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3 years ago

Which of these answers best describes a fossil?

Physics

2 answers:

Anettt [7]3 years ago

6 0

Answer:

Explanation:

The definition of a fossil is the preserved remains of a prehistoric organism or is slang for someone or something that is outdated. An example of a fossil is the preserved remains from a prehistoric organism that have been preserved inside rock.

Mekhanik [1.2K]3 years ago

3 0

C is the answer. Remains of a once living organism found in layers of rock, ice, or amber

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Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic colli

serious [3.7K]

Answer:

pf = 198.8 kg*m/s

θ = 46.8º N of E.

Explanation:

Since total momentum is conserved, and momentum is a vector, the components of the momentum along two axes perpendicular each other must be conserved too.
If we call the positive x- axis to the W-E direction, and the positive y-axis to the S-N direction, we can write the following equation for the initial momentum along the x-axis:

$p_{ox} = p_{oAx} + p_{oBx} (1)$

We can do exactly the same for the initial momentum along the y-axis:

$p_{oy} = p_{oAy} + p_{oBy} (2)$

The final momentum along the x-axis, since the collision is inelastic and both objects stick together after the collision, can be written as follows:

$p_{fx} = (m_{A} + m_{B} ) * v_{fx} (3)$

We can repeat the process for the y-axis, as follows:

$p_{fy} = (m_{A} + m_{B} ) * v_{fy} (4)$

Since (1) is equal to (3), replacing for the givens, and since p₀Bₓ = 0, we can solve for vfₓ as follows:

$v_{fx} = \frac{p_{oAx}}{(m_{A}+ m_{B)}} = \frac{m_{A}*v_{oAx} }{(m_{A}+ m_{B)}} =\frac{17.0kg*8.00m/s}{46.0kg} = 2.96 m/s (5)$

In the same way, we can find the component of the final momentum along the y-axis, as follows:

$v_{fy} = \frac{p_{oBy}}{(m_{A}+ m_{B)}} = \frac{m_{B}*v_{oBy} }{(m_{A}+ m_{B)}} =\frac{29.0kg*5.00m/s}{46.0kg} = 3.15 m/s (6)$

With the values of vfx and vfy, we can find the magnitude of the final speed of the two-object system, applying the Pythagorean Theorem, as follows:

$v_{f} = \sqrt{v_{fx} ^{2} + v_{fy} ^{2}} = \sqrt{(2.96m/s)^{2} + (3.15m/s)^{2}} = 4.32 m/s (7)$

The magnitude of the final total momentum is just the product of the combined mass of both objects times the magnitude of the final speed:

$p_{f} = (m_{A} + m_{B})* v_{f} = 46 kg * 4.32 m/s = 198.8 kg*m/s (8)$

Finally, the angle that the final momentum vector makes with the positive x-axis, is the same that the final velocity vector makes with it.
We can find this angle applying the definition of tangent of an angle, as follows:

$tg \theta = \frac{v_{fy}}{v_{fx}} = \frac{3.15 m/s}{2.96m/s} = 1.06 (9)$

⇒ θ = tg⁻¹ (1.06) = 46.8º N of E

8 0

3 years ago

Suppose you drop a tennis ball from a height of 6 feet. After the ball hits the floor, it rebounds to 80% of

sasho [114]

The formula to solve this is (.80)^3 X 6 and the answer would be 3.1 feet. That is how high the ball will rebound after its third bounce. Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help.

3 0

3 years ago

Read 2 more answers

During the push-up, the hips should never hit the ground and should move 1 point

Monica [59]

Hahahahahahahahahha true

4 0

3 years ago

Suppose a rocket ship accelerates upwards with acceleration equal in magnitude to twice the magnitude of g (we say that the rock

pashok25 [27]

Answer:

a) $s_a=98100\ m$ is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

b) $v_a=1962\ m.s^{-1}$ is speed of the rocket going when it stops accelerating.

c) $H=294300\ m$

d) $t_T=544.95\ s$

e) Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

Explanation:

Given:

acceleration of rocket, $a=2g=2\times 9.81=19.62\ m.s^{-2}$

time for which the rocket accelerates, $t_a=100\ s$

<u>For the course of upward acceleration:</u>

using eq. of motion,

$s_a=ut+\frac{1}{2}at_a^2$

where:

$u=$ initial velocity of the rocket at the launch $=0$

$s_a=$ height the rocket travels just before its fuel finishes off

so,

$s_a=0+\frac{1}{2}\times 19.62\times 100^2$

a) $s_a=98100\ m$ is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

<u>Now the velocity of the rocket just after the fuel is finished:</u>

$v_a=u+at_a$

$v_a=0+19.62\times 100$

b) $v_a=1962\ m.s^{-1}$ is speed of the rocket going when it stops accelerating.

After the fuel is finished the rocket starts to decelerates. So, we find the height of the rocket before it begins to fall back towards the earth.

Now the additional height the rocket ascends before it begins to fall back on the earth after the fuel is consumed completely, at this point its instantaneous velocity is zero:

using equation of motion,

$v^2=v_a^2-2gh$

where:

$g=$ acceleration due to gravity

$v=$ final velocity of the rocket at the top height

$0^2=1962^2-2\times 9.81\times h$

$h=196200\ m$

c) So the total height at which the rocket gets:

$H=h+s$

$H=196200+98100$

$H=294300\ m$

Time taken by the rocket to reach the top height after the fuel is over:

$v=v_a+g.t$

$0=1962-9.81t$

$t=200\ s$

Now the time taken to fall from the total height:

$H=v.t'+\frac{1}{2}\times gt'^2$

$294300=0+0.5\times 9.81\times t'^2$

$t'=244.95\ s$

Hence the total time taken by the rocket to strike back on the earth:

$t_T=t_a+t+t'$

$t_T=100+200+244.95$

$t_T=544.95\ s$

Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

8 0

3 years ago

A student measured the specific heat of water to be 4.29 J/g.Co. The

leonid [27]

Answer:

2.63 %.

Explanation:

Given that,

The calculated value of the specific heat of water is 4.29 J/g.C

Original value of specific heat of water is 4.18 J/g.C.

We need to find the student's percent error. The percentage error in any quantity is given by :

$P=\dfrac{|\text{original value-calculated value}|}{\text{original value}}\times 100\\\\P=\dfrac{4.29-4.18}{4.18}\times 100\\\\P=2.63\%$

So, the student's percent error is 2.63 %.

7 0

3 years ago