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Veronika [31]
3 years ago
7

How do scientist determine how old the world is and the fossils inside of it?

Physics
4 answers:
algol [13]3 years ago
8 0
Yeah they usually see how far down in the ground a fossil is and if it’s deep down they know it’s very old if it’s not deep down then it ain’t very old
valkas [14]3 years ago
6 0

The layers of the fossil the oldest is usual the bottom layer,and the top layer is the newest layer

yaroslaw [1]3 years ago
5 0
Yeah they usually see how far down in the ground a fossil is and if it’s deep down they know it’s very old if it’s not deep down then it ain’t very old
Mazyrski [523]3 years ago
5 0
Yeah they usually see how far down in the ground a fossil is and if it’s deep down they know it’s very old if it’s not deep down then it ain’t very old
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ANOTHER PHYSICS QUESTION PLEASE HELP!
Sergio039 [100]
Answer


= 60


Hope it helps:)
5 0
3 years ago
A roller-coaster track has six semicircular "dips" with different radii of curvature. The same roller-coaster cart rides through
inessss [21]

Answer:

E > A = B > C > D

Explanation:

On a semicircular curve, the magnitude of force is given by the formula as :

F=\dfrac{mv^2}{r}

Here, m is the mass of the vehicle

Case A : R = 60 m v = 16 m/s

F_1=\dfrac{m(16)^2}{60}=4.27m\ N

Case B : R = 15 m v = 8 m/s

F_2=\dfrac{m(8)^2}{15}=4.27m\ N

Case C : R = 30 m v = 4 m/s

F_3=\dfrac{m(4)^2}{30}=0.534m\ N

Case D : R = 45 m v = 4 m/s

F_4=\dfrac{m(4)^2}{45}=0.356m\ N

Case E : R = 30 m v = 16 m/s

F_5=\dfrac{m(16)^2}{30}=8.53m\ N

The order of the magnitude of the force of the roller-coaster track on the cart from largest to smallest is :

E > A = B > C > D

6 0
3 years ago
A 3.0-kilogram toy dump truck moving with a speed of 4.0 m/s starts up a ramp. How high does the truck roll before it stops?
baherus [9]

Before it reaches the ramp, the toy's kinetic energy is

                                        (1/2) (mass) (speed²)

                                   =   (1/2) (3 kg) (16 m²/s²)

                                   =          24 Joules .

On the way up the ramp, its kinetic energy will change to
gravitational potential energy, and it'll stop when it's height
gives it  24 joules of potential energy.

         Potential energy  =  (mass) (gravity) (height)

                    24  joules  =  (3 kg) (9.8 m/s²) (height)

Divide each side by  29.4 kg-m/s²  (that's the same as 29.4 newtons):

                    Height  =  24 joules / 29.4 newtons

                                =      0.816 meter higher than the bottom of the ramp.
7 0
3 years ago
A 2000 kg car moves along a horizontal road at speed vo
cluponka [151]

Answer:

The shortest possible stopping distance of the car is 175.319 meters.

Explanation:

In this case we see that driver use the brakes to stop the car by means of kinetic friction force. Deceleration of the car is directly proportional to kinetic friction coefficient and can be determined by Second Newton's Law:

\Sigma F_{x} = -\mu_{k}\cdot N = m \cdot a (Eq. 1)

\Sigma F_{y} = N-m\cdot g = 0 (Eq. 2)

After quick handling, we get that deceleration experimented by the car is equal to:

a = -\mu_{k}\cdot g (Eq. 3)

Where:

a - Deceleration of the car, measured in meters per square second.

\mu_{k} - Kinetic coefficient of friction, dimensionless.

g - Gravitational acceleration, measured in meters per square second.

If we know that \mu_{k} = 0.0735 and g = 9.807\,\frac{m}{s^{2}}, then deceleration of the car is:

a = -(0.0735)\cdot (9.807\,\frac{m}{s^{2}} )

a = -0.721\,\frac{m}{s^{2}}

The stopping distance of the car (\Delta s), measured in meters, is determined from the following kinematic expression:

\Delta s = \frac{v^{2}-v_{o}^{2}}{2\cdot a} (Eq. 4)

Where:

v_{o} - Initial speed of the car, measured in meters per second.

v - Final speed of the car, measured in meters per second.

If we know that v_{o} = 15.9\,\frac{m}{s}, v = 0\,\frac{m}{s} and a = -0.721\,\frac{m}{s^{2}}, stopping distance of the car is:

\Delta s = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(15.9\,\frac{m}{s} \right)^{2}}{2\cdot \left(-0.721\,\frac{m}{s^{2}} \right)}

\Delta s = 175.319\,m

The shortest possible stopping distance of the car is 175.319 meters.

8 0
4 years ago
Protons have ___charge; they have equal amounts of positive and negative ____charges?
Ray Of Light [21]
Protons have positive charges
5 0
3 years ago
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