclined upward

lbvjy [14]

Answer:

Final velocity at the top of the ramp is 6.58m/s

Explanation

Check the attachment

4 0

3 years ago

While standing at the edge of the roof of a building, a man throws a stone upward with an initial speed of 6.95 6.95 m/s. The st

qwelly [4]

Answer:

18.1347 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = a

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-6.95^2}{2\times -9.81}\\\Rightarrow s=2.4619\ m$

Total height the ball falls is 2.4619+14.3 = 16.7619 m

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 16.7619+0^2}\\\Rightarrow v=18.1347\ m/s$

The speed at which the stone reaches the ground is 18.1347 m/s

8 0

3 years ago

A particle with a charge of 3.00 elementary charges moves through a potential difference of 4.50 volts. What is the change in el

GuDViN [60]

Answer:

$7.2\cdot 10^{-19} J$

Explanation:

The change in electrical potential energy of a charged particle moving through a potential difference is given by

$\Delta U = q \Delta V$

where

q is the magnitude of the charge of the particle

$\Delta V$ is the potential difference

In this problem:

- the charge of the particle is 3.00 elementary charges, so

$q=3e=3\cdot 1.6\cdot 10^{-19} J=4.8\cdot 10^{-19}J$

- the potential difference is

$\Delta V=4.50 V$

So, the change in electrical potential energy is

$\Delta U=(1.6\cdot 10^{-19}C)(4.50 V)=7.2\cdot 10^{-19} J$

7 0

4 years ago