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3 years ago

Use the information from the article to answer the question.

Physics

2 answers:

Varvara68 [4.7K]3 years ago

6 0

Answer:

its AAAA Most asteroids can be found between Mars and Jupiter.

Explanation:

I got it right

lbvjy [14]3 years ago

3 0

Answer:

Most asteroids can be found between Mars and Jupiter.

Explanation:

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Bob is studying waves in science and learns that the energy of a wave is directly proportional to the square of the waves amplit

Mrrafil [7]

For a simple harmonic motion energy is given with:
$E=\frac{1}{2}kA^2$
Where k is a constant that depends on the type of the wave you are looking at and A is amplitude.
Let's calculate the energy of the wave using two different amplitudes given in the problem:
$E_1=\frac{1}{2}k(1)^2=\frac{1}{2}k\\E_2=\frac{1}{2}k(2)^2=\frac{4}{2}k=2k\\$
We can see that energy associated with the wave is 4 times smaller when we decrease its amplitude by half. So the answer should be C.

6 0

3 years ago

Read 2 more answers

You are taking an image of a patient who is in extreme discomfort while participating in the CT scanning process. Which of the f

brilliants [131]

Answer:

Interpersonal skills

Explanation:

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3 years ago

On a clear day at a certain location, a 119-V/m vertical electric field exists near the Earth's surface. At the same place, the

IrinaVladis [17]

Answer:

(a) 62.69 nJ/m^3

(b) 1015.22 μJ/m^3

Explanation:

Electric field, E = 119 V/m

Magnetic field, B = 5.050 x 10^-5 T

(a) Energy density of electric field = $\frac{1}{2}\varepsilon _{0}E^{2}$

$=\frac{1}{2}\times 8.854\times 10^{-12}\times 119\times 119$

= 6.269 x 10^-8 J/m^3 = 62.69 nJ/m^3

(b) energy density of magnetic field = $\frac{B^{2}}{2\mu _{0}}$

$=\frac{\left ( 5.05\times 10^{-5} \right )^{2}}{2\times 4\times 3.14\times 10^{-7}}$

= 1.01522 x 10^-3 J/m^3 = 1015.22 μJ/m^3

8 0

3 years ago

A 50.0 kg crate is pulled 375 N of force applied to a rope. The crate slides without friction.

LUCKY_DIMON [66]

Hi there!

We can use the work-energy theorem to solve.

Recall that:

$\large\boxed{W = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2}$

The initial kinetic energy is 0 J because the crate begins from rest, so we can plug in the given values for mass and final velocity:

$W = \frac{1}{2}(50)(5.61^2) = 786.8025 J$

Now, we can define work:

$\large\boxed{W = Fdcos\theta}}$

Now, plug in the values:

$786.8025 = Fdcos\theta\\\\786.8025 = (375)(3.07)cos\theta$

Solve for theta:

$cos\theta = .6834\\\theta = cos^{-1}(.6834) = \boxed{46.887^o}$

4 0

2 years ago

A train has a length of 81.1 m and starts from rest with a constant acceleration at time t = 0 s. At this instant, a car just re

arlik [135]

Answer: a) vcar= 7 m/s ; b) a train= 0.65 m/s^2

Explanation: By using the kinematic equation for the car and the train we can determine the above values of the car velocity and the acceletarion of the train, respectively.

We have for the car

distance = v car* t, considering the length of train (81.1 m) travel by the car during the first 11.6 s

the v car = distance/time= 81.1 m/11.6s= 7 m/s

In order to calculate the acceleration we have to use the kinematic equation for the train from the rest

distance train = (a* t^2)/2

distance train : distance travel by the car at constant speed

so distance train= (vcar*36.35)m=421 m

the a traiin= (2* 421 m)/(36s)^2=0.65 m/s^2

4 0

3 years ago