CAN ANYONE HELP WITH THESE QUESTIONS :,] PLS IS URGEN IM SO CLOSE TO FAILING ALSO ITS SCIENCE PHYSICS I THINK

Which example is a body fossil? a) dino bones b) shrimp burrow c)dino footprint d)an ornithopod track

Harrizon [31]

Answer:

The hard structure like dino bones, teeth and shells are the examples of body fossils.

Option: A

Explanation:

The body fossils are common type of fossils found in the world. These body fossils are formed by the remains of the ‘dead animals and plants’. The body fossils are hard parts of the dino teeth, bones, shells, woody trunks, branches, and stems. The body fossils are formed in different ways and they are the remains of the body parts of ancient animals, plants and other life forms.

They are rarely found in connected nature and are like large vertebrates like dinosaur bones, skulls and teeth’s.  The body fossils are also formed in different ways they are formed like a plants or animals dies in a watery environment and they can be buried in mud or silt they cannot decompose.

They build the top and hardens into the rocks at over time sedimentation. Two types of tissues are decomposed, they are soft and hard tissues.in the decomposition of there leaving hard bones, shells and teeth’s are behind and soft tissues are quickly decomposes.

7 0

3 years ago

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Jerome solves a problem using the law of conservation of momentum. What should Jerome always keep constant for each object after

s2008m [1.1K]

Jerome solves a problem using the law of conservation of momentum. What should Jerome always keep constant for each object after the objects collide and bounce apart?

a-velocity

b-mass

c-momentum

d-direction

Answer:

b. Mass

Explanation:

This question has to do with the principle of the law of conservation of momentum which states that the momentum of a system remains constant if no external force is acting on it.

As the question states, two objects collide with each other and eventually bounce apart, so their momentum may not be conserved but the mass of the objects is constant for each non-relativistic motion. Because of this, the mass of each object prior to the collision would be the same as the mass after the collision.

Therefore, the correct answer is B. Mass.

6 0

3 years ago

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A 27.0-m steel wire and a 48.0-m copper wire are attached end to end and stretched to a tension of 145 N. Both wires have a radi

algol13

Answer:

The time taken by the wave to travel along the combination of two wires is 458 ms.

Explanation:

Given that,

Length of steel wire= 27.0 m

Length of copper wire = 48.0 m

Tension = 145 N

Radius of both wires = 0.450 mm

Density of steel wire $\rho_{s}= 7.86\times10^{3}\ kg/m^{3}$

Density of copper wire $\rho_{c}=8.92\times10^{3}\ kg/m^3$

We need to calculate the linear density of steel wire

Using formula of linear density

$\mu_{s}=\rho_{s}A$

$\mu_{s}=\rho_{s}\times\pi r^2$

Put the value into the formula

$\mu_{s}=7.86\times10^{3}\times\pi\times(0.450\times10^{-3})^2$

$\mu_{s}=5.00\times10^{-3}\ kg/m$

We need to calculate the linear density of copper wire

Using formula of linear density

$\mu_{c}=\rho_{s}A$

$\mu_{c}=\rho_{s}\times\pi r^2$

Put the value into the formula

$\mu_{c}=8.92\times10^{3}\times\pi\times(0.450\times10^{-3})^2$

$\mu_{c}=5.67\times10^{-3}\ kg/m$

We need to calculate the velocity of the wave along the steel wire

Using formula of velocity

$v_{s}=\sqrt{\dfrac{T}{\mu_{s}}}$

$v_{s}=\sqrt{\dfrac{145}{5.00\times10^{-3}}}$

$v_{s}=170.3\ m/s$

We need to calculate the velocity of the wave along the steel wire

Using formula of velocity

$v_{c}=\sqrt{\dfrac{T}{\mu_{c}}}$

$v_{c}=\sqrt{\dfrac{145}{5.67\times10^{-3}}}$

$v_{c}=159.9\ m/s$

We need to calculate the time taken by the wave to travel along the combination of two wires

$t=t_{s}+t_{c}$

$t=\dfrac{l_{s}}{v_{s}}+\dfrac{l_{c}}{v_{c}}$

Put the value into the formula

$t=\dfrac{27.0}{170.3}+\dfrac{48.0}{159.9}$

$t=0.458\ sec$

$t=458\ ms$

Hence, The time taken by the wave to travel along the combination of two wires is 458 ms.

4 0

3 years ago

Because the time was the same for each segment you know the speed was the same for each segment

netineya [11]

Answer:

Market segmentation helps you send the right message, every time, by efficiently targeting specific groups of consumers.

Explanation:

3 0

2 years ago

A point on the string of a violin moves up and down in simple harmonic motion with an amplitude of 1.24 mm and frequency of 875

mario62 [17]

Using

V = Amplitude x angular frequency(omega)

But omega= 2πf

= 2πx875

=5498.5rad/s

So v= 1.25mm x 5498.5

= 6.82m/s

B. .Acceleration is omega² x radius= 104ms²

5 0

3 years ago

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