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2 years ago

High energy waves have what

Physics

2 answers:

Hitman42 [59]2 years ago

5 0

Answer:

It means that the frequency will be a lot higher with a shorter wave.

GarryVolchara [31]2 years ago

3 0

High energy waves have Gamma rays

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An 84.0 kg sprinter starts a race with an acceleration of 1.76 m/s2. If the sprinter accelerates at that rate for 11 m, and then

gulaghasi [49]

Answer:

t=17.838s

Explanation:

The displacement is divided in two sections, the first is a section with constant acceleration, and the second one with constant velocity. Let's consider the first:

The acceleration is, by definition:

$a=\frac{dv}{dt}=1.76$

So, the velocity can be obtained by integrating this expression:

$v=1.76t$

The velocity is, by definition: $v=\frac{dx}{dt}$ , so

$dx=1.76tdt\\x=1.76\frac{t^{2}}{2}$ .

Do x=11 in order to find the time spent.

$11=1.76\frac{t^2}{2}\\ t^2=\frac{2*11}{76} \\t=\sqrt{12.5}=3.5355s$

At this time the velocity is: $v=1.76t=1.76*3.5355s=6.2225\frac{m}{s}$

This velocity remains constant in the section 2, so for that section the movement equation is:

$x=v*t\\t=\frac{x}{v}$

The left distance is 89 meters, and the velocity is $6.2225\frac{m}{s}$ , so:

$t=\frac{89}{6.2225}=14.303s$

So, the total time is 14.303+3.5355s=17.838s

7 0

2 years ago

During a very quick stop, a car decelerates at 7.6 m/s2. Assume the forward motion of the car corresponds to a positive directio

Mashcka [7]

Answer:

24.57 revolutions

Explanation:

(a) If they do not slip on the pavement, then the angular acceleration is

$\alpha = a / r = 7.6 / 0.26 = 29.23 rad/s^2$

(b) We can use the following equation of motion to find out the angle traveled by the wheel before coming to rest:

$\omega^2 - \omega_0^2 = 2\alpha\Delta \theta$

where v = 0 m/s is the final angular velocity of the wheel when it stops, $\omega_0$ = 95rad/s is the initial angular velocity of the wheel, $\alpha = -29.23 rad/s^2$ is the deceleration of the wheel, and $\Delta \theta$ is the angle swept in rad, which we care looking for:

$0 - 95^2 = 2*29.23\Delta \theta$

$9025 = 58.46 \Delta \theta$

$\Delta \theta = 9025 / 58.46 = 154.375 rad$

As each revolution equals to 2π, the total revolution it makes before stop is

154.375 / 2π = 24.57 revolutions

8 0

2 years ago

You attach a 2.90 kg mass to a horizontal spring that is fixed at one end. You pull the mass until the spring is stretched by 0.

Murljashka [212]

Answer:

Explanation:

The spring is stretched by .5 m and then released that means its amplitude of oscillation A is 0.5 m .

A = 0.5 m

After the release at one extreme point , the mass comes to rest again at another extreme point after half the time period ie

T / 2 = .3 s

T = 0.6 s

Angular velocity

ω = $\frac{2\times \pi}{T}$

ω = $\frac{2\times \pi}{0.6}$

ω = 10.45

Maximum velocity = ω A

ω and A are angular velocity and amplitude of oscillation.

Maximum velocity = 10.45 x .5

= 5.23 m /s

7 0

3 years ago

a lump of putty and a rubber ball have equal mass. both are thrown with equal speed against a wall. the putty sticks to the wall

RUDIKE [14]

The bouncy ball experiences the greater momentum change.

To understand why, you need to remember that momentum is actually
a vector quantity ... it has a size AND it has a direction too.

The putty and the ball have the same mass, and you throw them
with the same speed. So, on the way from your hand to the wall,
they both have the same momentum.
Call it " M in the direction toward the wall ".

After they both hit the wall:

-- The putty has zero momentum.
Its momentum changed by an amount of M .

-- The ball has momentum of " M in the direction away from the wall ".
Its momentum changed by an amount of 2M .

5 0

3 years ago

Read 2 more answers

Last week we investigated the black hole at the center of our galaxy, with massMX= 8.5×1036kg. An object whose size is on the or

swat32

Answer:

$d=2.38*10^{13}m$