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dexar [7]
3 years ago
10

Which describes time dilation?

Physics
1 answer:
Trava [24]3 years ago
6 0

Answer:

A). The faster an object moves, the more stretched time becomes.    

Explanation:

'Time dilation' is described as the '<u>slowing of the passage of time experienced by objects in motion relative to an observer that is measurable only at the relative speeds</u>.' As per this particular relativity theory, the observer witnesses an elongated time with reference to the other setting. This reduction in speed of time(slowing down) becomes more clear and evident in the case of an object moving briskly near to the light's speed. Thus, <u>option A</u> is the correct answer.

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A 0.500 kg bullet is fired from a gun at 25.0 m/s, how much kinetic energy does it have?
slamgirl [31]

Considering the definition of kinetic energy, the bullet has a kinetic energy of 156.25 J.

<h3>Kinetic energy</h3>

Kinetic energy is a form of energy. It is defined as the energy associated with bodies that are in motion and this energy depends on the mass and speed of the body.

Kinetic energy is defined as the amount of work necessary to accelerate a body of a given mass and in a rest position, until it reaches a given speed. Once this point is reached, the amount of accumulated kinetic energy will remain the same unless there is a change in speed or the body returns to its rest state by applying a force to it.

The kinetic energy is represented by the following expression:

Ec= ½ mv²

Where:

  • Ec is the kinetic energy, which is measured in Joules (J).
  • m is the mass measured in kilograms (kg).
  • v is the speed measured in meters over seconds (m/s).

<h3>Kinetic energy of a bullet</h3>

In this case, you know:

  • m= 0.500 kg
  • v= 25 m/s

Replacing in the definition of kinetic energy:

Ec= ½ ×0.500 kg× (25 m/s)²

Solving:

<u><em>Ec= 156.25 J</em></u>

Finally, the bullet has a kinetic energy of 156.25 J.

Learn more about kinetic energy:

brainly.com/question/25959744

brainly.com/question/14028892

#SPJ1

5 0
2 years ago
Fill in the appropriate values for each blank as it refers to ATOM 1. The number of protons present in ATOM 1 is _________.​
wlad13 [49]
3, protons are positive and there are 3 positive atoms visible
6 0
3 years ago
A uniform magnetic field passes through a horizontal circular wire loop at an angle 19.5 ∘ from the vertical. The magnitude of t
nlexa [21]

To solve this problem, we will apply the concepts related to Faraday's law that describes the behavior of the emf induced in the loop. Remember that this can be expressed as the product between the number of loops and the variation of the magnetic flux per unit of time. At the same time the magnetic flux through a loop of cross sectional area is,

\Phi = BA Cos \theta

Here,

\theta = Angle between areal vector and magnetic field direction.

According to Faraday's law, induced emf in the loop is,

\epsilon= -N \frac{d\Phi }{dt}

\epsilon = -N \frac{(BAcos\theta)}{dt}

\epsilon = -NAcos\theta \frac{dB}{dt}

\epsilon = -N\pi r^2 cos\theta \frac{d}{dt} ( ( 3.75 T ) + ( 3.05T/s ) t + ( -6.95 T/s^2 ) t^2)

\epsilon = -N\pi r^2 cos\theta( (3.05T/s)-(13.9T/s)t )

At time t = 5.71s,  Induced emf is,

\epsilon = -(1) \pi (0.220m)^2 cos(19.5\°)(  (3.05T/s)-(13.9T/s)(5.71s))

\epsilon = 10.9V

Therefore the magnitude of the induced emf is 10.9V

4 0
3 years ago
Read 2 more answers
Which data set has the smallest standard deviation?
hjlf

Answer:

B. 22,22,23,23,22,22,23

Explanation:

The standard deviation is a measure of dispersion or variability of a data set. In order to determine the data set that has the smallest standard deviation, we shall investigate on the ranges of the data sets given. The range of a data set is simply the difference between the maximum and minimum values in a data set. A data set that has a smaller range also has a smaller standard deviation.

From the alternatives given, the data set given by alternative B has the smallest range and consequently the smallest standard deviation.

The maximum value is 23 while the minimum is 22. The range is 1.

5 0
3 years ago
A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra
Sedbober [7]

Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

sin(\alpha) = \frac{Vyi}{Vi}

Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}

Vy in this problem will follow this equation =

Vy(t) = Vyi -g.t

where g is the gravity acceleration

Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t

This is equation (1)

For Y(t) :

Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}

We suppose yi = 0

Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s

So in t = 0.675 s  → Vy = 0. Now we calculate the y in which this happen using (2)

Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} }  .(0.675s)^{2} \\Y(0.675s) =2.236 m

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

6 0
4 years ago
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