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1 year ago

The total energy of a 0.050 kg object travelling at 0.70 c is

1 answer:

4 0

Would presume the energy as kinetic energy.

E = (1/2)*mv²

But m = 0.05kg, velocity here = 0.70c, where c is the speed of light ≈ 3* 10⁸ m/s

Ke = (1/2)*mv² = 0.5*0.05*(0.7*3* 10⁸)² = 1.1025 * 10¹⁵ Joules

There is no exact match from the options.

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Which of the following statements is false?

agasfer [191]

A & B

Observe the path of the light ray as it passes through the lenses as shown in the attached images. Concave lenses diverge light rays while the convex lens converges the light rays.

Explanation:

Real images are formed where the rays converge, a property of images by convex lenses. Convex lenses can be used to magnify objects. If the image occurs before the focal point of the lens then the image will be upright but smaller. The images inverts and gets bigger past the focal point.

Virtual images are property of concave lenses. These images appear closer but smaller than the real object.

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6 0

2 years ago

Read 2 more answers

4. A 505-turn circular-loop coil with a diameter of 15.5 cm is initially aligned so that

Basile [38]

The strength of the magnetic field is $4.8\cdot 10^{-5} T$

Explanation:

According to Faraday's Law, the magnitude of the induced emf in the coil is equal to the rate of changeof the flux linkage through the coil:

$\epsilon = \frac{N\Delta \Phi}{\Delta t}$ (1)

where

N = 505 is the number of turns in the coil

$\Delta \Phi$ is the change in magnetic flux through the coil

$\Delta t = 2.77 ms = 2.77\cdot 10^{-3} s$ is the time interval

$\epsilon = 0.166 V$

The coil is rotated from a position perpendicular to the Earth's magnetic field to a position parallel to it, so the final flux is zero, and the magnitude of the flux change is simply equal to the initial flux:

$\Delta \Phi = B A cos \theta$

where

B is the strength of the magnetic field

A is the area of the coil

$\theta=0^{\circ}$ is the angle between the normal to the coil and the field

The area of the coil can be written as

$A=\pi r^2$

where

$r=\frac{15.5 cm}{2}=7.75 cm = 7.75\cdot 10^{-2} m$ is its radius

Substituting everything into eq.(1) and solving for B, we find:

$\epsilon= \frac{NB\pi r^2 cos \theta}{\Delta t}\\B=\frac{\epsilon \Delta t}{\pi r^2 cos \theta}=\frac{(0.166)(2.77\cdot 10^{-3})}{(505)\pi (7.75\cdot 10^{-2})^2(cos 0^{\circ})}=4.8\cdot 10^{-5} T$

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8 0

1 year ago

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You are riding on a roller coaster that starts from rest at a height of 25.0 m and moves along a frictionless track. however, af

djyliett [7]

I attached the missing picture.
We can figure this one out using the law of conservation of energy.
At point A the car would have potential energy and kinetic energy.
$A: mgh_1+\frac{mv_1^2}{2}$
Then, while the car is traveling down the track it loses some of its initial energy due to friction:
$W_f=F_f\cdot L$
So, we know that the car is approaching the point B with the following amount of energy:
$mgh_1+\frac{mv_1^2}{2}- F_fL$
The law of conservation of energy tells us that this energy must the same as the energy at point B.
The energy at point B is the sum of car's kinetic and potential energy:
$B: mgh_2+\frac{mv_2}{2}$
As said before this energy must be the same as the energy of a car approaching the loop:
$mgh_2+\frac{mv_2}{2}=mgh_1+\frac{mv_1^2}{2}- F_fL$
Now we solve the equation for $v_1$ :
$v_1^2=2g(h_2-h_1)+v_2^2+\frac{2F_fL}{m}\\
v_1^2=39.23\\
v_1=\sqrt{39.23}=6.26\frac{m}{s}$