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konstantin123 [22]
3 years ago
7

What types of elements are useful for dating materials?

Physics
2 answers:
Alekssandra [29.7K]3 years ago
4 0

the nswer is for sure radioactive

Semenov [28]3 years ago
3 0
Well im not sure if this is the correct dating materials but here are some examples of Fundamentals of radiometric dating<span>Radioactive decay.
Accuracy of radiometric dating.
Closure temperature.
The age equation.
Uranium–lead dating method.
Samarium–neodymium dating method.
Potassium–argon dating method.
<span>Rubidium–strontium dating method.</span></span>
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A car goes up an incline that makes an angle of 20 degrees with the horizontal. If the car has a mass of 1600 kg, and the engine
stepan [7]

Answer:

V = 5.96 m/s

Explanation:

First we find the force required to move the car up the inclined plane. This force shall be equal to the component of the weight of the car that is parallel to the inclined plane. Hence,

F = W Sin θ

F = mg Sin θ

where,

F = Force = ?

m = mass of car = 1600 kg

g = 9.8 m/s²

θ = Inclination Angle = 20°

Therefore,

F = (1600 kg)(9.8 m/s²) Sin 20°

F = 5362.87 N

Now, we use power formula:

P = FV

V = P/F

where,

P = Engine Power = 32000 W

V = Maximum Velocity of Car = ?

Therefore,

V = 32000 W/5362.87 N

<u>V = 5.96 m/s</u>

7 0
3 years ago
if you and your friend, ride bumper cars at the fair, what happens in terms of Newtons third law, when they collide?
belka [17]
Newton's third law says that for every action, there is an equal and opposite reaction. So if your friend's bumper car, both of you will go flying back at the same force in opposite directions.
Hope this helps.
3 0
3 years ago
Read 2 more answers
The average distance of the planet mercury from the sun is 0.39 times the average distance of the earth from the sun. How long i
Stels [109]

Answer:

T_1=0.24y

Explanation:

Using Kepler's third law, we can relate the orbital periods of the planets and their average distances from the Sun, as follows:

(\frac{T_1}{T_2})^2=(\frac{D_1}{D_2})^3

Where T_1 and T_2 are the orbital periods of Mercury and Earth respectively. We have D_1=0.39D_2 and T_2=1y. Replacing this and solving for

T_1^2=T_2^2(\frac{D_1}{D_2})^3\\T_1^2=(1y)^2(\frac{0.39D_2}{D_2})^3\\T_1^2=1y^2(0.39)^3\\T_1^2=0.059319y^2\\T_1=0.24y

5 0
3 years ago
MIT’s robot cheetah can jump over obstacles 46. cm high and has speed of 12.0 km/h. a) If the robot launches itself at an angle
labwork [276]

Answer:

(a)  y_{max}=0.423m

(b)  \alpha =64.3^{o}

Explanation:

Given data

v_{i}=12km/h=3.33m/s\\\alpha =60^{o}\\g=9.8m/s^{2}\\Required\\(a)y_{max}\\(b)Angle

Solution

For Part (a)

As the velocity component in direction of y is given by:

v_{yi}=v_{i}Sin\alpha \\v_{yi}=3.33Sin60\\v_{yi}=2.88m/s

The maximum displacement is given by:

v_{yf}^{2}=v_{yi}^{2}-2gy_{max}\\ y_{max}=\frac{(2.88)^{2}}{2(9.8)}\\ y_{max}=0.423m

For Part (b)

To reach y=46cm =0.46m apply:

0=v_{yi}^{2}-2(9.8)(0.46)\\v_{yi}=3m/s\\As\\Sin\alpha =\frac{v_{yi}}{v_{i}}\\\alpha  =Sin^{-1}(\frac{v_{yi}}{v_{i}})\\\alpha =Sin^{-1}(\frac{3}{3.33} )\\\alpha =64.3^{o}

5 0
3 years ago
Read 2 more answers
Which law best describes this behavior of gases in the Sun?
Anvisha [2.4K]

Answer:

The extreme pressure from the weight of the gases that make up the Sun raises the temperature of the core enough for the nuclear reactions to take place, Which law best describes this behavior of gases in the Sun? A. Charles law.

Explanation:

if it is helpful....... plzz like and follow

7 0
3 years ago
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