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3 years ago

Scientists find the same type of index fossil in two different locations on Earth. What can they conclude from this discovery?

Physics

2 answers:

Sergio [31]3 years ago

7 0

The index fossils are the same geological age. Sorry if i'm wrong

OverLord2011 [107]3 years ago

3 0

Answer:

The index fossils are the same geological age.

Explanation:

I just took the test and this was the correct answer

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Kalea throws a baseball directly upward at time t = 0 at an initial speed of 13.7 m/s. How high h does the ball rise above its r

Trava [24]

Answer:

h = 9.57 seconds

Explanation:

It is given that,

Initial speed of Kalea, u = 13.7 m/s

At maximum height, v = 0

Let t is the time taken by the ball to reach its maximum point. It cane be calculated as :

$v=u-gt$

$u=gt$

$t=\dfrac{u}{g}$

$t=\dfrac{13.7}{9.8}$

t = 1.39 s

Let h is the height reached by the ball above its release point. It can be calculated using second equation of motion as :

$h=ut+\dfrac{1}{2}at^2$

Here, a = -g

$h=ut-\dfrac{1}{2}gt^2$

$h=13.7\times 1.39-\dfrac{1}{2}\times 9.8\times (1.39)^2$

h = 9.57 meters

So, the height attained by the ball above its release point is 9.57 meters. Hence, this is the required solution.

7 0

3 years ago

What potential difference is needed to give a helium nucleus (q=2e) 85.0 kev of kinetic energy?

Vilka [71]

The kinetic energy K given to the helium nucleus is equal to its potential energy, which is
$E=q \Delta V$
where q=2e is the charge of the helium nucleus, and $\Delta V$ is the potential difference applied to it.
Since we know the kinetic energy, we have
$E=K=85~keV=q \Delta V$
and from this we can find the potential difference:
$\Delta V = \frac{K}{q}= \frac{85~keV}{2e}=42.5~kV$

6 0

2 years ago

jenny's model train is set up on a circular track. There are six telephone poles evenly spaced around the track. It takes the en

d1i1m1o1n [39]

Answer:

T = 60 s

Explanation:

There are 6 poles on the track which are equally spaced

so the angular separation between the poles is given as

$\theta = \frac{2\pi}{6}$

$\theta = \frac{\pi}{3}$

so the angular speed of the train is given as

$\omega = \frac{\theta}{t}$

$\omega = \frac{\pi}{30} rad/s$

now we have time period of the train given as

$T = \frac{2\pi}{\omega}$

$T = \frac{2\pi}{\frac{\pi}{30}}$

$T = 60 s$

5 0

3 years ago

A car moving at 30 m/s slows uniformly to a speed of 10 m/s in a time of 5 s. Determine 1. The acceleration of the car. 2. The d

ValentinkaMS [17]

Answer:

Explanation:

Initial velocity , u = 30 m/s

final velocity , v = 10 m/s

time , t = 5 seconds

1. Acceleration = v - u / t

= 10 - 30 / 5

= -20 / 5

= <u><em>- 4 m/s</em></u>

8 0

3 years ago

What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally h

zysi [14]

Complete question:

What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)

Answer:

The peak emf generated by the coil is 15.721 kV

Explanation:

Given;

Radius of coil, r = 0.250 m

Number of turns, N = 500-turn

time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s

magnetic field strength, B = 0.425 T

Induced peak emf = NABω

where;

A is the area of the coil

A = πr²

ω is angular velocity

ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s = 60 rev/s

Induced peak emf = NABω

= 500 x (π x 0.25²) x 0.425 x 376.738

= 15721.16 V

= 15.721 kV

Therefore, the peak emf generated by the coil is 15.721 kV

5 0

2 years ago