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2 years ago

11

Emma measured the maximum displacement of a wave that she made by moving the end of a string up and down what property of a wave

was she measured A. Period B. Frequency C. Amplitude D. Wavelength

1 answer:

Kamila [148]2 years ago

4 0

The correct answer is C. amplitude.

Have an awesome day!

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JOSEPH JOGS FROM END A TO OTHER END B OF A straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back

zzz [600]

(a) The average speed from A to B would be 1.76 metre per second and the average velocity from A to B would also be 1.76 metre per second

<span>(b) The average speed from A to C would be 1.73 metre per second and the average velocity from A to C would be 0.87 metre per second</span>

3 0

3 years ago

5.An ice skater pushes against a wall with a force of 59 N. Ignoring friction, if the ice skater has

natali 33 [55]

Answer:

<em>Answer: (A) 0.75 m/s^2</em>

Explanation:

The Second Newton's law states that an object acquires acceleration when an external unbalanced net force is applied to it.

That acceleration is proportional to the net force and inversely proportional to the mass of the object.

It can be expressed with the formula:

$\displaystyle a=\frac{F_n}{m}$

Where

Fn = Net force

m = mass

The ice skater pushes against a wall with a force of 59 N. The wall returns the force and the skater now has a net force of Fn=59 N that makes him accelerate. Being m=79 kg the mass of the skater, the acceleration is:

$\displaystyle a=\frac{59}{79}$

$a = 0.75\ m/s^2$

Answer: (A) 0.75 m/s^2

5 0

2 years ago

Which if the following statements is true? A. Instantaneous acceleration is always changing. B. An object at rest has an instant

ycow [4]

B is the right one

hope this helps you

3 0

3 years ago

Imagine that heike onnes cooled mercury down to absolute zero and there was still resistance. would he have discovered supercond

likoan [24]

No, because superconductivity cannot occur if there is resistance

In addition to explaining electrical resistance, equilibrium distance theory also foretells the existence of superconductivity. According to its postulates, electrical resistivity decreases with distance from the equilibrium. There is only superconductivity at zero distance, with no resistance

<h3>What is Superconductivity ?</h3>

The ability of some materials to transmit electric current with virtually little resistance is known as superconductivity.

This ability has intriguing and maybe beneficial ramifications. Low temperatures are necessary for a material to exhibit superconductor behaviour. H. K. made the initial discovery of superconductivity in 1911.
Aluminum, magnesium diboride, niobium, copper oxide, yttrium barium, and iron pnictides are a few well-known examples of superconductors.

Learn more about Superconductivity here:

brainly.com/question/17166152

#SPJ4

4 0

2 years ago

magine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet,

Mama L [17]

Answer:

$\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'} } )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g} } )$

Explanation:

Given:

height above which the rock is thrown up, $\Delta h=50\ m$

initial velocity of projection, $u=20\ m.s^{-1}$

let the gravity on the other planet be g'

The time taken by the rock to reach the top height on the exoplanet:

$v=u+g'.t'$

where:

$v=$ final velocity at the top height = 0 $m.s^{-1}$

$0=20-g'.t'$ (-ve sign to indicate that acceleration acts opposite to the velocity)

$t'=\frac{20}{g'}\ s$

The time taken by the rock to reach the top height on the earth:

$v=u+g.t$

$0=20-g.t$

$t=\frac{20}{g} \ s$

Height reached by the rock above the point of throwing on the exoplanet:

$v^2=u^2+2g'.h'$

where:

$v=$ final velocity at the top height = 0 $m.s^{-1}$

$0^2=20^2-2\times g'.h'$

$h'=\frac{200}{g'}\ m$

Height reached by the rock above the point of throwing on the earth:

$v^2=u^2+2g.h$

$0^2=20^2-2g.h$

$h=\frac{200}{g}\ m$

The time taken by the rock to fall from the highest point to the ground on the exoplanet:

$(50+h')=u.t_f'+\frac{1}{2} g'.t_f'^2$ (during falling it falls below the cliff)

here:

$u=$ initial velocity= 0 $m.s^{-1}$

$\frac{200}{g'}+50 =0+\frac{1}{2} g'.t_f'^2$

$t_f'^2=\frac{400}{g'^2}+\frac{100}{g'}$

$t_f'=\sqrt{\frac{400}{g'^2}+\frac{100}{g'} }$

Similarly on earth:

$t_f=\sqrt{\frac{400}{g^2}+\frac{100}{g} }$

Now the required time difference:

$\Delta t=(t'+t_f')-(t+t_f)$

$\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'} } )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g} } )$

3 0

2 years ago