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Liono4ka [1.6K]
3 years ago
10

What are the characteristics of an index fossil?

Physics
2 answers:
vichka [17]3 years ago
4 0
I think as a mold. when the Flood came in Genesis, i believe that when the fish were washed away, the kinda made a mold in a rock.
Assoli18 [71]3 years ago
4 0
D. Preserved as mold
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Donde esta la orilla del universo?
irinina [24]

Answer:

<em>El universo no tiene un borde finito</em>, continúa y se expande sin fin. La distancia adecuada desde la Tierra hasta el borde del universo observable es de aproximadamente 46.5 mil millones de años luz o 4.40 × 10 ^ 26 metros en cualquier dirección.

5 0
2 years ago
The electric field strength in the space between two closely spaced parallel disks is 1.0 105 N/C. This field is the result of t
balu736 [363]

Answer:

D=2.996\times 10^{-2} m

Explanation:

*Assume the parallel disks have equal diameters.

Given the electric strength as  1.0\times 10^5 N/C.  transferring 3.9\times 10^9 electrons, the disk's Area can be calculated using the formula:

E=\frac{\eta}{\epsilon_o}=\frac{Q}{A\epsilon_o}\\\\A=\frac{Q}{E\epsilon_o}\\\\=\frac{(3.9\times 10^9)\times (1.6\times10^{-19})}{(1.0\times 10^5 )\times (8.85\times10^{-12})}\\\\A=7.0508\times 10^{-4} \ m^2

#We now calculate the disks diameter:

A=\pi(D/2)^2\\\\2\sqrt{\frac{A}{\pi}}=D\\\\=2\sqrt{7.0508\times 10^{-4}/\pi}\\\\D=2.996\times 10^{-2} \ m

Hence, the diameter of the disks is D=2.996\times 10^{-2} m

8 0
3 years ago
A rocket car on a horizontal rail has an initial mass of 2500 kg and an additional fuel mass of 1000 kg. At time t0 the rocket m
slamgirl [31]

Answer: Acceleration of the car at time = 10 sec is 108 m/s^{2} and velocity of the car at time t = 10 sec is 918.34 m/s.

Explanation:

The expression used will be as follows.

M\frac{dv}{dt} = u\frac{dM}{dt}

\int_{t_{o}}^{t_{f}} \frac{dv}{dt} dt = u\int_{t_{o}}^{t_{f}} \frac{1}{M} \frac{dM}{dt} dt

       = u\int_{M_{o}}^{M_{f}} \frac{dM}{M}

v_{f} - v_{o} = u ln \frac{M_{f}}{M_{o}}

v_{o} = 0

As, v_{f} = u ln (\frac{M_{f}}{M_{o}})

u = -2900 m/s

M_{f} = M_{o} - m \times t_{f}

           = 2500 kg + 1000 kg - 95 kg \times t_{f}s

           = (3500 - 95t_{f})s

v_{f} = -2900 ln(\frac{3500 - 95 t_{f}}{3500}) m/s

Also, we know that

     a = \frac{dv_{f}}{dt_{f}} = \frac{u}{M} \frac{dM}{dt}

        = \frac{u}{3500 - 95 t} \times (-95) m/s^{2}

        = \frac{95 \times 2900}{3500 - 95t} m/s^{2}

At t = 10 sec,

v_{f} = 918.34 m/s

and,   a = 108 m/s^{2}

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